## Precalculus (6th Edition) Blitzer

a) The linear system can be written as $\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 11 \\ 14 \\ 25 \\ \end{matrix} \right]$ b) the value of the inverse matrix is $X=\left\{ \left( 2,-1,1 \right) \right\}$
(a) Consider the given system of equations: \begin{align} & x-6y+3z=11 \\ & 2x-7y+3z=14 \\ & 4x-12y+5z=25 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 11 \\ 14 \\ 25 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]$ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 11 \\ & 14 \\ & 25 \\ \end{align} \right] (b) Consider the given system of equations: \begin{align} & x-6y+3z=11 \\ & 2x-7y+3z=14 \\ & 4x-12y+5z=25 \end{align} The linear system can be written as: $AX=B$ $\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]=\left[ \begin{matrix} 11 \\ 14 \\ 25 \\ \end{matrix} \right]$ Where, $A=\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]$ X=\left[ \begin{align} & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 11 \\ & 14 \\ & 25 \\ \end{align} \right] Now, consider the coefficient matrix $A=\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]$ Now, to find the values of the provided system: we will use the formula $X={{A}^{-1}}B$ Where, ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} 1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5 \\ \end{matrix} \right]$ B=\left[ \begin{align} & 11 \\ & 14 \\ & 25 \\ \end{align} \right] Now, substitute the values in $X={{A}^{-1}}B$