If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$, $B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A+B)^2 = A^2 + B^2$, then the values of $a$ and $b$ are:
Options:
(A) $a = 4, \, b = 1$
(B) $a = 1, \, b = 4$
(C) $a = 0, \, b = 4$
(D) $a = 2, \, b = 4$
Official Answer Key:
B
Step-by-Step Solution & Explanation
(A+B)² = A²+AB+BA+B². The condition (A+B)²=A²+B² requires AB+BA = 0 (the zero matrix), i.e., matrices A and B must anticommute. Computing AB and BA in terms of a,b and setting AB+BA=0 element-wise gives a system of equations in a and b. Solving this system (verified symbolically) yields a = 1, b = 4. Substituting back confirms AB+BA=0 exactly for these values. Answer: (B) a=1, b=4.
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