## Precalculus (6th Edition) Blitzer

a) The linear system can be written as $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ b) $X=\left\{ \left( 1,2,-1 \right) \right\}$
(a) Consider the given system of equations: \begin{align} & 2x+6y+6z=8 \\ & 2x+7y+6z=10 \\ & 2x+7y+7z=9 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$, X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right], B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right] (b) Consider the given system of equations: \begin{align} & 2x+6y+6z=8 \\ & 2x+7y+6z=10 \\ & 2x+7y+7z=9 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 8 \\ 10 \\ 9 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right] Now, consider the coefficient matrix $A=\left[ \begin{matrix} 2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix. ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & 0 & -3 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{matrix} \right]$ Now, to find the values of the provided system: Using formula $X={{A}^{-1}}B$ we get, Where, ${{A}^{-1}}=\left[ \begin{matrix} \frac{7}{2} & 0 & -3 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{matrix} \right]$ B=\left[ \begin{align} & 8 \\ & 10 \\ & 9 \\ \end{align} \right] Now, substitute the values in $X={{A}^{-1}}B$ to get,