Q. 65 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 61– 66, show that each matrix has no inverse 

$\left[\begin{array}{rrr}-3& 1& -1\\ 1& -4& -7\\ 1& 2& 5\end{array}\right]$

Step-by-Step Solution

Answer

The given matrix has no inverse. 

Step 1. Given information

A matrix is given as,

$\left[\begin{array}{rrr}-3& 1& -1\\ 1& -4& -7\\ 1& 2& 5\end{array}\right]$

Step 2. Finding the inverse.

To find the inverse,

$$\begin{gathered} \left[A\mid I_3\right]=\left[\begin{array}{cccccc}-3& 1& -1& 1& 0& 0\\ 1& -4& -7& 0& 1& 0\\ 1& 2& 5& 0& 0& 1\end{array}\right] \\ {R}_1\leftrightarrow R_3 \\ \left[\begin{array}{cccccc}-3& 1& -1& 1& 0& 0\\ 1& -4& -7& 0& 1& 0\\ 1& 2& 5& 0& 0& 1\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 1& -4& -7& 0& 1& 0\\ -3& 1& -1& 1& 0& 0\end{array}\right] \\ {R}_2=r_2-r_1. \\ \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 1& -4& -7& 0& 1& 0\\ -3& 1& -1& 1& 0& 0\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 1-1& -4-2& -7-5& 0-0& 1-0& 0-1\\ -3& 1& -1& 1& 0& 0\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& -6& -12& 0& 1& -1\\ -3& 1& -1& 1& 0& 0\end{array}\right] \\ {R}_3=r_3+3r_1 \\ \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& -6& -12& 0& 1& -1\\ -3& 1& -1& 1& 0& 0\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& -6& -12& 0& 1& -1\\ -3+3& 1+6& -1+15& 1+0& 0+0& 0+3\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& -6& -12& 0& 1& -1\\ 0& 7& 14& 1& 0& 3\end{array}\right] \\ {R}_2=\frac{r_2}{-6}\text{ and }{R}_3=\frac{r_3}{7} \\ \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& -6& -12& 0& 1& -1\\ 0& 7& 14& 1& 0& 3\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ \frac{0}{-6}& \frac{-6}{-6}& \frac{-12}{-6}& \frac{0}{-6}& \frac{1}{-6}& \frac{-1}{-6}\\ \frac{0}{7}& \frac{7}{7}& \frac{14}{7}& \frac{1}{7}& \frac{0}{7}& \frac{3}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& 1& 2& 0& -\frac{1}{6}& \frac{1}{6}\\ 0& 1& 2& \frac{1}{7}& 0& \frac{3}{7}\end{array}\right] \\ {R}_3=r_3-r_2 \\ \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& 1& 2& 0& -\frac{1}{6}& \frac{1}{6}\\ 0& 1& 2& \frac{1}{7}& 0& \frac{3}{7}\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& 1& 2& 0& -\frac{1}{6}& \frac{1}{6}\\ 0-0& 1-1& 2-2& \frac{1}{7}-0& 0+\frac{1}{6}& \frac{3}{7}-\frac{1}{6}\end{array}\right]\to \left[\begin{array}{cccccc}1& 2& 5& 0& 0& 1\\ 0& 1& 2& 0& -\frac{1}{6}& \frac{1}{6}\\ 0& 0& 0& \frac{1}{7}0& \frac{1}{6}& \frac{11}{42}\end{array}\right] \end{gathered}$$

Since the left side does not represent an identity matrix, therefore, inverse does not exist.