2x + 4y + 6z = 22, 3x + 8y + 5z = 27,−x + y + 2z = 2
Reducing the augmented matrix to an equivalent row echelon form by using elementary row operations, we get
\(\left[ \begin{matrix} 2 & 4 & 6 \\ 3 & 8 & 5 \\ -1 & 1 & 2 \end{matrix}|\begin{matrix} 22 \\ 27 \\ 2 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 1 & 2 \\ 3 & 8 & 5 \\ 2 & 4 & 6 \end{matrix}|\begin{matrix} 2 \\ 27 \\ 22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+3R_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 1 & 2 \\ 0 & 11 & 11 \\ 0 & 6 & 10 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 26 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 11\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} -1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 3 & 5 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 13 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} -1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 2 \end{matrix} \right] \)
Writing the equivalent equations from the row echelon matrix we get,
-x + y + 2z = 2 .............(1)
y + z = 3 ..........(2)
z = 2 .............(3)
Substituting (3) in (2) we get, y + 2 =3
⇒ y = 3 - 2 = 1
Substituting y = 1 and z = 2 in (1) we get,
-x+1+2(2) = 2 ⇒ -x+1+4 = 2
⇒ -x+5 = 2 ⇒ -x = 2-5
⇒ -x = -3 ⇒ x = 3
∴ Solution set is {3, 1, 2}

2x + 2y + z = 5, x - y + z = 1, 3x + y + 2z = 4
The matrix form of the given system is AX = B where
A =\(\left[ \begin{matrix} 2 & 2 & 1 \\ 1 & -1 & 1 \\ 3 & 1 & 2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 5 \\ 1 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augmented matrix [A|B] we get,
\(\left[ \begin{matrix} 2 & 2 & 1 \\ 1 & -1 & 1 \\ 3 & 1 & 2 \end{matrix}|\begin{matrix} 5 \\ 1 \\ 4 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 2 & 1 \\ 3 & 1 & 2 \end{matrix}|\begin{matrix} 1 \\ 5 \\ 4 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 2 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 4 & -1 \\ 0 & 4 & -1 \end{matrix}|\begin{matrix} 1 \\ 3 \\ -1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 4 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ 3 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 [∵ There are 2 non-Zero rows]
and \(\rho \)[A|B] = 3 [∵ There are 3 non-zero rows]
Here, (A) ≠ p[A|B]
Hence, the given system is inconsistent and has no solution.

3x + y + z = 2, x - 3y + 2z = 1, 7x - y + 4z = 5
Then matrix form of the system in AX = B
where A = \(\left[ \begin{matrix} 3 & 1 & 1 \\ 1 & -3 & 2 \\ 7 & -1 & 4 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 2 \\ 1 \\ 5 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B] we get,
[A|B] = \(\left[ \begin{matrix} 3 & 1 & 1 \\ 1 & -3 & 2 \\ 7 & -1 & 4 \end{matrix}|\begin{matrix} 2 \\ 1 \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -3 & 2 \\ 3 & 1 & 1 \\ 7 & -1 & 4 \end{matrix}|\begin{matrix} 1 \\ 2 \\ 5 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-7{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -3 & 2 \\ 0 & 10 & -5 \\ 0 & 20 & -10 \end{matrix}|\begin{matrix} 1 \\ -1 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -3 & 2 \\ 0 & 10 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2, and \(\rho \)[A|B] = 2 [since there only two non-zero rows].
So, \(\rho \)(A)= \(\rho \)[A|B] = 2 < 3, the given system is consistent with one parameter family of solutions.
So, put z = t, x \(\in \) R Writing the f equivalent equations from the row echelon matrix we get,
x-3y+2z = 1.........(1)
10y-5z = -1..........(2)
z = t
(2) we comes 10y - St = -1
⇒ 10y = 5t - 1
⇒ y = \(\frac { 1 }{ 10 } \)[5t - 1]
Also, from (1), x- \(\frac { 3 }{ 10 } \)[5t-1]+2t = 1
⇒ x = \(\frac { 3 }{ 10 } \) [5t - 1]-2t + 1 = \(\frac { 15t-3-20t+10 }{ 10 } \)
⇒ x = \(\frac { 1 }{ 10 } \) [-5t + 7]
Hence the solution set is \( ( \frac { 7-5t }{ 10 } ,\frac { 5t-1 }{ 10 } ,t\ )\) where t \(\in \) R.

x - y + 2z = 2, 2x + y + 4z = 7, 4x - y + z = 4
The matrix forn n of the system is, AX = B
where A =\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 4 \\ 4 & -1 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 2 \\ 7 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B] we get,
[A|B] =\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 4 \\ 4 & -1 & 1 \end{matrix}|\begin{matrix} 2 \\ 7 \\ 4 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 4 \\ 4 & -1 & 1 \end{matrix}|\begin{matrix} 2 \\ 7 \\ 4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 3 & 0 \\ 0 & 3 & -7 \end{matrix}|\begin{matrix} 2 \\ 3 \\ -7 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) =\(\rho \)[A|B] = 3 = number of unknowns
Hence the system is consistent with unique solution.
Writing the equivalent equations from the rowechelon matrix, we get
x-y+2z = 2 ...........(1)
3y = 3 ⇒ y = 1.........(2)
-7z = -7 ⇒ z = \(\frac{-7}{-7}\) = 1 ............(3)
Solutions y = 1 and z = 1 in (1) we get,
x-1 + 2(1) = 2
⇒ x-1+2 = 2
⇒ x+1 = 2
⇒ x = 2-1 =1
⇒ x = 1
∴ x = 1, y = 1, z = 1
Thus, solution set is (1, 1, 1)

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0

x+y+z-2 = 0, 6x-4y+5z-31= 0, 5x+2y+2z = 13
The matrix form of the system is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 13 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right| =1\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| -1\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \)
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| & +\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ -4 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 6 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 6 & -4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(-8-10) & -(12-25) & +(12+20) \\ -(2-2) & +(2-5) & -(2-5) \\ +(5+4) & -(5-6) & +(-4-6) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 13 & 32 \\ 0 & -3 & 3 \\ 9 & 1 & -10 \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -36 & +0 & +117 \\ 26 & -93 & +13 \\ 64 & +93 & -130 \end{matrix} \right] =\frac { 1 }{ 27 } \left[ \begin{matrix} 81 \\ -54 \\ 27 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ 1 \end{matrix} \right] \)
∴ x = 3, y = -2, z = 1
∴ Solution set is {3, -2, 1}

