College Algebra, Chapter 7, 7.4, Section 7.4, Problem 20

Determine the determinant of the matrix $\displaystyle \left[ \begin{array}{ccc}
1 & 2 & 5 \\
-2 & -3 & 2 \\
3 & 5 & 3
\end{array} \right]$. State whether the matrix has an inverse, but don't calculate the inverse.

Let

$ A = \displaystyle \left[ \begin{array}{ccc}
1 & 2 & 5 \\
-2 & -3 & 2 \\
3 & 5 & 3
\end{array} \right]$

$\displaystyle \det (A) = \left[ \begin{array}{ccc}
1 & 2 & 5 \\
-2 & -3 & 2 \\
3 & 5 & 3
\end{array} \right] = 1 \left| \begin{array}{cc}
-3 & 2 \\
5 & 3
\end{array} \right| -2 \left| \begin{array}{cc}
-2 & 2 \\
3 & 3
\end{array} \right| + 5 \left| \begin{array}{cc}
-2 & -3 \\
3 & 5
\end{array} \right| = 1 (-3 \cdot 3 - 2 \cdot 5) - 2 (-2 \cdot 3 - 2 \cdot 3) + 5 (-2 \cdot 5 - (-3) \cdot 3)$

$\det (A) = -19 + 24 - 5$

$\det (A) = 0$

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