\( A=\begin{bmatrix}0&1&3\\1&2&x\\2&3&1\end{bmatrix} \) and \( 2B=\begin{bmatrix}1&-8&5\\-1&6&-3\\1&2y&1\end{bmatrix} \). If \( A=B^{-1} \) then \( x+y \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
\( A=B^{-1} \) means \( AB=I \), so \( A\,(2B)=2I \). The entries off the diagonal of the product must be 0.
Row 1 of A with column 2 of 2B:
\( \displaystyle 0(-8)+1(6)+3(2y)=0 \)
\( \displaystyle 6+6y=0 \)
\( \displaystyle y=-1 \)
Row 2 of A with column 1 of 2B:
\( \displaystyle 1(1)+2(-1)+x(1)=0 \)
\( \displaystyle x=1 \)
So \( x+y=0 \). The other entries agree; for example, row 2 with column 2 gives \( -8+12+2xy=2 \), which is the required diagonal entry.