2x + 3y - z = 9, x + y + z = 9, 3x - y - z = -1
The matrix form of the system is
\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ X = A^-1N
|A| = \(\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \)
= 2(-1+1)-3(-1-3)-1(-1-3)
= 0-3(-4) = 12+4 = 16
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & -1 \\ -1 & -1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ 3 & -1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & -1 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(-1+1) & -(-1-3) & +(-1-3) \\ -(-3-1) & +(2+3) & -(-2-9) \\ +(3+1) & -(2+1) & +(2-3) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 0 & 4 & -4 \\ 4 & 1 & 11 \\ 4 & -3 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
= \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0+36-4 \\ 36+9+3 \\ -36+99+1 \end{matrix} \right] =\frac { 1 }{ 16 } \left[ \begin{matrix} 32 \\ 48 \\ 64 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
∴ x = 2, y = 3, z = 4
∴ Solution set is {2, 3, 4}

\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₃] =\(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & -3 \\ 0 & -2 & 5 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & -3 \\ 0 & -2 & 5 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\\ \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & -3 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ -2 & 1 & 0 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 8 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 & 0 & 1 \\ 13 & -5 & -3 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { R_{ 1 }\rightarrow { R }_{ 1 }+8R_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ -5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }\times (-1) }{ \longrightarrow } \left[ \begin{matrix} 11 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{matrix} \right] \)
So we get A^-1 =\(\left[ \begin{matrix} -40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1 \end{matrix} \right] \).

\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₃] =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 0 & 4 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -6 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 & 1 & 0 \\ -1 & -4 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\\ \overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+R_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -1 & 1 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -3 & - & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
So, We get A^-1=\(\left[ \begin{matrix} -2 & -3 & 1 \\ -3 & -3 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \).

3x + 2y + 7z = 0, 4x − 3y − 2z = 0, 5x + 9y + 23z = 0
Reducing the augmented matrix to row-echelon form we,
[A|0] =\(\left[ \begin{matrix} 3 & 2 & 7 \\ 4 & -3 & -2 \\ 5 & 9 & 23 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 4 }{ 4 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & 9 & 23 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 5 }{ 3 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & \frac { 17 }{ 3 } & \frac { 34 }{ 3 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & -\frac { 17 }{ 3 } & \frac { -34 }{ 3 } \\ 5 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times \frac { -3 }{ 7 } }{ \longrightarrow } \left[ \begin{matrix} 3 & 2 & 7 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|0] = 2
So, \(\rho \)(A) = \(\rho \)(A|0]) = 2<3 = number of unknowns 
So put z = t where t \(\in \) R
writing the equations using the echelon form, we get
3x+2y+7z = 0 ............(1)
y+2z = 0 .............(2)
put z = t, (2) becomes
y+2t = 0
⇒ y = 2t
∴ (1) becomes, 3x+2(-2)+7t = 0
⇒ 3x-4t+7t = 0
⇒ 3x+3t = 0
⇒ 3x = -3t
⇒ x = -t
∴ Solution set is {-t, -2t, t} where t \(\in \) R

Here the number of unknowns is 3. So, if the system is consistent and has a non-trivial solution, then the rank of the coefficient matrix is equal to the rank of the augmented matrix and is less than 3.
So the determinant of the coefficient matrix should be 0.
Hence we get
\(\left| \begin{matrix} 3\lambda -8 & 3 & 3 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 or \(\left| \begin{matrix} 3\lambda -2 & 3\lambda -2 & 3\lambda -2 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by applying R₁ ➝ R₁ + R₂ + R₃)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -8 & 3 \\ 3 & 3 & 3\lambda -8 \end{matrix} \right| \) = 0 (by taking out (3λ − 2) from R₁)
or (3λ - 2) \(\left| \begin{matrix} 1 & 1 & 1 \\ 3 & 3\lambda -11 & 3 \\ 3 & 3 & 3\lambda -11 \end{matrix} \right| \) = 0 (by applying R₂ ➝ R₂ - 3R₁, R₃ ➝ R₃ - 3R₁)
or (3λ - 2)(3λ - 11)² 0. So λ = \(\frac { 2 }{ 3 } \) and λ = \(\frac { 11 }{ 3 } \).
We now give an application of system of linear homogeneous equations to chemistry. You are already aware of balancing chemical reaction equations by inspecting the number of atoms present on both sides.

Here the number of equations is 4 and the number of unknowns is 3. Reducing the augmented matrix to echelon-form, we get
[A | O] = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 6 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -3 \\ \begin{matrix} -7 \\ -9 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 1 \\ \begin{matrix} 10 \\ 10 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }-6{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -5 \\ \begin{matrix} -10 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 5 \\ \begin{matrix} 16 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }\div \left( -5 \right) \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -2 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 5 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -8 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }-5{ R }_{ 2 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }+15{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -3 \\ 7 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -3 \right) \\ { R }_{ 4 }\rightarrow { R }_{ 4 }\div 7 \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\rightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
So, ρ(A) = ρ([A | O]) = 3 =  number of unknowns
Hence the system has trivial solution only.

Here the number of unknowns is 3.
Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes
\(\left[ \begin{matrix} 1 & 3 & -2 \\ 2 & -1 & 4 \\ 1 & -11 & 14 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 3 & -2 \\ 0 & -7 & 8 \\ 0 & -14 & 14 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} R_{ 2 }\rightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -2 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 3 & -2 \\ 0 & 7 & -8 \\ 0 & 7 & -8 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 3 & -2 \\ 0 & 7 & -8 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
So, ρ(A) = ρ([A | 0]) = 2 < 3 = Number of unknowns.
Hence, the system has a one parameter family of solutions.
Writing the equations using the echelon form, we get
x + 3y - 2z = 0, 7y - 8z = 0, 0 = 0.
Taking z = t, where t is an arbitrary real number, we get by back substitution,
z = t,
7y - 8t = 0 ⇒ y = \(\frac { 8t }{ 7 } \),
x + 3\((\frac { 8t }{ 7 } )\) - 2t = 0 ⇒ x + \(\frac { 24t-14t }{ 7 } \) = 0 ⇒ x = \(-\frac { 10t }{ 7 } \).
So, the solution is (x = \(-\frac { 10t }{ 7 } \), y = \(\frac { 8t }{ 7 } \), z = t), where t is any real number.

