Chapter 11, Problem 19RE

To determine

To find: The inverse of the matrix if it is not singular.

Answer to Problem 19RE

The matrix is singular.

Explanation of Solution

Given information:

The matrix is

  $\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]$

Calculation:

The matrix A is of 2 by 2 order.

Obtain the matrix $\left[A|I_2\right]$ .

  $\left[A|I_2\right]=\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Perform the row operation $R_1=\frac{1}{4}{r}_1$ and then $R_2=r_2-r_1$ to obtain the row echelon matrix.

  $\begin{array}{c}\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -2\\ -1& 2\end{array}|\begin{array}{cc}\frac{1}{4}& 0\\ 0& 1\end{array}\right]\\ =\left[\begin{array}{cc}1& -2\\ 0& 0\end{array}|\begin{array}{cc}\frac{1}{4}& 0\\ \frac{1}{4}& 1\end{array}\right]\end{array}$

The above reduced matrix cannot contain the identity matrix $I_2$ on the left of the vertical bar. So the matrix A is singular and its inverse does not exist.

Therefore, the matrix is singular.