First slide

Applications of determinant

Question

$If A = [\begin{array}{ccc} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{array}], then {(A^{T})}^{- 1} =$

Moderate

A

$[\begin{matrix} - 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1 \end{matrix}]$
B

$[\begin{matrix} - 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1 \end{matrix}]$
C

$[\begin{matrix} 1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1 \end{matrix}]$
D

none

Solution

$\det A = 1 (- 1 - 8) + 2 (0 + 8) + 3 (0 - 2) = - 9 + 16 - 6 = 1$
$\begin{array}{l} A^{- 1} = \frac{A d j A}{\det A} = A d j A = [\begin{matrix} - 9 & 8 & - 5 \\ - 8 & 7 & - 4 \\ - 2 & 2 & - 1 \end{matrix}] \\ = [\begin{matrix} - 9 & - 8 & - 2 \\ 8 & 7 & 2 \\ - 5 & - 4 & - 1 \end{matrix}] {(A^{T})}^{- 1} = {(A^{- 1})}^{T} \end{array}$

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