Here the number of equations is equal to the number of unknowns.
Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -2 & -5 \\ 0 & -4 & -9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow R_{ 2 }\div \left( -1 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div 7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) }{ { \longrightarrow } } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ([A | O]) = 3 = Number of unknowns.
Hence, the system has a unique solution. Since x = 0, y = 0, z = 0 is always a solution of the homogeneous system, the only solution is the trivial solution x = 0, y = 0, z = 0.
Note
In the above example, we find that
|A| = \(\left| \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix} \right| \) = 1(48 - 40) - 2(36 - 28) + 3(30 - 28) = 8 - 16 + 6 = -2 ≠ 0.

2x+3y = 9, 7x+3y-5z = 8, 2x+3y+⋋z = μ
The matrix form of the system is AX = B where
A =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \)
Applying elementary row operations augmented matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 2 & 3 & 5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 8 \\ 9 \\ \mu \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 2 }{ 7 } { R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & \frac { 15 }{ 7 } & \frac { 45 }{ 7 } \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ \frac { 45 }{ 7 } \\ 4-9 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 7 }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ 47 \\ \mu -9 \end{matrix} \right] \)
Case (i): when λ = 5
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B]
Hence the system is inconsistent and has no solution
Case (ii) : When λ ≠ 5, μ ≠ 9
[A|B] =\(\\ \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} -8 \\ 47 \\ not\quad zero \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) = \(\rho \)[A|B] = 3 = number of unknowns
Hence, the system is consistent with solution
Case (iii) : When λ = 5, μ = 9
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2
∴ The system is consistent and has infinite number of solutions.

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

2x - y + z = 2, 6x - 3y + 3z = 6, 4x - 2y + 2z = 4
The matrix form of the given system is AX = B where
A =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B], we get,
[A|B] =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix}|\begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 2 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1 [∵ only one non zero row]
and \(\rho \)[A|B] = 1 [∵ only one-zero row]
∴ \(\rho \)(A) =\(\rho \)(A|B] = 1< 3 the given system is consistent and has two parameter family of solutions.
So, z = t and y = s where, t \(\in \)R
Writing the equivalent equations from the rowechelon matrix, we get
2x-y+z = 2 .............(1)
y = s
z = t
Substituting (2) and (3) In (1) we get
2x-s+t = 2
⇒ 2x-s+t = 2
⇒ x = \(\frac{1}{2}\)[s-t+2]
∴ Solution set is {\(\frac{1}{2}\)(s-t+2),s,t} here s, t \(\in \) R.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 5 & 7 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} a \\ b \\ c \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 5 & 7 \end{matrix}|\begin{matrix} a \\ b \\ c \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \end{matrix}|\begin{matrix} a \\ b-a \\ c-3a \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} a \\ b-a \\ \left( c-3a \right) -2\left( b-a \right) \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} a \\ b-a \\ \left( c-2b-a \right) \end{matrix} \right] \).
In order that the system should have one parameter family of solutions, we must have ρ(A) = ρ([A, B]) = 2. So, the third row in the echelon form should be a zero row.
So, c − 2b − a = 0 ⇒ c = a + 2b.

Here the number of unknowns is 3.
The matrix form of the system of equations is AX = B, where
A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 1 \\ 2 \end{matrix} \end{matrix} \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ \begin{matrix} 4 \\ \begin{matrix} 6 \\ 7 \end{matrix} \end{matrix} \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 4 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 1 \\ 2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} -9 \\ \begin{matrix} 4 \\ \begin{matrix} 6 \\ 7 \end{matrix} \end{matrix} \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \\ { R }_{ 4 }\longrightarrow { R }_{ 4 }-4{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} -9 \\ \begin{matrix} 22 \\ \begin{matrix} 33 \\ 43 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 }, \\ { R }_{ 4 }\longrightarrow { R }_{ 4 }-3{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} -9 \\ \begin{matrix} 22 \\ \begin{matrix} -11 \\ -23 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\leftrightarrow { R }_{ 4 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} -9 \\ \begin{matrix} 22 \\ \begin{matrix} -23 \\ -11 \end{matrix} \end{matrix} \end{matrix} \right] \)
So, ρ(A) = 3 and ρ([A | B]) = 4. Hence ρ(A) ≠ ρ([A | B]).
If we write the equivalent system of equations using the echelon form, we get
x − y + z = −9, y − z = 22, z = −23, 0 = −11. 
The last equation is a contradiction.
So the given system of equations is inconsistent and has no solution.

2x − 2y + 3z = 2, x + 2y − z = 3, 3x − y + 2z = 1
Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 2 & -2 & 3 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix}|\begin{matrix} 2 \\ 3 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \)
\(\left[ \begin{matrix} 1 & 2 & -1 \\ 2 & -2 & 3 \\ 3 & -1 & 2 \end{matrix}|\begin{matrix} 3 \\ 2 \\ 1 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & -7 & 5 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & -1 & 0 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -4 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { 6R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \)
\(\left[ \begin{matrix} 1 & 2 & -1 \\ 0 & -6 & 5 \\ 0 & 0 & -5 \end{matrix}|\begin{matrix} 3 \\ -4 \\ -20 \end{matrix} \right] \)
Writing the equivalent equations from the rowechelon matrix, we get,
x+2y-z = 3 .......... (1)
-6y + 5z =-4 ..........(2)
-5z = -20 ⇒ z =\(\frac{-20}{-5}\) = 4
Substituting z = 4 in 2 we ge
-6y + 5(4) = -4
⇒ -6y+20 = -4 ⇒ -4 - 20 = -20
⇒ y = \(\frac{-24}{-6}\) = 4
Substituting y = z = 4 in (1) we get
x+ 2(4) -4 = 3
⇒ x + 8 - 4 = 3
⇒ x + 4 = 3
⇒ x = 3 - 4 = -1
∴ Solution set is {-1, 4, 4}

The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \),X = \(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] \),B = \(\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \).
We find |A| = \(\left| \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right| \) = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 10 + 15 + 15 = 40 ≠ 0.
So, A⁻¹ exists and
A^-1 = \(\frac { 1 }{ \left| A \right| } \) (adj A) = \(\frac { 1 }{ 40 } { \left[ \begin{matrix} +\left( 4+1 \right) & -\left( -2-3 \right) & +\left( -1+6 \right) \\ -\left( -6+3 \right) & +\left( -4-9 \right) & -\left( -2-9 \right) \\ +\left( 3+6 \right) & -\left( 2-3 \right) & +\left( -4-3 \right) \end{matrix} \right] }^{ T }=\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
Then, applying X = A⁻¹B, we get
\(\left[ \begin{matrix} { x }_{ 1 } \\ { x }_{ 2 } \\ { x }_{ 3 } \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{matrix} \right] =\frac { 1 }{ 40 } \left[ \begin{matrix} 40 \\ 80 \\ -40 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
So, the solution is (x₁ = 1, x₂ = 2, x₃ = -1).

