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If A[1−214]=[6006]A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}, then matrices AA is

Solution

✅ Correct Option: 3

Let M=[1−214]M=\begin{bmatrix}1&-2\\1&4\end{bmatrix}, with ∣M∣=4+2=6|M|=4+2=6 and M−1=16[42−11]M^{-1}=\frac{1}{6}\begin{bmatrix}4&2\\-1&1\end{bmatrix}. Since AM=6IAM=6I, we get A=6M−1=[42−11]A=6M^{-1}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}.

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