Given A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)
\(\therefore\)A² = A.A =\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)\(\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 9+2 & 6+2 \\ 3+1 & 2+1 \end{bmatrix}=\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}\)
Given A² + aA + bI = 0
⇒ \(\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}+a\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}+b\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)= 0
⇒ \(\begin{bmatrix} 11+3a+b & 8+2a+0 \\ 4+a+0 & 3+a+b \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Equating the like terms we get,
4 + a = 0 ⇒ a = -4
3 + a + b = 0 ⇒ 3 - 4 + b = 0
-1 + b = 0 ⇒ b = 1

Given A \(=\begin{bmatrix} 3&2\\7&5 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 3&2\\7&5 \end{vmatrix}=15-14=1\Rightarrow{A}^{-1}\) exists.
A₁₁ = 5, A₁₂ = -7, A₂₁ = -2, A₂₂ = 3
\(\therefore\)  adj A \({=\begin{bmatrix} 5&-7 \\ -2 & 3 \end{bmatrix}}^{T}=\begin{bmatrix} 5 & -2 \\-7 & 3 \end{bmatrix}\)
\(\therefore\ {A}^{-1}={{1}\over{|A|}}adj\ A={{1}\over{1}}\begin{bmatrix} 5&-2 \\ -7&3 \end{bmatrix}=\begin{bmatrix} 5&-2 \\ -7&3 \end{bmatrix}\)
Now B \(=\begin{bmatrix} 4&6\\ 3&2 \end{bmatrix}\)
\(|B|=\begin{vmatrix}4 &6 \\ 3 &2 \end{vmatrix}=8-18-10\Rightarrow{B}^{-1}\) exists.
B₁₁ = 2, B₁₂ = -3, B₂₁ = -6, B₂₂ = 4
\(\therefore\ adj\ B={\begin{bmatrix} 2 & -3 \\ -6 &4 \end{bmatrix}}^{T}=\begin{bmatrix} 2&-6 \\ -3 &4 \end{bmatrix}\)
\(\therefore\ {B}^{-1}={{1}\over{|B|}}adj\ B={{1}\over{10}}\begin{bmatrix} 2&-6 \\ -3&4 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -2&6 \\ 3 &-4 \end{bmatrix}\)
\(\therefore\ {B}^{-1}{A}^{-1}={{1}\over{10}}\begin{bmatrix} -2 & 6 \\3 & -4 \end{bmatrix}\begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -10-42& 4+18\\ 15+28 & -6-12 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -52 & 22 \\ 43 & -18 \end{bmatrix}\)........(1)
\(AB=\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\begin{bmatrix} 4 & 6 \\ 3 & 2 \end{bmatrix}=\begin{bmatrix} 12+6 & 18+4\\ 28+15 & 42+10\end{bmatrix}=\begin{bmatrix} 18 & 22\\ 43 & 52 \end{bmatrix}\)
\(|AB|=\begin{vmatrix} 18&22 \\ 43 &52 \end{vmatrix}=936-946=-10\Rightarrow{(AB)}^{-1}\) exists.
\(adj\ AB={\begin{bmatrix} 52 &-43 \\ -22& 18 \end{bmatrix}}^{T}=\begin{bmatrix} 52 & -22 \\ -43 & 18 \end{bmatrix}\)
\(\therefore \ {(AB)}^{-1}adj\ (AB)={{-1}\over{10}}\begin{bmatrix} 52&-22 \\-43 & 18 \end{bmatrix}={{1}\over{10}}\begin{bmatrix} -52 & 22 \\ 43 &-18 \end{bmatrix}\)
From (1) and (2), (AB)^-1 = B^-1 A^-1.
Hence proved.

Let A =\(\left[ \begin{matrix} 2 & -1 & 3 \\ 0 & 5 & 1 \\ 3 & 6 & 8 \end{matrix} \right] \)
A₁₁ = 40 - 6 = 34
A₂₁ = -(-8 - 18) = 26
A₃₁ = -1 - 15 = -16
A₂₁ = -(0 - 3) = 3
A₂₂ = 16 - 9 = 7
A₃₂ = -(2 - 0) = -2
A₁₃ = 0 - 15 = -15
A₂₃ = -(12 + 3) = -15
A₃₃ = 10 - 0 = 10
\(\therefore\) adj A=\({ \left[ \begin{matrix} 34 & 3 & -15 \\ 26 & 7 & -15 \\ -16 & -2 & 10 \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 34 & 26 & -16 \\ 3 & 7 & -2 \\ -15 & -15 & 10 \end{matrix} \right] }\)

Let \(\triangle=\begin{vmatrix} 6&-1&5\\3&0&4\\-2&7&-3 \end{vmatrix}\)
Now M₁₂ = \(\begin{vmatrix} 3&4\\-2&-3 \end{vmatrix}=-9-(-8)=-1\)
M₂₂ = \(\begin{vmatrix} 6&5\\-2&-3 \end{vmatrix}=-18-(-10)=-8\)
M₃₂ = \(\begin{vmatrix} 6&5\\3&4 \end{vmatrix}=24-15=9\)
Now expansion of |A| using co-factors of elements of second column we get,
|A| = a₁₂ A₁₂ + a₂₁ A₂₁ + a₃₁A₃₁
A₁₂ = (-1)^{1 + 2} M₁₂ (-1) (-1) = 1
A₂₂ = (-1)^{2 + 2} M₂₂ = 1(-8) = -8
A₃₂ = (-1)^{3 + 2} M₃₂ = -(9) = -9
\(\therefore\) |A| = -1(1) + 0(-8) + 7(-9) = -1 - 63 = -64.

Let A = \(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| \)
Applying C₁➝C₁+9C₂ we get
A = \(\left| \begin{matrix} 2+63 & 7 & 65 \\ 3+72 & 7 & 65 \\ 5+81 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86 \end{matrix} \right| =0[ \because {C}_{1}\equiv{C}_{3}]\)