# Chapter 10 - 10.2 - Operations with Matrices - 10.2 Exercises - Page 724: 28

X = $\begin{bmatrix} -6 & -1 & 17\\ -9 & 0 & 10\\ \end{bmatrix}$

#### Work Step by Step

X = 3A - 2B First multiply matrix A by the scalar multiple 3: $\begin{bmatrix} -2(3) & 1(3) & 3(3)\\ -1(3) & 0(3) & 4(3)\\ \end{bmatrix}$ = $\begin{bmatrix} -6 & 3 & 9\\ -3 & 0 & 12\\ \end{bmatrix}$ Next, multiply matrix B by the scalar multiple 2: $\begin{bmatrix} 0(2) & 2(2) & -4(2)\\ 3(2) & 0(2) & 1(2)\\ \end{bmatrix}$ = $\begin{bmatrix} 0 & 4 & -8\\ 6 & 0 & 2\\ \end{bmatrix}$ Then perform X = 3A - 2B: X = $\begin{bmatrix} -6-0 & 3-4 & 9+8\\ -3-6 & 0-0 & 12-2\\ \end{bmatrix}$ = $\begin{bmatrix} -6 & -1 & 17\\ -9 & 0 & 10\\ \end{bmatrix}$

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