1. Tardigrade
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  4. If the product of the matrix B = [2&6&4 1&0&1 -1&1&-1] with a matrix A has the inverse C = [-1&0&1 1&1&3 2&0&2] then A-1 equals

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Q. If the product of the matrix $B = \begin{bmatrix}2&6&4\\ 1&0&1\\ -1&1&-1\end{bmatrix} $ with a matrix $A$ has the inverse $C = \begin{bmatrix}-1&0&1\\ 1&1&3\\ 2&0&2\end{bmatrix}$ then $A^{-1}$ equals

COMEDKCOMEDK 2014Matrices

Solution:

$ (BA)^{-1} = C$ (given)
or $A^-1 \: B^{-1} = C$
or $A^{-1} \begin{bmatrix}-3&-3&-5\\ 0&9&2 \\ 2&14& 6 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 &1 \\ 1 & 1 & 3 \\ 2& 0 & 2 \end{bmatrix}$
Multiply by B on both sides, we get
$A^{-1} (B^{-1} B) = \begin{bmatrix} - 1 & 0 &1 \\ 1 & 1 & 3 \\ 2& 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$
or $A^{-1} = \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2& 14 & 6 \end{bmatrix}_{3 \times 3}$