# Chapter 8 - Review Exercises - Page 951: 43

a) The required matrix equation is: $\left[ \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 3 & 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 7 \\ -2 \\ 0 \\ \end{matrix} \right]$ b) The solution set is: . $\left\{ -18,79,-27 \right\}$

#### Work Step by Step

(a) Note that the coefficient matrix $A$ contains all coefficients of the variables $x,y\text{ and }z$. The matrix $X$ contains all the variables and the matrix $B$ contains all the constant terms of the given linear system. Therefore, the matrix equation is given by: $\left[ \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 3 & 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 7 \\ -2 \\ 0 \\ \end{matrix} \right]$ Hence, the required matrix equation is given by: $\left[ \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 3 & 0 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 7 \\ -2 \\ 0 \\ \end{matrix} \right]$ (b) Consider the system $AX=B$ The solution of this system is $X={{A}^{-1}}B$ Since it is given that the 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 \\ 9 & -8 & -3 \\ -3 & 3 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 7 \\ -2 \\ 0 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -2\left( 7 \right)+2\left( -2 \right)+1\left( 0 \right) \\ 9\left( 7 \right)+\left( -8 \right)\left( -2 \right)+\left( -3 \right)\left( 0 \right) \\ \left( -3 \right)\left( 7 \right)+3\left( -2 \right)+1\left( 0 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -14-4 \\ 63+16 \\ -21-6 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -18 \\ 79 \\ -27 \\ \end{matrix} \right] \end{align} This implies that $x=-18,y=79\text{ and }z=-27$ Hence, the solution set is: $\left\{ -18,79,-27 \right\}$

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