If A = begin{bmatrix} 3 & 0 & 0 0 & -5 & 0 | vedclass

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(A) Step $1$: Calculate $2A = 2 \begin{bmatrix} 3 & 0 & 0 \ 0 & -5 & 0 \ 0 & 0 & 7 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \ 0 & -10 & 0 \ 0 & 0

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$\begin{bmatrix} 1/3 & 0 & 0 \ 0 & -1/4 & 0 \ 0 & 0 & 1/26 \end{bmatrix}$

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(A) Step $1$: Calculate $2A = 2 \begin{bmatrix} 3 & 0 & 0 \ 0 & -5 & 0 \ 0 & 0 & 7 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \ 0 & -10 & 0 \ 0 & 0 & 14 \end{bmatrix}$.Step $2$: Calculate $3B = 3 \begin{bmatrix} -1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} -3 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 12 \end{bmatrix}$.Step $3$: Calculate $2A + 3B = \begin{bmatrix} 6-3 & 0 & 0 \ 0 & -10+6 & 0 \ 0 & 0 & 14+12 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \ 0 & -4 & 0 \ 0 & 0 & 26 \end{bmatrix}$.Step $4$: The inverse of a diagonal matrix $D = 	ext{diag}(d_1, d_2, d_3)$ is $D^{-1} = 	ext{diag}(1/d_1, 1/d_2, 1/d_3)$.Step $5$: Thus, $(2A + 3B)^{-1} = \begin{bmatrix} 1/3 & 0 & 0 \ 0 & -1/4 & 0 \ 0 & 0 & 1/26 \end{bmatrix}$.

If $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix}$, then find $(2A + 3B)^{-1}$.

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