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If A=19[−8144471−84]A = \dfrac{1}{9}\begin{bmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{bmatrix}, then

(Where A−1A^{-1} and ATA^T represent inverse and transpose of matrix AA respectively)

Solution

✅ Correct Option: 3

Each row of the inner matrix has squared length 8181, and any two distinct rows have dot product 00.

So if MM is the inner matrix, MMT=81IMM^T = 81I, and since A=M9A = \dfrac{M}{9}, we get AAT=81I81=IAA^T = \dfrac{81I}{81} = I.

Hence AA is orthogonal and A−1=ATA^{-1} = A^T.

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