Let A = \(\left[ \begin{matrix} 6 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -9 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). Then A is a matrix of order 4 × 3 and ρ(A) ≤ 3.
The last two rows are zero rows. There are several second order minors.
We find that there is a second order minor, for example, \(\left| \begin{matrix} 6 & 0 \\ 0 & 2 \end{matrix} \right| \) = (6)(2) = 12 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third and fourth rows are zero rows.

Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

\(\left[ \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} -2 \\ -6 \end{matrix}\begin{matrix} -1 \\ -3 \end{matrix}\begin{matrix} 0 \\ 1 \end{matrix} \right] \)
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

\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 2
∴ \(\rho \)(A) ≤ min (3, 2) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} -1 & 3 \\ 4 & -7 \end{matrix} \right| \)= 7-12 = 5 ≠ 0
∴ \(\rho \)(A) = 2

\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
A is a matrix of order 2 \(\times\) 2
∴ \(\rho \)(A) ≤ min(2,2) = 2
The highest order of minor of A is 2
it is \(\left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| \)= 4 - 4 = 0
So, \(\rho \)(A)<2
Next consider the minor of order 1 |2| = 2 ≠ 0
∴ \(\rho \)(A) = 1

Let A = \(\left[ \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3
The third order minor |A| = \(\left| \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right| \) = (2)(3)(1) = 6 ≠ 0. So, ρ(A) = 3.
Note that there are three non-zero rows.

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

\(\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)
Let A = \(\left( \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right) \)
adj A = \(\left( \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right) \)
[Interchange the elements in the leading diagonal and change the sign of the elements in off diagonal]

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.

Suppose A is symmetric. Then, A^T = A and so, by theorem (vi), we get
adj(A^T) = (adj A)^T ⇒ adj A = (adj A)^T ⇒ adj A is symmetric.

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 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

\(\frac { 1 }{ 3 } \left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \) and λ = \(\frac { 1 }{ 3 } \)
Since adj (λA) = λ^n-1(adj A)
we get adj \(\left( \frac { 1 }{ 3 } \left[ \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2 \end{matrix} \right] \right) =\left( \frac { 1 }{ 3 } \right) ^{ 2 }\)
adj\(\left( \begin{matrix} 2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 1 \end{matrix} \right) \)
∴ Required adjoint matrix
\(\frac { 1 }{ 9 } \left[ \begin{matrix} +\left| \begin{matrix} 1 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} -2 & 2 \\ 1 & 2 \end{matrix} \right| & +\left| \begin{matrix} -2 & 1 \\ 1 & -2 \end{matrix} \right| \\ -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| \\ +\left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\frac { 1 }{ 9 } \left[ \begin{matrix} (2+4)-(-4-2)+(4-1) \\ -(4+2)+(4-1)+(-4-2) \\ +(4-1)-(4+2)+(2+4) \end{matrix} \right] ^{ T }\)
= \(\frac { 1 }{ 9 } \left[ \begin{matrix} 6 & 6 & 3 \\ -6 & 3 & 6 \\ 3 & -6 & 6 \end{matrix} \right] ^{ T }=\frac { 1 }{ 9 } \left[ \begin{matrix} 6 & -6 & 3 \\ 6 & 3 & -6 \\ 3 & 6 & 6 \end{matrix} \right] \)
= \(\frac { 3 }{ 9 } \left[ \begin{matrix} 2 & -2 & 1 \\ 2 & -1 & -2 \\ 1 & 2 & 2 \end{matrix} \right] \)
[Taking 3 common from each entry]
=\(\frac { 1 }{ 3 } \left[ \begin{matrix} 2 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 2 & 2 \end{matrix} \right] \).

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

The matrix form of the system is AX = B , where A = \(\left[ \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] \)
We find |A| = \(\left| \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right| \) = 10 - 6 = 4 ≠ 0. So, A⁻¹ exists and A⁻¹ = \(\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \)
Then, applying the formula X = A⁻¹B, we get
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} -4 \\ 16 \end{matrix} \right] =\left[ \begin{matrix} \frac { -4 }{ 4 } \\ \frac { 16 }{ 4 } \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 4 \end{matrix} \right] \).
So the solution is (x = −1, y = 4).

Let A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 0 & 5 \end{matrix} \right] \). Applying elementary row operations, we get
A \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & -6 & -4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -3 & -2 \\ 0 & 0 & 0 \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has two non-zero rows. So, ρ(A) = 2.

Let A = \(\left[ \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 3. So ρ(A) ≤ min {3, 3} = 3. 
The highest order of minors of A is 3. There is only one third order minor of A.
It is \(\left| \begin{matrix} 3 & 2 & 5 \\ 1 & 1 & 1 \\ 3 & 3 & 6 \end{matrix} \right| \) = 3(6 - 6) - 2(6 - 6) + 5(3 - 3) = 0. So, ρ(A) < 3.
Next consider the second - order minors of A.
We find that the second order minor \(\left| \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right| \) = 3 - 2 ≠ 0. So  ρ(A) = 2.

Given adj A =\(\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
Now adj(adj A) =\(\left[ \begin{matrix} +\left| \begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 2 \\ -1 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 0 \\ -1 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & 1 \\ 2 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(2-0) & -(0) & +(0+2) \\ -(0) & +(1+1) & -(0) \\ +(0+2) & -(0) & +(2-0) \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 2 & 0 & 2 \\ 0 & 2 & 0 \\ -2 & 0 & 2 \end{matrix} \right] ^{ T }\)
adj(adj A) =\(\left[ \begin{matrix} 2 & 0 & -2 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{matrix} \right] \)

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

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.

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
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 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).