1. **Problem statement:** Given matrices $$A=\begin{pmatrix}-3 & 0 & 5 \\ -2 & 1 & -1 \\ -1 & 4 & 0\end{pmatrix},\quad B=\begin{pmatrix}5 & -1 & 2 \\ 3 & 2 & 1 \\ -1 & 4 & 1\end{pmatrix}$$ Find the matrices $2A - 3B$, $A \cdot B$, and $A^{-1}$. 2. **Formula and rules:** - Scalar multiplication: multiply each element by the scalar. - Matrix addition/subtraction: add/subtract corresponding elements. - Matrix multiplication: element $(i,j)$ of $A \cdot B$ is the dot product of row $i$ of $A$ and column $j$ of $B$. - Inverse of a $3\times3$ matrix $A$ exists if $\det(A) \neq 0$ and is given by $A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$. 3. **Calculate $2A$ and $3B$:** $$2A = 2 \times \begin{pmatrix}-3 & 0 & 5 \\ -2 & 1 & -1 \\ -1 & 4 & 0\end{pmatrix} = \begin{pmatrix}-6 & 0 & 10 \\ -4 & 2 & -2 \\ -2 & 8 & 0\end{pmatrix}$$ $$3B = 3 \times \begin{pmatrix}5 & -1 & 2 \\ 3 & 2 & 1 \\ -1 & 4 & 1\end{pmatrix} = \begin{pmatrix}15 & -3 & 6 \\ 9 & 6 & 3 \\ -3 & 12 & 3\end{pmatrix}$$ 4. **Calculate $2A - 3B$ element-wise:** $$2A - 3B = \begin{pmatrix}-6 & 0 & 10 \\ -4 & 2 & -2 \\ -2 & 8 & 0\end{pmatrix} - \begin{pmatrix}15 & -3 & 6 \\ 9 & 6 & 3 \\ -3 & 12 & 3\end{pmatrix} = \begin{pmatrix}-6-15 & 0-(-3) & 10-6 \\ -4-9 & 2-6 & -2-3 \\ -2-(-3) & 8-12 & 0-3\end{pmatrix} = \begin{pmatrix}-21 & 3 & 4 \\ -13 & -4 & -5 \\ 1 & -4 & -3\end{pmatrix}$$ 5. **Calculate $A \cdot B$:** Row 1 of $A$ times columns of $B$: - $(1,1): (-3)(5) + (0)(3) + (5)(-1) = -15 + 0 - 5 = -20$ - $(1,2): (-3)(-1) + (0)(2) + (5)(4) = 3 + 0 + 20 = 23$ - $(1,3): (-3)(2) + (0)(1) + (5)(1) = -6 + 0 + 5 = -1$ Row 2 of $A$ times columns of $B$: - $(2,1): (-2)(5) + (1)(3) + (-1)(-1) = -10 + 3 + 1 = -6$ - $(2,2): (-2)(-1) + (1)(2) + (-1)(4) = 2 + 2 - 4 = 0$ - $(2,3): (-2)(2) + (1)(1) + (-1)(1) = -4 + 1 - 1 = -4$ Row 3 of $A$ times columns of $B$: - $(3,1): (-1)(5) + (4)(3) + (0)(-1) = -5 + 12 + 0 = 7$ - $(3,2): (-1)(-1) + (4)(2) + (0)(4) = 1 + 8 + 0 = 9$ - $(3,3): (-1)(2) + (4)(1) + (0)(1) = -2 + 4 + 0 = 2$ So, $$A \cdot B = \begin{pmatrix}-20 & 23 & -1 \\ -6 & 0 & -4 \\ 7 & 9 & 2\end{pmatrix}$$ 6. **Calculate $\det(A)$ to check invertibility:** $$\det(A) = -3 \times \begin{vmatrix}1 & -1 \\ 4 & 