Chapter 7, Problem 32RE

Solution

To calculate: The solution of the system by using inverse matrix.

  $\begin{array}{c}x-2y+z-w=2\\ 2x+y-z-w=-1\\ x-y+2z-w=-1\\ x+3y-z+w=4\end{array}$

The solution is $\left(\frac{13}{3},-\frac{8}{3},-\frac{1}{3},\frac{22}{3}\right)$ .

Given Information:

The system of equation is defined as,

  $\begin{array}{c}x-2y+z-w=2\\ 2x+y-z-w=-1\\ x-y+2z-w=-1\\ x+3y-z+w=4\end{array}$

Calculation:

Consider the given equation,

  $\begin{array}{c}x-2y+z-w=2\\ 2x+y-z-w=-1\\ x-y+2z-w=-1\\ x+3y-z+w=4\end{array}$

Rewrite the given system as matrix form.

  $\begin{array}{c}AX=B\\ \left[\begin{array}{cccc}1& -2& 1& -1\\ 2& 1& -1& -1\\ 1& -1& 2& -1\\ 1& 3& -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ w\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ -1\\ 4\end{array}\right]\end{array}$

Now, solve the equation $AX=B$ by multiply $A^{-1}$ on both sides.

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ X=A^{-1}B\\ ={\left[\begin{array}{cccc}1& -2& 1& -1\\ 2& 1& -1& -1\\ 1& -1& 2& -1\\ 1& 3& -1& 1\end{array}\right]}^{-1}\cdot \left[\begin{array}{c}2\\ -1\\ -1\\ 4\end{array}\right]\end{array}$

Now, find the inverse of the matrix.

  $A^{-1}=\frac{1}{\left|A\right|}\text{adj}\left(A\right)$

First find the determinant of the matrix.

  $\begin{array}{c}\left|\begin{array}{cccc}1& -2& 1& -1\\ 2& 1& -1& -1\\ 1& -1& 2& -1\\ 1& 3& -1& 1\end{array}\right|=1\left|\begin{array}{ccc}1& -1& -1\\ -1& 2& -1\\ 3& -1& 1\end{array}\right|-2\left|\begin{array}{ccc}2& -1& -1\\ 1& 2& -1\\ 1& -1& 1\end{array}\right|+1\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& -1\\ 1& 3& 1\end{array}\right|-1\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& 2\\ 1& 3& -1\end{array}\right|\\ =8-14+1\left(1\right)+11\\ =9\end{array}$

Substitute all the values in the equation $X=A^{-1}B$ .

  $\begin{array}{c}X=\frac{1}{\left|A\right|}\text{adj}\left(A\right)\cdot B\\ =\frac{1}{9}\left[\begin{array}{cccc}8& -1& -2& 5\\ -7& 2& 4& -1\\ -2& -2& 5& 1\\ 11& -7& -5& 8\end{array}\right]\cdot \left[\begin{array}{c}2\\ -1\\ -1\\ 4\end{array}\right]\\ =\frac{1}{9}\left[\begin{array}{c}39\\ -24\\ -3\\ 66\end{array}\right]\\ =\left[\begin{array}{c}\frac{13}{3}\\ -\frac{8}{3}\\ -\frac{1}{3}\\ \frac{22}{3}\end{array}\right]\end{array}$

Hence, the obtained solution is $\left(\frac{13}{3},-\frac{8}{3},-\frac{1}{3},\frac{22}{3}\right)$ .