Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

Given A³ = I ⇒ IA³I = |I|
⇒ |A.A.A| = 1
⇒ |A|. |A|·|A| = 1
⇒ |A|³ = 1
∴ |A| ≠ 0
Hence, A is non-singular.

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

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Let A be a non-singular matrix of order 2m+1, where m = 0, 1, 2,... Then, we get |A| ≠ 0 and, by property (ii), we have |adj A| = |A|^(2m+1) − 1 = |A|^2m.
Since |A|^2m is always positive, we get that |adj A| is positive.

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

Given rank of \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \) is 2
⇒ The value of the third order determinant is zero
⇒ \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \)=0
⇒ λ\(\left| \begin{matrix} \lambda & -1 \\ 0 & \lambda \end{matrix} \right| +1\left| \begin{matrix} 0 & -1 \\ -1 & \lambda \end{matrix} \right| \)+0 = 0
⇒ λ(λ² - 0) + 1(0 - 1) = 0
⇒ λ³ - 1 = 0 ⇒ λ³ = 1 ⇒ λ = 1
∴ λ = 1

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.

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
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 = \(\frac{1}{2} \left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

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

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 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

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) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

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

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=\(\left[ \begin{matrix} -2 & -3 & 1 \\ -3 & -3 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \).

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution.
So, we have \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0, Applying R₂ ➝ R₂ - R₁ and R₃ ➝ R₃ - R₁ in the above equation,
we get \(\left| \begin{matrix} p & b & c \\ a-p & q-b & c \\ a-p & b & r-c \end{matrix} \right| \) = 0. That is,  \(\left| \begin{matrix} p & b & c \\ -\left( p-a \right) & q-b & c \\ -\left( p-a \right) & b & r-c \end{matrix} \right| \) = 0.
Since p ≠ a, q ≠ b, r ≠ c, we get (p - a)(q - b)(r - c) \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
So, we have \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
Expanding the determinant, we get \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 0.
That is, \(\frac { p }{ p-a } +\frac { q-\left( q-b \right) }{ q-b } +\frac { r-\left( r-c \right) }{ r-c } \) = 0
⇒ \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 2.

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.

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 = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix[A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix}|\begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -9 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ ([A | B]) = 1 < 3.
From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.
The equivalent system has one non-trivial equation and three unknowns.
Taking y = s, z = t arbitrarily, we get x - s + t = -9; x = -9 + s - t.
So, the solution is (x = -9 + s - t, y = s, z = t), where s and t are parameters.
The above solution set is a two-parameter family of solutions.
Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

Let the time by one man alone be x days and one woman alone be y days
∴ By the given data,
\(\frac { 4 }{ x } +\frac { 4 }{ y } =\frac { 1 }{ 3 } \)
and \(\frac { 2 }{ x } +\frac { 5 }{ y } =\frac { 1 }{ 4 } \)
put \(\frac { 1 }{ x } \) = s and \(\frac { 1 }{ y } \) = t
∴ 4s + 4t = \(\frac { 1 }{ 3 } \)
and 2s + 5t = \(\frac { 1 }{ 4 } \)
The matrix form of the system of equation is
\(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} s \\ t \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \) ⇒ AX = B where
A = \(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \)
X = A^-1B
Now |A| = \(\left| \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right| \) = 20 - 8 =12 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \left[ \frac { \begin{matrix} 1 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 4 \end{matrix} } \right] \)
=\(\frac { 1 }{ 12 } \left[ \begin{matrix} \frac { 5 }{ 3 } & -1 \\ \frac { -2 }{ 3 } & +1 \end{matrix} \right] \)
= \(\frac { 1 }{ 12 } \left[ \frac { \begin{matrix} 2 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 3 \end{matrix} } \right] =\left[ \begin{matrix} \frac { 2 }{ 3 } \times \frac { 1 }{ 12 } \\ \frac { 1 }{ 3 } \times \frac { 1 }{ 12 } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 18 } \\ \frac { 1 }{ 36 } \end{matrix} \right] \)
∴ \(\frac { 1 }{ 18 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 18 } \Rightarrow \)x = 18
t = \(\frac { 1 }{ 36 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 36 } \Rightarrow \)y = 36
one man can do 18 days
one woman can do 36 days.

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Applying Gauss-Jordan method, we get
[A | I₂] = \(\left[ \begin{matrix} 0 & 5 \\ -1 & 6 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \frac { 1 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 1 \end{matrix}|\begin{matrix} 0 & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+6{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \).
So, we get A^-1 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).

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".