Question:

If $A = \left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right]$, $B= \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$, then verify that

 (i)  $(AB)^{t} = B^{t}A^{t}$ (ii) $(AB)^{-1} = B^{-1}A^{-1}$

Difficulty: Easy

Solution:

(i)  $(AB)^{t} = B^{t}A^{t}$

Left Hand Side: $(AB)^{t}$

$AB$ = $\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right] \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 2+2\times \left( -3 \right) & 3\times 4+2\times \left( -5 \right) \\ 1\times 2+\left( -1 \right)\times \left( -3 \right) & 1\times 4+\left( -1 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 12-10 \\ 2+3 & 4+5 \\ \end{matrix} \right]$

$AB$= $\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{t}}$ = ${{\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]}^{t}}$

${{\left( \text{AB} \right)}^{t}}$ = $\left[ \begin{matrix} 0 & 5 \\ 2 & 9 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Right Hand Side: $B^{t}A^{t}$

${{\text{A}}^{t}}$ = ${{\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right]}^{t}}$ = $\left[ \begin{matrix} 3 & 1 \\ 2 & -1 \\ \end{matrix} \right]$

${{\text{B}}^{t}}$ = ${{\left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]}^{t}}$ = $\left[ \begin{matrix} 2 & -3 \\ 4 & -5 \\ \end{matrix} \right]$

${{\text{B}}^{t}}{{\text{A}}^{t}}$ = $\left[ \begin{matrix} 2 & -3 \\ 4 & -5 \\ \end{matrix} \right]$ $\left[ \begin{matrix} 3 & 1 \\ 2 & -1 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 2\times 3+\left( -3 \right)\times 2 & 2\times 1+\left( -3 \right)\times \left( -1 \right) \\ 4\times 3+\left( -5 \right)\times 2 & 4\times 1+\left( -5 \right)\times \left( -1 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 2~+~3 \\ 12-10 & 4+5 \\ \end{matrix} \right]$

${{\text{B}}^{t}}{{\text{A}}^{t}}$ = $\left[ \begin{matrix} 0 & 5 \\ 2 & 9 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that $(AB)^{t} = B^{t}A^{t}$

(ii) $(AB)^{-1} = B^{-1}A^{-1}$

Left Hand Side: $(AB)^{-1}$

$AB$ = $\left[ \begin{matrix} 3 & 2 \\ 1 & -1 \\ \end{matrix} \right] \left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 3\times 2+2\times \left( -3 \right) & 3\times 4+2\times \left( -5 \right) \\ 1\times 2+\left( -1 \right)\times \left( -3 \right) & 1\times 4+\left( -1 \right)\times \left( -5 \right) \\ \end{matrix} \right]$

= $\left[ \begin{matrix} 6-6 & 12-10 \\ 2+3 & 4+5 \\ \end{matrix} \right]$

$AB$= $\left[ \begin{matrix} 0 & 2 \\ 5 & 9 \\ \end{matrix} \right]$

$\left| \text{AB} \right|=0\times 9-2~\times 5=~0-10=~-10\ne 0$

${{\left( \text{AB} \right)}^{-1}}$= $\frac{1}{\left| \text{AB} \right|}Adj~\text{AB}$

= $\frac{1}{-10}\left[ \begin{matrix} 9 & -2 \\ -5 & 0 \\ \end{matrix} \right]$

${{\left( \text{AB} \right)}^{-1}}$ = $\left[ \begin{matrix} \frac{-9}{10} & \frac{1}{5} \\ \frac{1}{2} & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(1)$

Right Hand Side: $B^{-1}A^{-1}$

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

$\left| \text{A} \right|=3\times \left( -1 \right)-1~\times 2=~-3-2=~-5\ne 0$

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

= $\frac{1}{-5}\left[ \begin{matrix} -1 & -2 \\ -1 & 3 \\ \end{matrix} \right]$

${{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{1}{5} & \frac{2}{5} \\ \frac{1}{5} & -\frac{3}{5} \\ \end{matrix} \right]$

$\text{B}=\left[ \begin{matrix} 2 & 4 \\ -3 & -5 \\ \end{matrix} \right]$

$\left| \text{B} \right|=2\times \left( -5 \right)-4~\times \left( -3 \right)=~-10+12=~2\ne 0$

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

= $\frac{1}{2}\left[ \begin{matrix} -5 & -4 \\ 3 & 2 \\ \end{matrix} \right]$

${{\text{B}}^{-1}}$ = $\left[ \begin{matrix} -\frac{5}{2} & -2 \\ \frac{3}{2} & 1 \\ \end{matrix} \right]$

Now solving ${{\text{B}}^{-1}}{{\text{A}}^{-1}}$

= $\left[ \begin{matrix} -\frac{5}{2} & -2 \\ \frac{3}{2} & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} \frac{1}{5} & \frac{2}{5} \\ \frac{1}{5} & -\frac{3}{5} \\ \end{matrix} \right]$

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

= $\left[ \begin{matrix} \frac{-5}{10}+\frac{-2}{5} & \frac{-10}{10}+\frac{6}{5} \\ \frac{3}{10}+\frac{1}{5} & \frac{6}{10}-\frac{3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-1}{2}-\frac{2}{5} & -1+\frac{6}{5} \\ \frac{3}{10}+\frac{1}{5} & \frac{3}{5}-\frac{3}{5} \\ \end{matrix} \right]$

= $\left[ \begin{matrix} \frac{-5-4}{10} & \frac{-5+6}{5} \\ \frac{3+2}{10} & \frac{3-3}{5} \\ \end{matrix} \right]$

${{\text{B}}^{-1}}{{\text{A}}^{-1}}$ = $\left[ \begin{matrix} \frac{-9}{10} & \frac{1}{5} \\ \frac{1}{2} & 0 \\ \end{matrix} \right] \quad\quad\quad\quad\quad...~(2)$

From $(1)$ and $(2)$, it is proved that $(AB)^{-1} = B^{-1}A^{-1}$