The correct option is A [ 1 amp; 1; 1 amp; 0 ][ 2 amp; 1; 3 amp; 2 ]A[ 3 amp; 2; 5 amp; 3 ]=[ 1 amp; 0; 0 amp; 1 ] ...

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Inverse of a Matrix

beginbmatrix ...

Question

If $[\begin{matrix} 2 & 1 3 & 2 \end{matrix}] A [\begin{matrix} - 3 & 2 5 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ , then A =

[\begin{matrix} 1 & 1 1 & 0 \end{matrix}]

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[\begin{matrix} 1 & 1 0 & 1 \end{matrix}]

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[\begin{matrix} 1 & 0 1 & 1 \end{matrix}]

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- [\begin{matrix} 1 & 1 1 & 0 \end{matrix}]

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Solution

The correct option is A $[\begin{matrix} 1 & 1 1 & 0 \end{matrix}]$
$[\begin{matrix} 2 & 1 3 & 2 \end{matrix}] A [\begin{matrix} - 3 & 2 5 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] A = [\begin{matrix} 2 & - 1 - 3 & 2 \end{matrix}] [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] [\begin{matrix} - 3 & - 2 - 5 & - 3 \end{matrix}] (- 1) = [\begin{matrix} 2 & - 1 - 3 & 2 \end{matrix}] [\begin{matrix} 3 & 2 5 & 3 \end{matrix}] = [\begin{matrix} 1 & 1 1 & 0 \end{matrix}]$

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If [2132]A[−325−3]=[1001], then A =

The correct option is A [1110][2132]A[−325−3]=[1001]A=[2−1−32][1001][−3−2−5−3](−1)=[2−1−32][3253]=[1110]

If $[\begin{matrix} 2 & 1 3 & 2 \end{matrix}] A [\begin{matrix} - 3 & 2 5 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ , then A =