Applying Gauss-Jordan method, we get
[A | I₃] =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \frac { 1 }{ 2 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & \left( 1/2 \right) & -\left( 1/2 \right) \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -\left( 3/2 \right) & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 & -1 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 1 }\longrightarrow { R }_{ 1 }-{ R }_{ 3 } \\ { R }_{ 2 }\longrightarrow { R }_{ 2 }+{ R }_{ 3 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).
So, A^-1 = \(\left[ \begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).

Let the encryption matrix be A =\(\left[ \begin{matrix} -1 & -1 \\ 2 & 1 \end{matrix} \right] \)
|A| = -1 + 2 = 1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1 } \left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)
Hence the decryption matrix is \(\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)

So, the sequence of decoded row matrices is [8 5], [12 16]
Now the 8^th English alphabet is H.
5^th English alphabet is E.
12^th English alphabet is L.
and the 16^th English alphabet is P.
Thus the receiver reads the message as "HELP".

Given F (\(\alpha\)) = \(\left[ \begin{matrix} \cos { \alpha } & 0 & \sin { \alpha } \\ 0 & 1 & 0 \\ -\sin { \alpha } & 0 & \cos { \alpha } \end{matrix} \right] \)
Expanding along R₁ we get,
|F(\(\alpha\))| = cos \(\alpha\) \(\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| -0+sin\alpha \left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \)
= cos \(\alpha\) (cos - 0) + sin \(\alpha\) (0 + sin \(\alpha\))
= cos² + sin² \(\alpha\) = 1 ≠ 0
Since F (\(\alpha\)) is a non-singular matrix, [F(\(\alpha\))]^-1 exists
Now, adj (F(\(\alpha\))) = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ sin\alpha & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & sin\alpha \\ 0 & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & 0 \\ sin\alpha & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & sin\alpha \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & sin\alpha \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & 0 \\ 0 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(cos\alpha -0) & -(0) & +(0+sin\alpha ) \\ -(0) & +(cos^{ 2 }\alpha +sin^{ 2 }\alpha & -(0) \\ +(0-sin\alpha ) & -(0) & +(cos-0) \end{matrix} \right] ^{ T }\)
\(\left[ \begin{matrix} cos\alpha & 0 & +sin\alpha \\ 0 & 1 & 0 \\ -sin\alpha & 0 & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
∴ F(\(\alpha\))^-1 = \(\frac { 1 }{ |F(\alpha )| } \) adj (F(\(\alpha\)))
[F(\(\alpha\))]^-1 = \(\frac { 1 }{ 1 } \left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(1)
Now, F(-\(\alpha\))=\(\left[ \begin{matrix} cos(-\alpha ) & 0 & sin(-\alpha ) \\ 0 & 1 & 0 \\ -s9n(-\alpha ) & 0 & cos(-\alpha ) \end{matrix} \right] \)
=\(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(2)
[∵ cos \(\alpha\) is an even function, cos (-\(\alpha\)) = cos \(\alpha\) and sin \(\alpha\) is an odd function, sin (-\(\alpha\)) = -sin\(\alpha\)]
From (1) and (2)
[F(\(\alpha\))]^-1 = F (-\(\alpha\))

We find that |A| = \(\left| \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & 4 \\ 2 & -4 & 3 \end{matrix} \right| \) = 8(21 - 16) + 6(-18 + 8) + 2(24 - 14) = 40 - 60 + 20 = 0
By the definition of adjoint, we get
adj A = \({ \left[ \begin{matrix} \left( 21-16 \right) & -\left( -18+8 \right) & \left( 24-14 \right) \\ -\left( -18+8 \right) & \left( 24-4 \right) & -\left( 32+12 \right) \\ \left( 24-14 \right) & -\left( -32+12 \right) & \left( 56-36 \right) \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
So, we get
A(adj A) = \(\left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & 80-120+40 & 80-120+40 \\ -30+70-40 & -60+140-80 & -60+140-80 \\ 10-40+30 & 20-80+60 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃,
Similarly, we get
(adj A)A = \(\left[ \begin{matrix} 5 & 10 & 10 \\ 10 & 20 & 20 \\ 10 & 20 & 20 \end{matrix} \right] \left[ \begin{matrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 40-60+20 & -30+70-40 & 10-40+30 \\ 80-120+40 & -60+140-80 & 20-80+60 \\ 80-120+40 & -60+140-80 & 20-80+60 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) = 0I₃ = |A|I₃.
Hence, A(adj A) = (adj A)A = |A|I₃.

Let us consider the matrix \(A=\left[\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array}\right]\)
Then, \(|A|\) =12 + 3 =15 \(\neq\) 0. So, A is non-singular. Let us transform A into I₂ by a sequence of elementary row operations. First, we search for a row operation to make a₁₁ of A as 1. The elementary row operation needed for this is \(R_{1} \rightarrow\left(\frac{1}{2}\right) R_{1}\) corresponding elementary matrix is \(E_{1}=\left[\begin{array}{ll} \frac{1}{2} & 0 \\ 0 & 1 \end{array}\right]\)
Then, we get \(E_{1} A=\left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{-1}{2} \\ 3 & 4 \end{array}\right]\)
Next, let us make all elements below a₁₁ of E₁ A as 0. There is only one element a₂₁ .
The elementary row operation needed for this is \(R_{2} \rightarrow R_{2}+(-3) R_{1}\)
The corresponding elementary matrix is \(E_{2}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right]\)
Then, we get \(E_{2}\left(E_{1} A\right)=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 0 & \frac{11}{2} \end{array}\right]\)
Next, let us make a₂₂ of E₂ (E₁A) ( ) as 1. The elementary row operation needed for this is \(R_{2} \rightarrow\left(\frac{2}{11}\right) R_{2}\)
The corresponding elementary matrix is \(E_{3}=\left[\begin{array}{cc} 1 & 0 \\ 0 & \frac{2}{11} \end{array}\right]\)
Then, we get \(E_{3}\left(E_{2}\left(E_{1} A\right)\right)=\left[\begin{array}{cc} 1 & 0 \\ 0 & \frac{2}{11} \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 0 & \frac{11}{2} \end{array}\right]=\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 0 & 1 \end{array}\right]\)
Finally, let us find an elementary row operation to make a₁₂ of E₃(E₂(E₁A)) as 0. The elementary row operation needed for this is \(R_{1} \rightarrow R_{1}+\left(\frac{1}{2}\right) R_{2}\) . The corresponding elementary matrix is \(E_{4}=\left[\begin{array}{ll} 1 & \frac{1}{2} \\ 0 & 1 \end{array}\right]\)
Then, we get \(E_{4}\left(E_{3}\left(E_{2}\left(E_{1} A\right)\right)\right)=\left[\begin{array}{ll} 1 & \frac{1}{2} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I_{2}\)

The rank of a non-zero matrix is equal to the number of non-zero rows in a row-echelon form of the matrix

The rank of a matrix which is not in a row-echelon form, can be found by applying the following result which is stated without proof.

