1. Statement of the problem. We are given the matrices A and B and are asked to compute $2A-3B$. $$A = \begin{bmatrix}-1 & 2 & -5 \\ 3 & 0 & 5 \\ 4 & 7 & 6\end{bmatrix}, \quad B = \begin{bmatrix}1 & -1 & 5 \\ -5 & 3 & 3 \\ 0 & 2 & 4\end{bmatrix}$$ 2. Formula and rules. For a scalar $c$ and matrix $M$, the scalar multiple is computed entrywise as $cM=(c m_{ij})$. For matrices of equal size, subtraction is entrywise: $(A-B)_{ij}=a_{ij}-b_{ij}$. 3. Compute $2A$ by multiplying each entry of $A$ by 2. $$2A = \begin{bmatrix}-2 & 4 & -10 \\ 6 & 0 & 10 \\ 8 & 14 & 12\end{bmatrix}$$ 4. Compute $3B$ by multiplying each entry of $B$ by 3. $$3B = \begin{bmatrix}3 & -3 & 15 \\ -15 & 9 & 9 \\ 0 & 6 & 12\end{bmatrix}$$ 5. Compute $2A-3B$ entrywise and show intermediate arithmetic. Row 1: $-2-3=-5$, $4-(-3)=7$, $-10-15=-25$. Row 2: $6-(-15)=21$, $0-9=-9$, $10-9=1$. Row 3: $8-0=8$, $14-6=8$, $12-12=0$. Therefore $$2A-3B = \begin{bmatrix}-5 & 7 & -25 \\ 21 & -9 & 1 \\ 8 & 8 & 0\end{bmatrix}$$ 6. Final answer. The result of $2A-3B$ is the matrix shown above.