If \(A = \begin{bmatrix}1&2\\-1&-1\end{bmatrix}\), \(B = \begin{bmatrix}a&1\\b&-1\end{bmatrix}\) and \((A + B)^2 = A^2 + B^2\), then the values of a and b are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(a = 1,\ b = -\displaystyle\frac{1}{2}\)
Matrix multiplication is not commutative, so
\((A + B)^2 = A^2 + AB + BA + B^2\)
The given condition therefore means
\(AB + BA = O\)
Multiplying row by column:
\( \displaystyle AB = \begin{bmatrix}a + 2b&-1\\-a - b&0\end{bmatrix} \)
\( \displaystyle BA = \begin{bmatrix}a - 1&2a - 1\\b + 1&2b + 1\end{bmatrix} \)
\( \displaystyle AB + BA = \begin{bmatrix}2a + 2b - 1&2a - 2\\-a + 1&2b + 1\end{bmatrix} \)
Every element must be zero (equality of matrices):
\(2a - 2 = 0\), so \(a = 1\)
\( \displaystyle 2b + 1 = 0, \text{ so } b = -\frac{1}{2} \)
The first element agrees:
\( \displaystyle 2(1) + 2\left(-\frac{1}{2}\right) - 1 = 0 \)