0\end{vmatrix} - 0 \times \begin{vmatrix}-2 & -1 \\ -1 & 0\end{vmatrix} + 5 \times \begin{vmatrix}-2 & 1 \\ -1 & 4\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}1 & -1 \\ 4 & 0\end{vmatrix} = (1)(0) - (-1)(4) = 0 + 4 = 4$$ $$\begin{vmatrix}-2 & 1 \\ -1 & 4\end{vmatrix} = (-2)(4) - (1)(-1) = -8 + 1 = -7$$ So, $$\det(A) = -3 \times 4 + 0 + 5 \times (-7) = -12 - 35 = -47 \neq 0$$ 7. **Calculate $A^{-1}$:** Since $\det(A) \neq 0$, $A$ is invertible. Calculate the matrix of cofactors $C$: - $C_{11} = + \begin{vmatrix}1 & -1 \\ 4 & 0\end{vmatrix} = 4$ - $C_{12} = - \begin{vmatrix}-2 & -1 \\ -1 & 0\end{vmatrix} = -((-2)(0) - (-1)(-1)) = - (0 - 1) = 1$ - $C_{13} = + \begin{vmatrix}-2 & 1 \\ -1 & 4\end{vmatrix} = -7$ - $C_{21} = - \begin{vmatrix}0 & 5 \\ 4 & 0\end{vmatrix} = - (0*0 - 5*4) = - (0 - 20) = 20$ - $C_{22} = + \begin{vmatrix}-3 & 5 \\ -1 & 0\end{vmatrix} = (-3)(0) - (5)(-1) = 0 + 5 = 5$ - $C_{23} = - \begin{vmatrix}-3 & 0 \\ -1 & 4\end{vmatrix} = -((-3)(4) - 0*(-1)) = -(-12 - 0) = 12$ - $C_{31} = + \begin{vmatrix}0 & 5 \\ 1 & -1\end{vmatrix} = 0*(-1) - 5*1 = 0 - 5 = -5$ - $C_{32} = - \begin{vmatrix}-3 & 5 \\ -2 & -1\end{vmatrix} = -((-3)(-1) - 5*(-2)) = - (3 + 10) = -13$ - $C_{33} = + \begin{vmatrix}-3 & 0 \\ -2 & 1\end{vmatrix} = (-3)(1) - 0*(-2) = -3 - 0 = -3$ Cofactor matrix: $$C = \begin{pmatrix}4 & 1 & -7 \\ 20 & 5 & 12 \\ -5 & -13 & -3\end{pmatrix}$$ Transpose $C$ to get adjugate matrix $\mathrm{adj}(A)$: $$\mathrm{adj}(A) = C^T = \begin{pmatrix}4 & 20 & -5 \\ 1 & 5 & -13 \\ -7 & 12 & -3\end{pmatrix}$$ Finally, $$A^{-1} = \frac{1}{-47} \begin{pmatrix}4 & 20 & -5 \\ 1 & 5 & -13 \\ -7 & 12 & -3\end{pmatrix} = \begin{pmatrix}-\frac{4}{47} & -\frac{20}{47} & \frac{5}{47} \\ -\frac{1}{47} & -\frac{5}{47} & \frac{13}{47} \\ \frac{7}{47} & -\frac{12}{47} & \frac{3}{47}\end{pmatrix}$$ **Final answers:** $$2A - 3B = \begin{pmatrix}-21 & 3 & 4 \\ -13 & -4 & -5 \\ 1 & -4 & -3\end{pmatrix}$$ $$A \cdot B = \begin{pmatrix}-20 & 23 & -1 \\ -6 & 0 & -4 \\ 7 & 9 & 2\end{pmatrix}$$ $$A^{-1} = \begin{pmatrix}-\frac{4}{47} & -\frac{20}{47} & \frac{5}{47} \\ -\frac{1}{47} & -\frac{5}{47} & \frac{13}{47} \\ \frac{7}{47} & -\frac{12}{47} & \frac{3}{47}\end{pmatrix}$$