Replacing A by AB in adj(A) = \(|A| A^{-1}\) we get
\(\operatorname{adj}(A B)=|A B|(A B \mid)^{-1}=\left(|B| B^{-1}\right)\left(|A| A^{-1}\right)=\operatorname{adj}(B) \operatorname{adj}(A)\)

Since A is a non-singular square matrix, we have \(|A| \neq 0\) and so, we get
\(\text { (i) } A^{-1}=\frac{1}{|A|}(\operatorname{adj} A) \Rightarrow \operatorname{adj} A=|A| A^{-1} \Rightarrow(\operatorname{adj} A)^{-1}=\left(|A| A^{-1}\right)^{-1}=\frac{1}{|A|}\left(A^{-1}\right)^{-1}=\frac{1}{|A|} A \text {. }\)
Replacing A by A⁻¹ in adj \(A=|A| A^{-1}, \text { we get } \operatorname{adj}\left(A^{-1}\right)=\left|A^{-1}\right|\left(A^{-1}\right)^{-1}=\frac{1}{|A|} A\)
Hence, we get \((\operatorname{adj} A)^{-1}=\operatorname{adj}\left(A^{-1}\right)=\frac{1}{|A|} A\)
\(\text { (ii) } A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I_{n} \Rightarrow \operatorname{det}(A(\operatorname{adj} A))=\operatorname{det}((\operatorname{adj} A) A)=\operatorname{det}\left(|A| I_{n}\right)\)
\(\left.\Rightarrow|A||\operatorname{adj}| A|=| A\right|^{n} \Rightarrow|\operatorname{adj} A|=|A|^{n-1}\)
For any non-singular matrix B of order n, we have B (adj B ) = (adj B)B = \(|B| I_{n}\)
Put B = adj A. Then, we get \((\operatorname{adj} A)(\operatorname{adj}(\operatorname{adj} A))=|\operatorname{adj} A| I_{n}\)
So, since \(|\operatorname{adj} A|=|A|^{n-1} \text {, we get }(\operatorname{adj} A)(\operatorname{adj}(\operatorname{adj} A))=|A|^{n-1} I_{n}\)
Pre-multiplying both sides by A, we get A \(((\operatorname{adj} A)(\operatorname{adj}(\operatorname{adj} A)))=A\left(|A|^{n-1} I_{n}\right)\)
Using the associative property of matrix multiplication, we get
\((A(\operatorname{adj} A)) \operatorname{adj}(\operatorname{adj} A)=A\left(|A|^{n-1} I_{n}\right)\)
Hence, we get \(\left(|A| I_{n}\right)(\operatorname{adj}(\operatorname{adj} A))=|A|^{n-1} A \text {. That is, } \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A\)
(iv) Replacing A by \(\lambda A \text { in adj }(A)=|A| A^{-1} \text { where } \lambda\) a non-zero scalar, we get
\(\operatorname{adj}(\lambda A)=|\lambda A|(\lambda A)^{-1}=\lambda^{n}|A| \frac{1}{\lambda} A^{-1}=\lambda^{n-1}|A| A^{-1}=\lambda^{n-1} \operatorname{adj}(A)\)
(v) By (iii), we have adj \((\operatorname{adj} A)=|A|^{n-2} A\). So, by taking determinant on both sides, we get
\(|\operatorname{adj}(\operatorname{adj} A)|=\left.\left.|| A\right|^{n-2} A\left|=\left(|A|^{(n-2)}\right)^{n}\right| A|=| A\right|^{n^{2}-2 n+1}=|A|^{(n-1)^{2}} .\)
(vi) Replacing A by A^T in \(A^{-1}=\frac{1}{|A|} \operatorname{adj} A, \text { we get }\left(A^{T}\right)^{-1}=\frac{1}{\left|A^{T}\right|} \operatorname{adj}\left(A^{T}\right)\) and hence, we get \(\operatorname{adj}\left(A^{T}\right)=\left|A^{T}\right|\left(A^{T}\right)^{-1}=|A|\left(A^{-1}\right)^{T}=\left(|A| A^{-1}\right)^{T}=\left(|A| \frac{1}{|A|} \operatorname{adj} A\right)^{T}=(\operatorname{adj} A)^{T}\)

Assume that A is non-singular. Then \(|A| \neq 0, \text { and } A^{-1}\) exists.
Now \(\left|A^{-1}\right|=\frac{1}{|A|} \neq 0 \Rightarrow A^{-1}\) is also non-singular, and \(A A^{-1}=A^{-1} A=I\).
Now, \(A A^{-1}=I \Rightarrow\left(A A^{-1}\right)^{-1}=I \Rightarrow\left(A^{-1}\right)^{-1} A^{-1}=I\)       ......(1)
Post-multiplying by A on both sides of equation (1), we get \(\left(A^{-1}\right)^{-1}=A\)

Assume that A and B are non-singular matrices of same order n. Then \(,|A| \neq 0,|B| \neq 0\) both A⁻¹ and B⁻¹ exist and they are of order n. The products AB and B⁻¹A⁻¹ can be found and they are also of order n. Using the product rule for determinants, we get \(|A B|=|A \| B| \neq 0\). So, AB is non-singular and
\( (A B)\left(B^{-1} A^{-1}\right)=\left(A\left(B B^{-1}\right)\right) A^{-1}=\left(A I_{n}\right) A^{-1}=A A^{-1}=I_{n} \)
\(\left(B^{-1} A^{-1}\right)(A B)=\left(B^{-1}\left(A^{-1} A\right)\right) B=\left(B^{-1} I_{n}\right) B=B^{-1} B=I_{n} \)
Hence (AB)⁻¹ = B⁻¹A⁻¹.

