For two 3 times 3 matrices A and B,let A + B | vedclass

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(B) Given equations are:$(1) \ A + B = 2B^T$$(2) \ 3A + 2B = I_3$Taking the transpose of equation $(1)$:$(A + B)^T = (2B^T)^T \Rightarrow A^T + B^T =

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$10A + 5B = 3I_3$

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(B) Given equations are:$(1) \ A + B = 2B^T$$(2) \ 3A + 2B = I_3$Taking the transpose of equation $(1)$:$(A + B)^T = (2B^T)^T \Rightarrow A^T + B^T = 2B$From $(1)$,$A = 2B^T - B$. Substitute this into $(2)$:$3(2B^T - B) + 2B = I_3 \Rightarrow 6B^T - 3B + 2B = I_3 \Rightarrow 6B^T - B = I_3 \Rightarrow B = 6B^T - I_3$Substitute $B$ into $(1)$:$A + (6B^T - I_3) = 2B^T \Rightarrow A = I_3 - 4B^T$Taking transpose of $A = I_3 - 4B^T$:$A^T = I_3 - 4B$From $A^T + B^T = 2B$,we have $B^T = 2B - A^T$. Substitute $A^T = I_3 - 4B$:$B^T = 2B - (I_3 - 4B) = 6B - I_3$Substitute $B^T$ back into $B = 6B^T - I_3$:$B = 6(6B - I_3) - I_3 = 36B - 6I_3 - I_3 = 36B - 7I_3$$35B = 7I_3 \Rightarrow B = \frac{1}{5}I_3$Now find $A$:$A = I_3 - 4B^T = I_3 - 4(6B - I_3) = I_3 - 24B + 4I_3 = 5I_3 - 24(\frac{1}{5}I_3) = \frac{25I_3 - 24I_3}{5} = \frac{1}{5}I_3$Check the options:$10A + 5B = 10(\frac{1}{5}I_3) + 5(\frac{1}{5}I_3) = 2I_3 + I_3 = 3I_3$Thus,the correct option is $B$.

For two $3 \times 3$ 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 $3 \times 3$ identity matrix. Then:

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