1. **Problem:** Perform the matrix operations and multiplication: $$\left(3 \cdot \begin{pmatrix} -1 & 2 \\ -2 & 4 \\ 5 & 0 \end{pmatrix} - 2 \cdot \begin{pmatrix} -1 & 0 \\ 5 & -1 \\ 3 & -3 \end{pmatrix}\right) \cdot \begin{pmatrix} -3 & 1 & 2 \\ 4 & -1 & 0 \end{pmatrix}$$ 2. **Formula and rules:** - Scalar multiplication: multiply each element by the scalar. - Matrix subtraction: subtract corresponding elements. - Matrix multiplication: if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$ with elements $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$. 3. **Step 1: Scalar multiplications** $$3 \cdot \begin{pmatrix} -1 & 2 \\ -2 & 4 \\ 5 & 0 \end{pmatrix} = \begin{pmatrix} -3 & 6 \\ -6 & 12 \\ 15 & 0 \end{pmatrix}$$ $$2 \cdot \begin{pmatrix} -1 & 0 \\ 5 & -1 \\ 3 & -3 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 10 & -2 \\ 6 & -6 \end{pmatrix}$$ 4. **Step 2: Matrix subtraction** $$\begin{pmatrix} -3 & 6 \\ -6 & 12 \\ 15 & 0 \end{pmatrix} - \begin{pmatrix} -2 & 0 \\ 10 & -2 \\ 6 & -6 \end{pmatrix} = \begin{pmatrix} -3 - (-2) & 6 - 0 \\ -6 - 10 & 12 - (-2) \\ 15 - 6 & 0 - (-6) \end{pmatrix} = \begin{pmatrix} -1 & 6 \\ -16 & 14 \\ 9 & 6 \end{pmatrix}$$ 5. **Step 3: Multiply the result by the third matrix** $$\begin{pmatrix} -1 & 6 \\ -16 & 14 \\ 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} -3 & 1 & 2 \\ 4 & -1 & 0 \end{pmatrix}$$ Calculate each element: - Row 1: - $(-1)(-3) + 6(4) = 3 + 24 = 27$ - $(-1)(1) + 6(-1) = -1 - 6 = -7$ - $(-1)(2) + 6(0) = -2 + 0 = -2$ - Row 2: - $(-16)(-3) + 14(4) = 48 + 56 = 104$ - $(-16)(1) + 14(-1) = -16 - 14 = -30$ - $(-16)(2) + 14(0) = -32 + 0 = -32$ - Row 3: - $9(-3) + 6(4) = -27 + 24 = -3$ - $9(1) + 6(-1) = 9 - 6 = 3$ - $9(2) + 6(0) = 18 + 0 = 18$ 6. **Final answer:** $$\begin{pmatrix} 27 & -7 & -2 \\ 104 & -30 & -32 \\ -3 & 3 & 18 \end{pmatrix}$$ This completes the matrix operation.