## Precalculus (6th Edition) Blitzer

b) The value of the inverse matrix is $X=\left\{ \left( 2,1,-1,3 \right) \right\}$
a) Consider the given system of equations: \begin{align} & 2w+2y+z=6 \\ & 3w+z=9 \\ & -w+x-2y+z=4 \\ & 4w-x+y=6 \end{align} The linear system can be written as: $AX=B$ Where, $A=\left[ \begin{matrix} 2 & 0 & 1 & 1 \\ 3 & 0 & 0 & 1 \\ -1 & 1 & -2 & 1 \\ 4 & -1 & 1 & 0 \\ \end{matrix} \right]$ X=\left[ \begin{align} & w \\ & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 6 \\ & 9 \\ & 4 \\ & 6 \\ \end{align} \right] (b) Consider the given system of equations: \begin{align} & 2w+y+z=6 \\ & 3w+z=9 \\ & -w+x-2y+z=4 \\ & 4w-x+y=6 \end{align} The linear system can be written as: $AX=B$ Where, $A=\left[ \begin{matrix} 2 & 0 & 1 & 1 \\ 3 & 0 & 0 & 1 \\ -1 & 1 & -2 & 1 \\ 4 & -1 & 1 & 0 \\ \end{matrix} \right]$ X=\left[ \begin{align} & w \\ & x \\ & y \\ & z \\ \end{align} \right] B=\left[ \begin{align} & 6 \\ & 9 \\ & 4 \\ & 6 \\ \end{align} \right] Now, we will consider the coefficient matrix is $A=\left[ \begin{matrix} 2 & 0 & 1 & 1 \\ 3 & 0 & 0 & 1 \\ -1 & 1 & -2 & 1 \\ 4 & -1 & 1 & 0 \\ \end{matrix} \right]$ Use the inverse of the coefficient matrix to get ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} -1 & 2 & -1 & -1 \\ -4 & 9 & -5 & -6 \\ 0 & 1 & -1 & -1 \\ 3 & -5 & 3 & 3 \\ \end{matrix} \right]$ Now, to find the values of the provided system: we will use the formula $X={{A}^{-1}}B$ Where, ${{\left[ A \right]}^{-1}}=\left[ \begin{matrix} -1 & 2 & -1 & -1 \\ -4 & 9 & -5 & -6 \\ 0 & 1 & -1 & -1 \\ 3 & -5 & 3 & 3 \\ \end{matrix} \right]$ B=\left[ \begin{align} & 6 \\ & 9 \\ & 4 \\ & 6 \\ \end{align} \right] Now, substitute the values in $X={{A}^{-1}}B$