Chapter 7, Problem 29RE

Solution

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

  $\begin{array}{c}x+2y+z=-1\\ x-3y+2z=1\\ 2x-3y+z=5\end{array}$

The solution is $\left(\frac{9}{4},-\frac{3}{4},-\frac{7}{4}\right)$ .

Given Information:

The system of equation is defined as,

  $\begin{array}{c}x+2y+z=-1\\ x-3y+2z=1\\ 2x-3y+z=5\end{array}$

Calculation:

Consider the given equation,

  $\begin{array}{c}x+2y+z=-1\\ x-3y+2z=1\\ 2x-3y+z=5\end{array}$

Rewrite the given system as matrix form.

  $\begin{array}{c}AX=B\\ \left[\begin{array}{ccc}1& 2& 1\\ 1& -3& 2\\ 2& -3& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 1\\ 5\end{array}\right]\end{array}$

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

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ X=A^{-1}B\\ ={\left[\begin{array}{ccc}1& 2& 1\\ 1& -3& 2\\ 2& -3& 1\end{array}\right]}^{-1}\cdot \left[\begin{array}{c}-1\\ 1\\ 5\end{array}\right]\end{array}$

Now, find the inverse of the matrix.

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

First find the determinant of the matrix.

  $\begin{array}{c}\left|\begin{array}{ccc}1& 2& 1\\ 1& -3& 2\\ 2& -3& 1\end{array}\right|=1\left[-3+6\right]-2\left[1-4\right]+1\left[-3+6\right]\\ =1\left[3\right]-2\left[-3\right]+1\left[3\right]\\ =12\end{array}$

And,

  $\begin{array}{c}\text{adj}A=\text{adj}\left[\begin{array}{ccc}1& 2& 1\\ 1& -3& 2\\ 2& -3& 1\end{array}\right]\\ =\left[\begin{array}{ccc}3& -5& 7\\ 3& -1& -1\\ 3& 7& -5\end{array}\right]\end{array}$

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

  $\begin{array}{c}X=\frac{1}{\left|A\right|}\text{adj}A\cdot B\\ =\frac{1}{12}\left[\begin{array}{ccc}3& -5& 7\\ 3& -1& -1\\ 3& 7& -5\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 1\\ 5\end{array}\right]\\ =\frac{1}{12}\left[\begin{array}{c}27\\ -9\\ -21\end{array}\right]\\ =\left[\begin{array}{c}\frac{9}{4}\\ -\frac{3}{4}\\ -\frac{7}{4}\end{array}\right]\end{array}$

Hence, the obtained solution is $\left(\frac{9}{4},-\frac{3}{4},-\frac{7}{4}\right)$ .