If \( A=\begin{bmatrix}1 & 2 & 2\\2 & 1 & -2\\\alpha & 2 & \beta\end{bmatrix} \) and \( AA^T=9I \), where \( I \) is the identity matrix of order 3, then \( (\beta-\alpha) \) equals:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
The (i, j) entry of \( AA^T \) is the dot product of row i and row j of A. \( AA^T=9I \) means that different rows have dot product 0 and each row has dot product 9 with itself.
Row 1 with row 3, and row 2 with row 3:
\( \displaystyle \alpha+4+2\beta=0 \)
\( \displaystyle 2\alpha+2-2\beta=0 \)
The second equation gives \( \alpha=\beta-1 \). Substituting in the first:
\( \displaystyle (\beta-1)+4+2\beta=0 \)
\( \displaystyle 3\beta=-3 \)
\( \displaystyle \beta=-1,\ \alpha=-2 \)
Check row 3 with itself: \( \alpha^2+4+\beta^2=4+4+1=9 \).
\( \displaystyle \beta-\alpha=-1-(-2)=1 \)
Shortcut: The second equation alone, \( 2\alpha+2-2\beta=0 \), gives \( \beta-\alpha=1 \) directly.