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Derivative from First Principle

If A=[ 2 3;...

Question

If $A = [\begin{matrix} 2 & 3 1 & - 4 \end{matrix}]$ and $B = [\begin{matrix} 1 & - 2 - 1 & 3 \end{matrix}]$ , then verify that ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ .

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Solution

$\begin{matrix} A = [\begin{matrix} 2 & 3 1 & - 4 \end{matrix}] a d j A = [\begin{matrix} - 4 & - 3 - 1 & 2 \end{matrix}] | A | = - 8 - 3 = - 11 s o, A^{- 1} = \frac{- 1}{11} [\begin{matrix} - 4 & - 3 - 1 & 2 \end{matrix}] = \frac{1}{11} [\begin{matrix} 4 & 3 1 & - 2 \end{matrix}] B = [\begin{matrix} 1 & - 2 - 1 & 3 \end{matrix}] a d j B = [\begin{matrix} 3 & 2 1 & 1 \end{matrix}] | B | = 3 - 2 = 1 s o, B^{1 - 1} = [\begin{matrix} 2 & 3 1 & - 4 \end{matrix}] [\begin{matrix} 1 & - 2 - 1 & 3 \end{matrix}] . = [\begin{matrix} 2 - 3 & - 4 + 9 1 + 4 & - 2 - 12 \end{matrix}] = [\begin{matrix} - 1 & 5 5 & - 14 \end{matrix}] a d j A B = [\begin{matrix} - 14 & - 5 - 5 & - 1 \end{matrix}] | A B | = 14 - 25 = - 11 \end{matrix} \begin{matrix} {(A B)}^{- 1} = \frac{- 1}{11} [\begin{matrix} - 14 & - 5 - 5 & - 1 \end{matrix}] = \frac{1}{11} [\begin{matrix} 14 & 5 5 & 1 \end{matrix}] - e q (1) B^{- 1} A^{- 1} = \frac{1}{11} [\begin{matrix} 3 & 2 1 & 1 \end{matrix}] [\begin{matrix} 4 & 3 1 & - 2 \end{matrix}] = \frac{1}{11} [\begin{matrix} 12 + 2 & 9 - 4 4 + 1 & 3 - 2 \end{matrix}] = \frac{1}{11} [\begin{matrix} 14 & 5 5 & 1 \end{matrix}] - e q 2 \end{matrix}$
Therefore, eq 1 and 2
${(A B)}^{- 1} = B^{- 1} A^{- 1}$
Proved

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