Chapter 7, Problem 31RE

Solution

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

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

The solution does not exist.

Given Information:

The system of equation is defined as,

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

Calculation:

Consider the given equation,

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

Rewrite the given system as matrix form.

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

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

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

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.

  $\begin{array}{c}\left|\begin{array}{cccc}2& 1& 1& -1\\ 2& -1& 1& -1\\ -1& 1& -1& 1\\ 1& -2& 1& -1\end{array}\right|=2\left|\begin{array}{ccc}-1& 1& -1\\ 1& -1& 1\\ -2& 1& -1\end{array}\right|-1\left|\begin{array}{ccc}2& 1& -1\\ -1& -1& 1\\ 1& 1& -1\end{array}\right|+1\left|\begin{array}{ccc}2& -1& -1\\ -1& 1& 1\\ 1& -2& -1\end{array}\right|+1\left|\begin{array}{ccc}2& -1& 1\\ -1& 1& -1\\ 1& -2& 1\end{array}\right|\\ =2\left(0\right)-1\left(0\right)+1\left(1\right)+1\left(-1\right)\\ =0\end{array}$

As the determinate of the matrix is zero. So this indicate that the inverse of the matrix does not exist. This implies the system of equation have no solution.

Hence, the solution does not exist.