Chapter 10 - Systems of Equations and Inequalities - Chapter Review - Review Exercises - Page 814: 14

$\begin{bmatrix} 6 & 0\\ 12 & 24 \\ -6 & 12 \end{bmatrix}$

Given $A=\begin{bmatrix} 1 & 0\\ 2 & 4 \\ -1 & 2 \end{bmatrix}$, $B=\begin{bmatrix} 4 & -3 & 0 \\ 1 & 1 & -2 \end{bmatrix}$, $C=\begin{bmatrix} 3 & -4\\ 1 & 5 \\ 5 & 2 \end{bmatrix}$, We have $6A=6\times\begin{bmatrix} 1 & 0\\ 2 & 4 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 6 & 0\\ 12 & 24 \\ -6 & 12 \end{bmatrix}$