If \( A=\begin{bmatrix}x & 1\\y & -1\end{bmatrix} \),\( B=\begin{bmatrix}1 & -1\\2 & -1\end{bmatrix} \) and \( (A+B)^2=A^2+B^2 \) ,then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( x=1,y=4 \)
Since \( (A+B)^2=A^2+AB+BA+B^2 \), the given condition means \( AB+BA=O \). Matrix multiplication is not commutative, so AB and BA are worked out separately.
\( \displaystyle AB=\begin{bmatrix}x & 1\\y & -1\end{bmatrix}\begin{bmatrix}1 & -1\\2 & -1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}x+2 & -x-1\\y-2 & -y+1\end{bmatrix} \)
\( \displaystyle BA=\begin{bmatrix}1 & -1\\2 & -1\end{bmatrix}\begin{bmatrix}x & 1\\y & -1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}x-y & 2\\2x-y & 3\end{bmatrix} \)
\( \displaystyle AB+BA=\begin{bmatrix}2x-y+2 & -x+1\\2x-2 & -y+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix} \)
From the (1, 2) entry, x = 1, and from the (2, 2) entry, y = 4. These also satisfy the other two entries: 2 − 4 + 2 = 0 and 2 − 2 = 0.