If A and B are square matrices of same order such that AB = BA, then $A^2 + B^2$ is equal to :

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

Given the condition $AB = BA$, the expression $A^2 + B^2$ can be simplified. \[A^2 + B^2 = (A + B)^2 - 2AB\] Due to the commutativity of $A$ and $B$ (i.e., $AB = BA$), the simplification proceeds as follows:\[A^2 + B^2 = (A + B)^2 - 2AB = 2(A + B)\] Therefore, the final simplified form is $2(A + B)$.

If A and B are square matrices of same order such that AB = BA, then $A^2 + B^2$ is equal to :

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The Correct Option is C

Solution and Explanation

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