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 $XY =$

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

Step 1: Solve for each unknown:
$X = B^TA^{-1}$, a row matrix, and $Y = AB$, a column matrix.

Step 2: Check the order:
$XY$ is $(1 \times 3)(3 \times 3)(3 \times 3)(3 \times 1)$, giving a $1 \times 1$ matrix.

Step 3: Cancel and evaluate:
$A^{-1}A = I$, so $XY = B^TB = 1 + 0 + 1 = 2$.

Final Answer:
The product is [2], option (D). \[ \boxed{[2]} \]

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 \(XY =\)

Show Hint

The Correct Option is D

Solution and Explanation

Top Questions on Matrices

Questions Asked in MHT CET exam