Since A is non-singular, A⁻¹ exists and \(A A^{-1}=A^{-1} A=I_{n}\). Taking BA = CA and post-multiplying both sides by A⁻¹, we get (BA)A⁻¹ = (CA)A⁻¹. By using the associative property of matrix multiplication and property of inverse matrix, we get B = C.

Since A is non-singular, A⁻¹ exists and \(A A^{-1}=A^{-1} A=I_{n}\). Taking AB = AC and pre-multiplying both sides by A⁻¹, we get \(A^{-1}(A B)=A^{-1}(A C)\). By using the associative property of matrix multiplication and property of inverse matrix, we get B = C

Let A be non-singular. Then \(|A| \neq 0\) and A⁻¹ exists. By definition
\(A A^{-1}=A^{-1} A=I_{n}\)
(i) By (1), we get \(\left|A A^{-1}\right|=\left|A^{-1} A\right|=\left|I_{n}\right|\)
Using the product rule for determinants, we get \(|A|\left|A^{-1}\right|=\left|I_{n}\right|=1\)
Hence, \(\left|A^{-1}\right|=\frac{1}{|A|}\)
(ii) From (1), we get \(\left(A A^{-1}\right)^{T}=\left(A^{-1} A\right)^{T}=\left(I_{n}\right)^{T} .\)
Using the reversal law of transpose, we get \(\left(A^{-1}\right)^{T} A^{T}=A^{T}\left(A^{-1}\right)^{T}=I_{n}\). Hence \(\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}\)
(iii) Since λ is a non-zero scalar, from (1), we get \((\lambda A)\left(\frac{1}{\lambda} A^{-1}\right)=\left(\frac{1}{\lambda} A^{-1}\right)(\lambda A)=I_{n}\)
\(\text { So, }(\lambda A)^{-1}=\frac{1}{\lambda} A^{-1}\)

Suppose that A⁻¹ exists. Then \(A A^{-1}=A^{-1} A=I_{n}\)
By the product rule for determinants, we get
\(\operatorname{det}\left(A A^{-1}\right)=\operatorname{det}(A) \operatorname{det}\left(A^{-1}\right)=\operatorname{det}\left(A^{-1}\right) \operatorname{det}(A)=\operatorname{det}\left(I_{n}\right)=1 . \text { So, }|A|=\operatorname{det}(A) \neq 0\)
Hence A is non-singular.
Conversely, suppose that A is non-singular.
Then \(|A| \neq 0\) By Theorem, we get
\(A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I_{n}\)
So, dividing by \(|A| \text {, we get } A\left(\frac{1}{|A|} \operatorname{adj} A\right)=\left(\frac{1}{|A|} \operatorname{adj} A\right) A=I_{n} \text {. }\)
Thus, we are able to find a matrix \(B=\frac{1}{|A|} \operatorname{adj} A\) such that AB = BA I_n.
Hence, the inverse of A exists and it is given by \(A^{-1}=\frac{1}{|A|} \mathbf{a d j} A\)

Let A be a square matrix order n such that an inverse of A exists. If possible, let there be two inverses B and C of A. Then, by definition, we have AB = BA = I_n and AC = CA = I_n  
Using these equations, we get
\(C=C I_{n}=C(A B)=(C A) B=I_{n} B=B\)
Hence the uniqueness follows.
Notation The inverse of a matrix A is denoted by A⁻¹.

For simplicity, we prove the theorem for n = 3 only
Consider A \(=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]\) Then, we get
\(\begin{array}{lll} a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}=|A|, & a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=0, & a_{11} A_{31}+a_{12} A_{32}+a_{13} A_{33}=0 \\ a_{21} A_{11}+a_{22} A_{12}+a_{23} A_{13}=0, & a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23}=|A|, & a_{21} A_{31}+a_{22} A_{32}+a_{23} A_{33}=0 \\ a_{31} A_{11}+a_{32} A_{12}+a_{33} A_{13}=0, & a_{31} A_{21}+a_{32} A_{22}+a_{33} A_{23}=0, & a_{31} A_{31}+a_{32} A_{32}+a_{33} A_{33}=|A| . \end{array}\) 
By using the above equations, we get
\(A(\operatorname{adj} A)=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]\left[\begin{array}{ccc} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}\right]=\left[\begin{array}{ccc} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{array}\right]=|A|\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=|A| I_{3}\) ............... (1)
\((\operatorname{adj} A) A=\left[\begin{array}{lll} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}\right]\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]=\left[\begin{array}{ccc} |A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A| \end{array}\right]=|A|\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=|A| I_{3},\) ............... (2)
where I₃ is the identity matrix of order 3.
So, by equations (1) and (2), we get \(A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I_{3}\) 

Let A = \(\left[ \begin{matrix} 4 & 3 \\ -3 & -1 \\ 6 & 7 \end{matrix}\begin{matrix} 1 & -2 \\ -2 & 4 \\ -1 & 2 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 4. So ρ(A)  ≤ min {3, 4} = 3.
The highest order of minors of A is 3. We search for a non-zero third-order minor of A. But we find that all of them vanish. In fact, we have
\(\left| \begin{matrix} 4 & 3 & 1 \\ -3 & -1 & -2 \\ 6 & 7 & -1 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 3 & -2 \\ -3 & -1 & 4 \\ 6 & 7 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 1 & -2 \\ -3 & -2 & 4 \\ 6 & -1 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 3 & 1 & -2 \\ -1 & -2 & 4 \\ 7 & -1 & 2 \end{matrix} \right| \) = 0.
So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A.
We find that \(\left| \begin{matrix} 4 & 3 \\ -3 & -1 \end{matrix} \right| \) = -4 + 9 = 5 ≠ 0. So, ρ(A) = 2.

