College Algebra, Chapter 7, 7.3, Section 7.3, Problem 24

Determine the inverse of the matrix $\left[ \begin{array}{cccc}
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1
\end{array} \right]$ if it exists.

First, let's add the identity matrix to the right of our matrix

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

By using Gauss-Jordan Elimination

$\displaystyle R_3 - R_1 \to R_3 $

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

$\displaystyle R_4 - R_1 \to R_4 $

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

$\displaystyle R_3 - R_2 \to R_3 $

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

$\displaystyle R_4 - R_2 \to R_4 $

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

$\displaystyle - R_3 $

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

$\displaystyle - R_4 $

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

$\displaystyle R_3 - R_4 \to R_3 $

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

$\displaystyle R_1 - R_4 \to R_1 $

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

$\displaystyle R_2 - R_3 \to R_2 $

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


The given matrix doesn't have an inverse because the left half of the matrix can not convert to identity matrix.