Q65 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
If $A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ such that $XA = B^T$ and $A^{-1}Y = B$, then $XY = $
- A$[-1]$
- B$[1]$
- C$[-2]$
- Correct answerD$[2]$
Solution
Given $XA = B^T$
Post-multiplying by $A^{-1}$, we get $X = B^T A^{-1}$
Given $A^{-1}Y = B$
Pre-multiplying by $A$, we get $Y = AB$
Now, $XY = (B^T A^{-1})(AB)$
Using the associative property of matrix multiplication, $XY = B^T (A^{-1}A) B$
$XY = B^T I B = B^T B$
Substituting the given matrix $B$, we get:
$XY = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$
$XY = [1 \times 1 + 0 \times 0 + 1 \times 1] = [2]$
Answer: $[2]$
Post-multiplying by $A^{-1}$, we get $X = B^T A^{-1}$
Given $A^{-1}Y = B$
Pre-multiplying by $A$, we get $Y = AB$
Now, $XY = (B^T A^{-1})(AB)$
Using the associative property of matrix multiplication, $XY = B^T (A^{-1}A) B$
$XY = B^T I B = B^T B$
Substituting the given matrix $B$, we get:
$XY = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$
$XY = [1 \times 1 + 0 \times 0 + 1 \times 1] = [2]$
Answer: $[2]$
Hence, the correct answer is option: 'D'.