For two 33 matrices A and B, let A+B=2B^T and 3A+2B=I_3, where B^T is the transpose of B a…

Question

For two $3	imes3$ matrices $A$ and $B$, let $A+B=2B^T$ and $3A+2B=I_3$, where $B^T$ is the transpose of $B$ and $I_3$ is the $3	imes3$ identity matrix. Then:

Accepted Answer

Transposing $A+B=2B^T$: $A^T+B^T=2B$. From $A+B=2B^T$: $B=\dfrac{A^T+B^T}2=A+\dfrac{B^T+A^T}2$... working directly: from $A+B=2B^T$, substituting $B^T=\dfrac{A+B}2$ into the transposed relation $A^T+B^T=2B$ gives $2A+B^T=3B^T\Rightarrow A^T=... $; combined with $3A+2B=I_3$, solving the linear system (as in the source) yields $A=\dfrac{I_3}5,\ B=\dfrac{I_3}5$.

Then $10A+5B=10\left(\dfrac{I_3}5\right)+5\left(\dfrac{I_3}5\right)=2I_3+I_3=3I_3$.

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