Solution:

(i) $\text{A}\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{= }\!\!~\!\!\text{ }\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A=}\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

Matrix A is given as

A = $\left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]$

 

Adj A = $\left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]$

 

det A = $\left| \text{A} \right|$

= $\left| \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right|$

= $1\times 6-2\times 4$

= $6-8$

$\left| \text{A} \right|$ = $-2$

 

Now

$\text{A}\left( Adj\text{ }\!\!~\!\!\text{ A} \right)$

= $\left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]\times \left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1~\times 6+2~\times \left( -4 \right) & 1~\times \left( -2 \right)+2~\times 1 \\ 4~\times 6+6~\times \left( -4 \right) & 4\times \left( -2 \right)+6~\times 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-8 & -2+2 \\ 24-24 & -8+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$ 

 

$\left( Adj\text{ }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A}$

= $\left[ \begin{matrix} 6 & -2 \\ -4 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 1 & 2 \\ 4 & 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6~\times 1+\left( -2 \right)~\times 4 & 6~\times 2+\left( -2 \right)~\times 6 \\ -4~\times 1+1~\times 4 & -4~\times 2+1~\times 6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-8 & 12-12 \\ -4+4 & -8+6 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

 

$\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

= $~\left( -2 \right)\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} -2 & 0 \\ 0 & -2 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(3)$

 

From $(1)$, $(2)$ and $(3)$ it is clear that:

$\text{A}\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{= }\!\!~\!\!\text{ }\left( \text{Adj }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ A=}\left( \text{det }\!\!~\!\!\text{ A} \right)\text{ }\!\!~\!\!\text{ I}$

 

 

(ii) $\text{B}{{\text{B}}^{-1}}=I=~{{\text{B}}^{-1}}\text{B}$

The multiplicative inverse of matrix B is calculated as:

${{\text{B}}^{-1}}$ = $\frac{Adj~\text{B}}{\left| \text{B} \right|}$

 

B = $\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]$

$\text{Adj }\!\!~\!\!\text{ B}$ = $\left[ \begin{matrix} -2 & 1 \\ -2 & 3 \\ \end{matrix} \right]$

 

$\left| \text{B} \right|$ = $\left| \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right|$

$\left| \text{B} \right|$ = $3\times \left( -2 \right)-\left( -1 \right)\times 2$

$\left| \text{B} \right|$ = $-6+2$

$\left| \text{B} \right|$ = $-4\ne 0$

 

Since B is a non-singular matrix therefore ${{\text{B}}^{-1}}$ is possible

 

${{\text{B}}^{-1}}$ = $\frac{\left[ \begin{matrix} -2 & 1 \\ -2 & 3 \\ \end{matrix} \right]}{-4}$

= $\left[ \begin{matrix} \frac{-2}{-4} & \frac{1}{-4} \\ \frac{-2}{-4} & \frac{3}{-4} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & -\frac{3}{4} \\ \end{matrix} \right]$

 

Now

$\text{B}{{\text{B}}^{-1}}$

= $\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{2} & \frac{-1}{4} \\ \frac{1}{2} & \frac{-3}{4} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times \left( \frac{1}{2} \right)+\left( -1 \right)\times \left( \frac{1}{2} \right) & 3~\times \left( -\frac{1}{4} \right)+\left( -1 \right)\times \left( -\frac{3}{4} \right) \\ 2~\times \left( \frac{1}{2} \right)+\left( -2 \right)\times \left( \frac{1}{2} \right) & 2\times \left( -\frac{1}{4} \right)+\left( -2 \right)\times \left( -\frac{3}{4} \right) \\ \end{matrix} \right]$

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

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I \quad\quad\quad\quad\quad...~(1)$

 

${{\text{B}}^{-1}}\text{B}$

= $\left[ \begin{matrix} \frac{1}{2} & -\frac{1}{4} \\ \frac{1}{2} & -\frac{3}{4} \\ \end{matrix} \right]~\times \text{ }\!\!~\!\!\text{ }\left[ \begin{matrix} 3 & -1 \\ 2 & -2 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{1}{2}\times 3+\left( -\frac{1}{4} \right)\times 2 & \frac{1}{2}\times \left( -1 \right)+\left( -\frac{1}{4} \right)\times \left( -2 \right) \\ \frac{1}{2}\times 3+\left( -\frac{3}{4} \right)\times 2 & \frac{1}{2}\times \left( -1 \right)+\left( -\frac{3}{4} \right)\times \left( -2 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{3}{2}-\frac{1}{2} & -\frac{1}{2}+\frac{1}{2} \\ \frac{3}{2}-\frac{3}{2} & -\frac{1}{2}+\frac{3}{2} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ = $I \quad\quad\quad\quad\quad...~(2)$

 

From $(1)$ and $(2)$, it is clear that:

$\text{B}{{\text{B}}^{-1}}=I=~{{\text{B}}^{-1}}\text{B}$