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

\(\left[ \begin{matrix} 3 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -8 \\ \begin{matrix} -5 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 5 \\ \begin{matrix} 1 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ -2 \end{matrix} \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} 3 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -8 \\ \begin{matrix} -5 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 5 \\ \begin{matrix} 1 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ -2 \end{matrix} \end{matrix} \right] \)
A = \(\overset { { { R } }_{ 3 }\leftrightarrow { { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 2 \\ 3 \end{matrix}\begin{matrix} 2 \\ -5 \\ -8 \end{matrix}\begin{matrix} 3 \\ 1 \\ 5 \end{matrix}\begin{matrix} -2 \\ 4 \\ 2 \end{matrix} \right] \)
\(\begin{matrix} { { R } }_{ 2 }\rightarrow { { R } }_{ 2 }+2{ { R } }_{ 1 } \\ \longrightarrow \\ { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }+2{ { R } }_{ 1 } \end{matrix}\left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ -2 \end{matrix}\begin{matrix} 3 \\ 7 \\ 14 \end{matrix}\begin{matrix} -2 \\ 0 \\ -4 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ -1 \end{matrix}\begin{matrix} 3 \\ 7 \\ 7 \end{matrix}\begin{matrix} -2 \\ 0 \\ -2 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 2 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 7 \\ 0 \end{matrix}\begin{matrix} -2 \\ 0 \\ -2 \end{matrix} \right] \)
The last equivalent matrix is in row-echelon form. It has three non-zero rows.
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 1 \\ 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \right] \overset { { { R } }_{ 2 }\rightarrow { { R } }_{ 2 }3{ { R } }_{ 1 }\\ { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-{ { R } }_{ 1 } }{ \underset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-{ { R } }_{ 1 } }{ \longrightarrow } } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} -1 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow { { R } }_{ 4 }-3{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 1 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow 7{ { R } }_{ 3 }-{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ -1 \end{matrix} \end{matrix} \right] \)
\(\overset { { { R } }_{ 4 }\rightarrow 2{ { R } }_{ 4 }-{ { R } }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 2 \\ -7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -1 \\ 5 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix} \right] \)
The last equivalent matrix is in row echelon form It has three non-zero rows.
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 4
∴ \(\rho \)(A) ≤ min(3, 4) = 3 
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{matrix} \right| \) = 0+0-8(4-4) = 0
[Expanded along C₁]
Also, \(\left| \begin{matrix} 0 & 2 & 1 \\ 0 & 4 & 3 \\ 8 & 0 & 2 \end{matrix} \right| =0+0-8\left| \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right| \)
[Expanded along C₁]
= -8(6-4) = -8(2) = -16 ≠ 0
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 3
∴ \(\rho \)(A) ≤ min(3, 3) = 2
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right| =1\left| \begin{matrix} 4 & -6 \\ 1 & -1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} \right| \)
[Expanded along R₁]
= 1(-4+6)+2(-2+30)+3(2-20)
= 1(2)+2(28)+3(-18)
= 2+56-54 = 58-54 = 4 ≠ 0
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Expanding along R₁ we get,
|A| = \(2\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| -3\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| +1\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \)
= 2(8-7) -3 (6-3) +1(21-12)
= 2(1) - 3(3) + 1(9)
Since A is a non-singular matrix, A^-1 exis
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)+(14-9) \\ +(3-4)+(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)
Now, A-1 = \(\frac { 1 }{ |A| } \)adj A
⇒ A-1 = \(\frac{1}{2} \left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Expending along R₁,
|A| = \(5\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| +1\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \)
= 5 (25 - 1)-1 (5 - 1)+ 1 (1 - 5)
= 5 (24) - 1(4) + 1(- 4)
= 120 - 4 - 4 = 120 - 8 = 112 ≠ 0
Since A is non singular, A^-1 exit
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 5 & 1 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(25-1)-(5-1)+(1-5) \\ -(5-1)+(25-1)-(5-1) \\ +(1-5)+(5-1)+(25-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -24 \\ -4 & -4 & 24 \end{matrix} \right] \)
Taking 4 common from every entry we get,
adj A = \(4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 112 } .4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
= \(\frac { 1 }{ 28 } \left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \).

5x − 2y + 16 = 0, x + 3y − 7 = 0
Given Δ = \(\left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| \) = 15+2 = 17
Δ₁ = \(\left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| \) = -48+14 = -34
Δ₂ = \(\left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| \) = 35+16 = 51
∴ x = \(\frac { \triangle _{ 1 } }{ \triangle } =\frac { -34 }{ 7 } \) = -2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 51 }{ 17 } \)
∴ Solution set is {-2, 3}

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
A =\(\left[ \begin{matrix} 1 \\ 2 \\ 5 \end{matrix}\begin{matrix} 1 \\ -1 \\ -1 \end{matrix}\begin{matrix} 1 \\ 3 \\ 7 \end{matrix}\begin{matrix} 3 \\ 4 \\ 11 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { { R } }_{ 2 }-2{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 5 \end{matrix}\begin{matrix} 1 \\ -3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 7 \end{matrix}\begin{matrix} 3 \\ -2 \\ 11 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-5{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -6 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\begin{matrix} 3 \\ -2 \\ -4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { { R } }_{ 3 }-2{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 3 \\ -2 \\ 0 \end{matrix} \right] \)
The last equivalent matrix is in row echelon form it ha two non-zero row \(\rho \)(A) = 2

Let A = \(\left[ \begin{matrix} 3 & 1 & 4 \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \). Then, |A| = 3(0 - 2) - 1(2 + 5) + 4(4 - 0) = 6 - 7 + 16 = 15 ≠ 0. So, A is non-singular.
Keeping the identity matrix as our goal, we perform the row operations sequentially on A as follows:
\(\left[ \begin{matrix} 3 & 1 & 4 \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \frac { 1 }{ 3 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 2 & 0 & -1 \\ 5 & 2 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 0 & -\frac { 2 }{ 3 } & -\frac { 11 }{ 3 } \\ 0 & \frac { 1 }{ 3 } & -\frac { 17 }{ 3 } \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \left( \frac { 3 }{ 2 } \right) { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \frac { 1 }{ 3 } & \frac { 4 }{ 3 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & \frac { 1 }{ 3 } & -\frac { 17 }{ 3 } \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }-\frac { 1 }{ 3 } { R }_{ 2 },{ R }_{ 3 }-\frac { 1 }{ 3 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -\frac { 1 }{ 2 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & 0 & -\frac { 15 }{ 2 } \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -\frac { 2 }{ 15 } \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -\frac { 1 }{ 2 } \\ 0 & 1 & \frac { 11 }{ 2 } \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 3 }.{ R }_{ 2 }-\frac { 11 }{ 2 } { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

\(\left[ \begin{matrix} 0 \\ -1 \\ 4 \end{matrix}\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}\begin{matrix} 6 \\ 5 \\ 0 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 4 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 0 & 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 8 & 20 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 2 }{ 3 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ \frac { 22 }{ 3 } & 16 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 3R_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 22 & 48 \end{matrix} \right] \)

