If \(A = \displaystyle\frac{1}{3}\begin{bmatrix}1&2&2\\2&1&-2\\x&2&y\end{bmatrix}\) satisfies \(AA' = I\), where I is the identity matrix of order 3, then the value of \(x + y\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-3\)
Let M be the matrix inside the brackets, so \(A = \frac{1}{3}M\) and
\( \displaystyle AA' = \frac{1}{9}MM' \)
So \(AA' = I\) means \(MM' = 9I\). The element in row i, column j of \(MM'\) is row i of M times row j of M.
Row 3 with row 1 must give 0:
\(x + 4 + 2y = 0\)
Row 3 with row 2 must give 0:
\(2x + 2 - 2y = 0\)
\(x = y - 1\)
Substituting in the first equation:
\(y - 1 + 4 + 2y = 0\)
\(y = -1,\quad x = -2\)
Check: row 3 with itself gives \(4 + 4 + 1 = 9\), and rows 1 and 2 give \(1 + 4 + 4 = 9\) and \(2 + 2 - 4 = 0\), so \(MM' = 9I\) holds.
Hence
\(x + y = -2 - 1 = -3\)
Why the other options are incorrect
Option (1) \(-1\) — Incorrect. This is \(x - y = -2 + 1\), not \(x + y\).
Option (3) 3 — Incorrect. This is \(-(x + y)\); the only values are \(x = -2,\ y = -1\), whose sum is negative.
Option (4) 1 — Incorrect. This is \(y - x = -1 + 2\), not \(x + y\).