College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 50

Determine the determinant of the matrix $\displaystyle A = \left[
\begin{array}{cccc}
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
1 & 1 & 1 & 2 \\
1 & 2 & 1 & 2
\end{array}
\right]$ and if possible, the inverse of the matrix.

Using rows and columns transformation,

If we add $-1$ times column 1 to column 3, we get

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

So,

$\displaystyle \det (A) = 1 \left|
\begin{array}{ccc}
1 & 0 & 1 \\
1 & 0 & 2 \\
2 & 0 & 2
\end{array}
\right|$

Now, adding $-1$ times column 1 to column 3 in this determinant gives us


$
\begin{equation}
\begin{aligned}

\det (A) =& 1 \left| \begin{array}{ccc}
1 & 0 & 0 \\
1 & 0 & 1 \\
2 & 0 & 0
\end{array} \right| \qquad \text{Expand this by column 3}
\\
\\
\det (A) =& 1 (1) \left| \begin{array}{cc}
1 & 0 \\
2 & 0
\end{array} \right|
\\
\\
\det (A) =& 1(1 \cdot 0 - 0 \cdot 2)
\\
\\
\det (A) =& 0

\end{aligned}
\end{equation}
$


Since the determinant of $A$ is zero, $A$ can't have an inverse, by the invertibility criterion.