## Elementary Linear Algebra 7th Edition

(a) $x=1$, $y=1$, $z=1$. (b) $x=1$, $y=0$, $z=1$.
(a) The coefficient matrix $A$ is given by $$A=\left[ \begin {array}{ccc} 1&1&-2\\ 1&-2&1 \\ 1&-1&-1\end {array} \right] .$$ Using Gauss-Jordan elimination, one can calculate $A^{-1}$ as follows $$A^{-1}=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right] .$$ The solution is given by $$\left[ \begin {array}{cc} {x}\\ {y}\\{z} \end {array} \right]=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right]\left[ \begin {array}{cc} {0}\\ {0}\\{-1} \end {array} \right]=\left[ \begin {array}{c} 1\\ 1\\1\end {array} \right].$$ That is, $x=1$, $y=1$, $z=1$. (b) The coefficient matrix $A$ is given by $$A=\left[ \begin {array}{ccc} 1&1&-2\\ 1&-2&1 \\ 1&-1&-1\end {array} \right] .$$ Using Gauss-Jordan elimination, one can calculate $A^{-1}$ as follows $$A^{-1}=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right] .$$ The solution is given by $$\left[ \begin {array}{cc} {x}\\ {y}\\{z} \end {array} \right]=\left[ \begin {array}{ccc} 1&1&-1\\ \frac{ 2}{3}&\frac{1}{3}&-1 \\ \frac{ 1}{3}&\frac{2}{3}&-1 \end {array} \right]\left[ \begin {array}{cc} {-1}\\ {2}\\{0} \end {array} \right]=\left[ \begin {array}{c} 1\\ 0\\1\end {array} \right].$$ That is, $x=1$, $y=0$, $z=1$.