Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 933: 40

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 $