First slide

Applications of determinant

Question

If $A = [\begin{array}{ll} 3 2 \\ 0 1 \end{array}]$ , then $A^{- 3}$ is

Moderate

A

$\frac{1}{27} [\begin{array}{cc} 1 & - 26 \\ 0 & - 27 \end{array}]$
B

$\frac{1}{27} [\begin{array}{cc} - 1 & - 26 \\ 0 & - 27 \end{array}]$
C

$\frac{1}{27} [\begin{array}{cc} 1 & - 26 \\ 0 & 27 \end{array}]$
D

$\frac{1}{27} [\begin{array}{cc} - 1 & 26 \\ 0 & - 27 \end{array}]$

Solution

We have $A^{- 1} = \frac{1}{3} [\begin{array}{cc} 1 & - 2 \\ 0 & 3 \end{array}]$
$\begin{aligned} \Rightarrow A^{- 3} = \frac{1}{27} [\begin{array}{cc} 1 & - 2 \\ 0 & 3 \end{array}] [\begin{array}{cc} 1 & - 2 \\ 0 & 3 \end{array}] [\begin{array}{cc} 1 & - 2 \\ 0 & 3 \end{array}] \\ = \frac{1}{27} [\begin{array}{cc} 1 & - 8 \\ 0 & 9 \end{array}] [\begin{array}{cc} 1 & - 2 \\ 0 & 3 \end{array}] = \frac{1}{27} [\begin{array}{cc} 1 & - 26 \\ 0 & 27 \end{array}] \end{aligned}$

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