Chapter 11.4, Problem 64AYU

To determine

To show: The matrix has no inverse.

Answer to Problem 64AYU

Solution:

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

Explanation of Solution

Given:

$A=\left[\begin{array}{cc}-3& 0\\ 4& 0\end{array}\right]$

Calculation:

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

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

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

$=[\begin{array}{cc}-3& 0\\ 4& 0\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}]$

$\underset{R_1=\left(- \frac{1}{3}\right)r_1}{\to }[\begin{array}{cc}1& 0\\ 4& 0\end{array}|\begin{array}{cc}-\frac{1}{3}& 0\\ 0& 1\end{array}]$

$\underset{R_2=r_2-4r_1}{\to }[\begin{array}{cc}1& 0\\ 0& 0\end{array}|\begin{array}{cc}-\frac{1}{3}& 0\\ \frac{4}{3}& 1\end{array}]$

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.