Chapter 10.5, Problem 24E

To determine

The inverse matrix of $\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 1& 1& 0\\ 1& 2& 1& 1\end{array}\right]$ is to be estimated.

Answer to Problem 24E

There is no inverse matrix for $\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 1& 1& 0\\ 1& 2& 1& 1\end{array}\right]$ .

Explanation of Solution

Given:

  $\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 1& 1& 0\\ 1& 2& 1& 1\end{array}\right]$

Calculations:

The given matrix is defined as matrix $A$

  $A=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 1& 1& 0\\ 1& 2& 1& 1\end{array}\right]$

Know that:

  $A\times A^{-1}=I$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

  $\begin{array}{l}A\times A^{-1}=I\\ \left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 1& 1& 1& 0\\ 1& 2& 1& 1\end{array}\right]\times A^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]\\ R_3\to R_3-R_1,R_4\to R_4-R_1\\ \left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 1& 0& 0\\ 0& 1& 0& 1\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -1& 0& 1& 0\\ -1& 0& 0& 1\end{array}\right]\\ R_3\to R_3-R_2,R_4\to R_4-R_2\\ \left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 1& 0& -1\\ 0& 0& 0& 0\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ -1& -1& 1& 0\\ -1& -1& 0& 1\end{array}\right]\end{array}$

Since row $4$ consists solely of zeros, then the determinant of the matrix equals $0$ . This means that the matrix is singular, i.e. not invertible.

Conclusion:

Hence, matrix is singular.