## Precalculus (6th Edition) Blitzer

a) The required matrix equation is: $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 12 \\ -5 \\ 10 \\ \end{matrix} \right]$ b) The solution set is: $\left\{ 4,-2,3 \right\}$
(a) Note that the coefficient matrix $A$ contains all coefficients of the variables $x,y\text{ and }z$. The matrix $X$ contains all the variable and the matrix $B$ contains all the constant terms of the provided linear system. Therefore, the matrix equation is given by: $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 12 \\ -5 \\ 10 \\ \end{matrix} \right]$ Hence, the required matrix equation is: $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 12 \\ -5 \\ 10 \\ \end{matrix} \right]$ (b) Consider the given system $AX=B$ The solution of this system is $X={{A}^{-1}}B$ Since it is provided that inverse of the matrix $A=\left[ \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 3 & 0 & -2 \\ \end{matrix} \right]$ is $\left[ \begin{matrix} -2 & 2 & 1 \\ 9 & -8 & -3 \\ -3 & 3 & 1 \\ \end{matrix} \right]$. Therefore, \begin{align} & X={{A}^{-1}}B \\ & =\left[ \begin{matrix} 2 & 2 & -1 \\ -1 & 0 & 1 \\ -1 & -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 12 \\ -5 \\ 10 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2\left( 12 \right)+2\left( -5 \right)-1\left( 10 \right) \\ -1\left( 12 \right)+0\left( -5 \right)+1\left( 10 \right) \\ -1\left( 12 \right)-1\left( -5 \right)+1\left( 10 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4 \\ -2 \\ 3 \\ \end{matrix} \right] \end{align} This implies that $x=4,y=-2\text{ and }z=3$.