If the product of the matrix $B = [\begin{matrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 4 \end{matrix}]$ with a matrix A has inverse $C = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}], then A^{- 1} =$

Moderate

Solution

$We have (B A)^{- 1} = C$
$\begin{array}{l} \Rightarrow A^{- 1} B^{- 1} = C \\ \Rightarrow A^{- 1} = C B \\ ∴ A^{- 1} = [\begin{array}{ccc} - 1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{array}] [\begin{array}{ccc} 2 & 6 & 4 \\ 1 & 0 & 1 \\ - 1 & 1 & - 4 \end{array}] \\ = [\begin{array}{ccc} - 3 & - 5 & - 8 \\ 0 & 9 & - 7 \\ 2 & 14 & 0 \end{array}] \end{array}$

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