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

Given A =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] =\left[ \begin{matrix} -3+10 & -9+4 \\ -7+25 & -21+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 & -5 \\ 18 & -11 \end{matrix} \right] \)
|AB| = -77+90 = 13 ≠ 0 ⇒ (AB)^-1 exists
|A| = 15-14 = 1 ≠ 0 ⇒ A^-1 exists
|B| = -2+15 = 13 ≠ 0 ⇒ B^-1 exists
(AB)^-1 = \(\frac { 1 }{ |AB| } adj(AB)=\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ...............(1)
B^-1 = \(\frac { 1 }{ |B| } adj(B)=\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \)
A^-1 = \(\frac { 1 }{ |A| } \)(adj A)
= \(\frac { 1 }{ 1 } \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) =\left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
∴ B^-1A^-1 = \(\frac { 1 }{ 13 } \left( \begin{matrix} 2 & 3 \\ -5 & -1 \end{matrix} \right) \left( \begin{matrix} 5 & -2 \\ -7 & 3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} 10-21 & -4+9 \\ -25+7 & 10-3 \end{matrix} \right) \)
= \(\frac { 1 }{ 13 } \left( \begin{matrix} -11 & 5 \\ -18 & 7 \end{matrix} \right) \) ..............(2)
 From (1) or (2) it is proved that
(AB)^-1 = B^-1 A^-1

Given A =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂

Given A = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \)
A^T = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)...............(1)
We know that (\(\lambda\)A) -1 = \(\frac { 1 }{ \lambda } \)A-1
A-1 =\(\left\{ \frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \right\} ^{ -1 }=\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] ^{ T }\)
where \(\lambda\) = \(\frac{1}{9}\)
A^-1= 9B^-1 where B =\(\left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \) ............(2)
Now, |B|=\(-8\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| -1\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| +4\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \)
= -8(16+56)-1(16-7)+4(-32-4)
= -8(72)-1(9)+4(-36) = -576-9-144
= -729
adj B =\(\left[ \begin{matrix} +\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| & -\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| & +\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 4 \\ -8 & 4 \end{matrix} \right| & +\left| \begin{matrix} -8 & 4 \\ 1 & 4 \end{matrix} \right| & -\left| \begin{matrix} -8 & 1 \\ 1 & -8 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 4 \\ 4 & 7 \end{matrix} \right| & -\left| \begin{matrix} -8 & 4 \\ 4 & 7 \end{matrix} \right| & +\left| \begin{matrix} -8 & 1 \\ 4 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(16+56)-(16-7)+(-32-3) \\ -(4+32)+(-32-4)+(64-1) \\ +(7-16)-(-56-16)+(-3-4) \end{matrix} \right] \)
=\(\left[ \begin{matrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{matrix} \right] \)
= \(9\left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
∴ B^-1=\(\frac { 1 }{ |B| } adjB=\frac { -9 }{ 729 } \left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
= \(\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)
Substituting this in (2) we get,
A^-1 = \(9.\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] =\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \) ...............(3)
From (1) and (3)we get,
A^T = A^-1

Given A =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \)
A² = \(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22 & 9 \\ -2 & 1 \end{matrix} \right] \)
∴ A²- 3A - 7I₂
=\(\left[ \begin{matrix} 22 & 9 \\ -3 & 1 \end{matrix} \right] -3\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22-15-7 & 9-9+0 \\ -3+3+0 & 1+6-7 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)= O₂
Hence proved.
∴ A²-3A-7I₂ = O₂
Postmultiplying by A^-1 we get,
A²-A^-1-3AA^-1-7I₂A^-1 = 0.A^-1
⇒ A(AA^-1)-3(AA^-1)-7(A^-1) = 0
[∵ I₂A^-1 = A^-1 and | (0)A^-1= 0]
⇒ AI-3I-7A-1 = 0  [∵ AA^-1= 1]
⇒ AI-3I = 7A^-1
⇒ A^-1 = \(\frac { 1 }{ 7 } \)[A - 3I] [∴ AI = A]
⇒ A^-1 = \(\frac { 1 }{ 7 } =\left[ \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right] \)
⇒ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

For the given A, We get |A| = (2)(7) - (9)(1) = 14 - 9 = 5. So, A^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -9 \\ -1 & 2 \end{matrix} \right] =\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 9 }{ 5 } \\ -\frac { 1 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \).
Then, (A^-1)^T = \(\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 1 }{ 5 } \\ -\frac { 9 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).           ....(1)
For the given A, We get A^T = \(\left[ \begin{matrix} 2 & 1 \\ 9 & 7 \end{matrix} \right] \). So |A^T| = (2)(7) - (1)(9) = 5
Then, (A^T)^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).             ...(2)
From (1) and (2), we get (A^-1)^T = (A^T)^-1. Thus, we have verified the given property.

First, we find |adj (A)| = \(\left| \begin{matrix} 7 & 7 & -7 \\ -1 & 11 & 7 \\ 11 & 5 & 7 \end{matrix} \right| \) =  7(77 - 35) - 7(-7 - 77) - 7(-5 - 121) = 1764 > 0
So, we get
A = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(adj A) = \(\pm \frac { 1 }{ \sqrt { 1764 } } { \left[ \begin{matrix} +\left( 77-35 \right) & -\left( -7-77 \right) & +\left( -5-121 \right) \\ -\left( 49+35 \right) & +\left( 49+77 \right) & -\left( 35-77 \right) \\ +\left( 49+77 \right) & -\left( 49-7 \right) & +\left( 77+7 \right) \end{matrix} \right] }^{ T }\)
= \(\pm \frac { 1 }{ 42 } { \left[ \begin{matrix} 42 & 84 & -126 \\ -84 & 126 & 42 \\ 126 & -42 & 84 \end{matrix} \right] }^{ T }=\pm \left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2 \end{matrix} \right] \).

Let A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \). Then |A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0.
Therefore, A⁻¹ exists. Now, we get
adj A = \({ \left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \).
Hence, A^-1 = \(\frac { 1 }{ \left| A \right| } \)(adj A) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

Distance from the orgin (0, 0, 0) to the plane is 
\(=\left|\frac{3(0)-6(0)+2(0)+7}{\sqrt{9+36+4}}\right|\\ =\frac{7}{\sqrt{49}}\\ =\frac{7}{7}=1\)

\(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right| \)= 6 - 4 = 2 ≠ 0
Since A is nonsingular, A^-1 exists
A^-1 = \(\frac { 1 }{ |A| } \)
Now, adj A = \(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \)
[Interchange the entries in leading diagonal and change the sign of elements in the off diagonal]
∴ A^-1 = \(\frac{1}{2}\)\(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \).