Question

If \(A = \left[ \begin{array}{ccc}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array} \right]\), \(B = \left[ \begin{array}{c}1 \\ 0 \\ 1\end{array} \right]\) such that \(XA = B^T\) and \(A^{-1}Y = B\), then \(XY =\)

• A. \([-1]\)
• B. \([1]\)
• C. \([-2]\)
• D. \([2]\)

Hint:

Express X and Y using the inverse of A and watch A and its inverse cancel.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
From \(XA = B^T\), multiply on the right by \(A^{-1}\): \(X = B^T A^{-1}\). From \(A^{-1}Y = B\), multiply on the left by \(A\): \(Y = AB\).

Step 2: Multiply:
\[ XY = B^T A^{-1} A B = B^T (A^{-1}A) B = B^T B \]
The matrix \(A\) cancels out, so we do not need to compute \(A^{-1}\) at all.

Step 3: Compute:
\(B = [1, 0, 1]^T\), so \(B^T B = 1\cdot1 + 0\cdot0 + 1\cdot1 = 2\), the \(1 \times 1\) matrix \([2]\).

Step 4: Why the other options are wrong.
\([-1]\), \([1]\) and \([-2]\) would need a different sum of squares. \(B^TB\) is a sum of squares, so it can never be negative.

Final Answer:
\(XY = [2]\), option (D). \[ \boxed{[2]} \]