If \( A=\displaystyle\frac{1}{9}\begin{bmatrix}-8 & 1 & 4\\4 & 4 & 7\\1 & -8 & 4\end{bmatrix} \), then
(Where \( A^{-1} \) and \( A^{T} \) represent inverse and transpose of matrix A respectively)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( A^{-1}=A^{T} \)
Consider \( AA^T \). Its (i, j) entry is the dot product of row i with row j of A. Write M for the matrix without the factor \( \frac{1}{9} \), so \( AA^T=\frac{1}{81}MM^T \).
Each row of M with itself:
\( \displaystyle (-8)^2+1^2+4^2=64+1+16=81 \)
\( \displaystyle 4^2+4^2+7^2=16+16+49=81 \)
\( \displaystyle 1^2+(-8)^2+4^2=1+64+16=81 \)
Different rows:
\( \displaystyle R_1\cdot R_2=-32+4+28=0 \)
\( \displaystyle R_1\cdot R_3=-8-8+16=0 \)
\( \displaystyle R_2\cdot R_3=4-32+28=0 \)
So \( MM^T=81I \) and
\( \displaystyle AA^T=\frac{1}{81}(81I)=I \)
Hence \( A^{-1}=A^{T} \).
Tip: When every row of a matrix has length 1 and the rows are mutually perpendicular (an orthogonal matrix), its inverse is simply its transpose; check the dot products instead of computing the adjoint.