## Precalculus (6th Edition) Blitzer

The solution is, $\underline{x=-3,y=0.5,z=1}$.
Consider the given system of equations: \begin{align} & x+2y-z=-3 \\ & 2x-4y+z=-7 \\ & -2x+2y-3z=4 \end{align} Therefore, in matrix form, the system of equations can be written as below: $AX=b$ Where, $A=\left[ \begin{array}{*{35}{r}} 1 & 2 & -1 \\ 2 & -4 & 1 \\ -2 & 2 & -3 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} -3 \\ -7 \\ 4 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Therefore, the solution of the system of equations is given by: $X={{A}^{-1}}b$ Consider the determinant of the matrix \begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 1 & 2 & -1 \\ 2 & -4 & 1 \\ -2 & 2 & -3 \\ \end{array} \right| \\ & =\left( 12-2 \right)-2\left( -6+2 \right)-1\left( 4-8 \right) \\ & =10+8+4 \\ & =22 \end{align} Consider the adjoint of the matrix $\text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} +\left| \begin{array}{*{35}{r}} -4 & 1 \\ 2 & -3 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 2 & 1 \\ -2 & -3 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 2 & -4 \\ -2 & 2 \\ \end{array} \right| \\ -\left| \begin{array}{*{35}{r}} 2 & -1 \\ 2 & -3 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & -1 \\ -2 & -3 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & 2 \\ -2 & 2 \\ \end{array} \right| \\ +\left| \begin{array}{*{35}{r}} 2 & -1 \\ -4 & 1 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & 2 \\ 2 & -4 \\ \end{array} \right| \\ \end{array} \right]}^{t}}$ It can be further solved as below: \begin{align} & \text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} \left( 12-2 \right) & -\left( -6+2 \right) & \left( 4-8 \right) \\ -\left( -6+2 \right) & \left( -3-2 \right) & -\left( 2+4 \right) \\ \left( 2-4 \right) & -\left( 1+2 \right) & \left( -4-4 \right) \\ \end{array} \right]}^{t}} \\ & ={{\left[ \begin{array}{*{35}{r}} 10 & 4 & -4 \\ 4 & -5 & -6 \\ -2 & -3 & -8 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right] \end{align} Therefore \begin{align} & {{A}^{-1}}=\frac{1}{\left| A \right|}\text{adj}\left( A \right) \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right] \end{align} Therefore the solution of the system of equations is \begin{align} & X=\frac{1}{22}\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} -3 \\ -7 \\ 4 \\ \end{array} \right] \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} -30-28-8 \\ -12+35-12 \\ 12+42-32 \\ \end{array} \right] \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} -66 \\ 11 \\ 22 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -3 \\ 0.5 \\ 1 \\ \end{array} \right] \end{align} The solutions of the system is $x=-3,y=0.5,z=1$.