\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

\(A=\left[ \begin{matrix} 2 & 3 \\ 1 & 4 \end{matrix} \right] \)
adj A = [A_ij]^T
\(=\left[ \begin{matrix} 4 & -3 \\ -1 & 2 \end{matrix} \right] \)

Minor of 1 = M₁₁ = 3
Minor of − 2 = M₁₂ = 4
Minor of 4 = M₂₁ = − 2
Minor of 3 = M₂₂ = 1
Cofactor of 1 = A₁₁ = (-1)¹⁺¹M₁₁ = 3
Cofactor of -2 = A₁₂ = (-1)¹⁺²M₁₂ = -4
Cofactor of 4 = A₂₁ = (-1)²⁺¹M₂₁ = 2
Cofactor of 3 = A₂₂ = (-1)²⁺²M₂₂ = 1

\(\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] \)
\(A=\left(\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right)\)
\(|A|=1(10-0)=10 \neq 0(\text {expanding along} \ C_1 )\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(A_{11}=\text {Cofactor of } 1=10-0=10 \)
\(A_{12}=\text {Cofactor of } 2=-(0-0)=0 \)
\(A_{13}=\text {Cofactor of } 3=0-0=0 \)
\(A_{21}=\text {Co-factor of } 0=-(10-0)=-10 \)
\(A_{22}=\text {Co-factor of } 2=5-0=5 \)
\(A_{23}=\text {Co-factor of } 4=-(0-0)=0\)
\(A_{31}=\text {Co-factor of } 0=8-6=2 \)
\(A_{32}=\text {Co-factor of } 0=-(4-0)=-4 \)
\(A_{33}=\text {Co-factor of } 5=2-0=2\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2 \end{array}\right)\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{10}\left(\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right)\)

2x + 3y = 5
x - 2y = -1
The given system can be written as
\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\)\(\begin{bmatrix} X\\Y \end{bmatrix}\)=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
AX = B \(\Rightarrow \) \(X=A^{-1} B\)
Where A =\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\), \(X= \left[ \begin{matrix} x \\ y \end{matrix} \right] \),B=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
\(|A|=-4-3=-7 \neq 0\)
\(\therefore\) A^-1 exists.
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\)
\(X=A^{-1} B\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\left(\begin{array}{c} 5 \\ -1 \end{array}\right)\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -10 & +3 \\ -5 & -2 \end{array}\right)=\frac{-1}{7}\left(\begin{array}{l} -7 \\ -7 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)\)
\(\text {Solution is } x=1, y=1\)