Chapter 11.4, Problem 65AYU

To determine

To show: The matrix has no inverse.

Answer to Problem 65AYU

Solution:

$A$ is a singular matrix and has no inverse.

Explanation of Solution

Given:

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

Calculation:

Step 1: Construction of $\left[A|I_n\right]$.

$\left[A|I_3\right]=[\begin{array}{ccc}-3& 1& -1\\ 1& -4& -7\\ 1& 2& 5\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

Step 2: Sequentially transforming the matrix $\left[A|I_n\right]$ into reduced row echelon form.

$=[\begin{array}{ccc}-3& 1& -1\\ 1& -4& -7\\ 1& 2& 5\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}]$

By definition, if we interchange the rows of a matrix, then the value of the matrix remains unchanged (i.e.) interchanging Rows 1 and 3.

$[\begin{array}{ccc}1& 2& 5\\ 1& -4& -7\\ -3& 1& -1\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0 \end{array}]$

$\underset{\begin{array}{c}{R}_2=r_2-r_1\\ R_3=r_3+3r_1\end{array}}{\to }[\begin{array}{ccc}1& 2& 5\\ 0& -6& -12\\ 0& 7& 14\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& 1& -1\\ 1& 0& 3\end{array}]$

$\underset{\begin{array}{c}{R}_2=\left(- \frac{1}{6}\right)r_2\\ R_3=\left(\frac{1}{7}\right)r_3\end{array}}{\to }[\begin{array}{ccc}1& 2& 5\\ 0& 1& 2\\ 0& 1& 2\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{7}& 0& \frac{3}{7}\end{array}]$

$\underset{R_3=r_3-r_2}{\to }[\begin{array}{ccc}1& 2& 5\\ 0& 1& 2\\ 0& 0& 0\end{array}|\begin{array}{ccc}0& 0& 1\\ 0& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{7}& \frac{1}{6}& \frac{11}{42}\end{array}]$

The matrix $\left[A|I_n\right]$ is sufficiently reduced for it to be clear that the identity matrix cannot appear to the left of the vertical bar. So $A$ is a singular matrix and has no inverse.