## College Algebra (6th Edition)

a. $\left[\begin{array}{lll} 2 & 6 & 6\\ 2 & 7 & 6\\ 2 & 7 & 7 \end{array}\right]\left[\begin{array}{l} x\\ y\\ z \end{array}\right]=\left[\begin{array}{l} 8\\ 10\\ 9 \end{array}\right]$ b. Solution set: $\{( 1$ , $2$ , $-1 ) \}$
$a.$ $A=\left[\begin{array}{lll} 2 & 6 & 6\\ 2 & 7 & 6\\ 2 & 7 & 7 \end{array}\right],\qquad B=\left[\begin{array}{l} 8\\ 10\\ 9 \end{array}\right]$ $AX=B$ $\left[\begin{array}{lll} 2 & 6 & 6\\ 2 & 7 & 6\\ 2 & 7 & 7 \end{array}\right]\left[\begin{array}{l} x\\ y\\ z \end{array}\right]=\left[\begin{array}{l} 8\\ 10\\ 9 \end{array}\right]$ $b.\qquad X=A^{-1}B$ Find the inverse, $A^{-1}:$ $\left[\begin{array}{lllllll} 2 & 6 & 6 & | & 1 & 0 & 0\\ 2 & 7 & 6 & | & 0 & 1 & 0\\ 2 & 7 & 7 & | & 0 & 0 & 1 \end{array}\right]\left\{\begin{array}{l} \times\frac{1}{2}\\ -R_{1}\\ -R_{1} \end{array}\right.$ $\left[\begin{array}{lllllll} 1 & 3 & 3 & | & 1/2 & 0 & 0\\ 0 & 1 & 0 & | & -1 & 1 & 0\\ 0 & 1 & 1 & | & -1 & 0 & 1 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ -R_{2}. \end{array}\right.$ $\left[\begin{array}{lllllll} 1 & 3 & 3 & | & 1/2 & 0 & 0\\ 0 & 1 & 0 & | & -1 & 1 & 0\\ 0 & 0 & 1 & | & 0 & -1 & 1 \end{array}\right]\left\{\begin{array}{l} -3R_{2}\\ .\\ . \end{array}\right.$ $\left[\begin{array}{lllllll} 1 & 0 & 3 & | & 7/2 & -3 & 0\\ 0 & 1 & 0 & | & -1 & 1 & 0\\ 0 & 0 & 1 & | & 0 & -1 & 1 \end{array}\right]\left\{\begin{array}{l} -3R_{3}\\ .\\ . \end{array}\right.$ $\left[\begin{array}{lllllll} 1 & 0 & 0 & | & 7/2 & 0 & -3\\ 0 & 1 & 0 & | & -1 & 1 & 0\\ 0 & 0 & 1 & | & 0 & -1 & 1 \end{array}\right]$ $A^{-1}=\left[\begin{array}{lll} 7/2 & 0 & -3\\ -1 & 1 & 0\\ 0 & -1 & 1 \end{array}\right]$ $X=A^{-1}B$ $X=\left[\begin{array}{lll} 7/2 & 0 & -3\\ -1 & 1 & 0\\ 0 & -1 & 1 \end{array}\right]\left[\begin{array}{l} 8\\ 10\\ 9 \end{array}\right]=\left[\begin{array}{l} 28+0-27\\ -8+10+0\\ 0-10+9 \end{array}\right]$ $X=\left[\begin{array}{l} 1\\ 2\\ -1 \end{array}\right]$ Solution set: $\{( 1$ , $2$ , $-1 ) \}$