First slide

Applications of determinant

Question

$If A = [\begin{array}{cc} 2 & 2 \\ - 3 & 2 \end{array}], B = [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}], then {(B^{- 1} A^{- 1})}^{- 1}$

Moderate

A

$[\begin{matrix} 2 & - 2 \\ 2 & 3 \end{matrix}]$
B

$[\begin{matrix} 3 & - 2 \\ 2 & 2 \end{matrix}]$
C

$\frac{1}{10} [\begin{matrix} 2 & 2 \\ - 2 & 3 \end{matrix}]$
D

$\frac{1}{10} [\begin{matrix} 3 & 2 \\ - 2 & 2 \end{matrix}]$

Solution

$\begin{array}{l} {(B^{- 1} A^{- 1})}^{- 1} \\ = {(A^{- 1})}^{- 1} {(B^{- 1})}^{- 1} \\ = A B = [\begin{array}{cc} 2 & 2 \\ - 3 & 2 \end{array}] [\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array}] \\ = [\begin{array}{cc} 0 + 2 & - 2 + 2 \\ 0 + 2 & 3 + 0 \end{array}] \\ = [\begin{array}{cc} 2 & - 2 \\ 2 & 3 \end{array}] \end{array}$

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