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If $A$ and $B$ are two square matrices of the same order such that $A B=B$ and $B A=A$, then $\mathrm{A}^{2}+\mathrm{B}^{2}$ is always equal to
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$\mathrm{A}+\mathrm{B}$
$\begin{array}{lr}\text { Given, } \quad \mathrm{AB}=\mathrm{B}, \mathrm{BA}=\mathrm{A} & \ldots \text { (i) } \\ \text { Then, } \quad \mathrm{A}^{2}+\mathrm{B}^{2}=\mathrm{A} \cdot \mathrm{A}+\mathrm{B} \cdot \mathrm{B} & \\ =\mathrm{A}(\mathrm{BA})+\mathrm{B}(\mathrm{AB}) & \text { [from Eq. (i) }] \\ =(\mathrm{AB}) \mathrm{A}+(\mathrm{BA}) \mathrm{B} & \text { (by commutative law) } \\ =\mathrm{BA}+\mathrm{AB} & \text { [from Eq. (i)] } \\ =\mathrm{A}+\mathrm{B} & \text { [from Eq. (i) }\end{array}$
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