Precalculus (6th Edition) Blitzer

a) The linear system can be written as $\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 3 \\ 3 \\ \end{matrix} \right]$ b) the value of the inverse matrix is $X=\left\{ \left( -2,-3,2 \right) \right\}$
(a) Consider the given system of equations: \begin{align} & x+2y+5z=2 \\ & 2x+3y+8z=3 \\ & -x+y+2z=3 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 3 \\ 3 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]$ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 2 \\ & 3 \\ & 3 \\ \end{align} \right] (b) Consider the given system of equations: \begin{align} & x+2y+5z=2 \\ & 2x+3y+8z=3 \\ & -x+y+2z=3 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 2 \\ 3 \\ 3 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]$ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 2 \\ & 3 \\ & 3 \\ \end{align} \right] Now, consider the coefficient matrix $A=\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 8 \\ -1 & 1 & 2 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} 2 & -1 & -1 \\ 12 & -7 & -2 \\ -5 & 3 & 1 \\ \end{matrix} \right]$ Now, to find the values of the provided system: We will use the formula $X={{A}^{-1}}B$ Where, ${{A}^{-1}}=\left[ \begin{matrix} 2 & -1 & -1 \\ 12 & -7 & -2 \\ -5 & 3 & 1 \\ \end{matrix} \right]$ B=\left[ \begin{align} & 2 \\ & 3 \\ & 3 \\ \end{align} \right] Now, substitute the values in $X={{A}^{-1}}B$