College Algebra, Chapter 7, 7.4, Section 7.4, Problem 22

Determine the determinant of the matrix $\displaystyle \left[ \begin{array}{ccc}
-2 & \displaystyle \frac{-3}{2} & \displaystyle \frac{1}{2} \\
2 & 4 & 0 \\
\displaystyle \frac{1}{2} & 2 & 1
\end{array} \right]$. State whether the matrix has an inverse, but don't calculate the inverse.

Let

$ A = \displaystyle \left[ \begin{array}{ccc}
-2 & \displaystyle \frac{-3}{2} & \displaystyle \frac{1}{2} \\
2 & 4 & 0 \\
\displaystyle \frac{1}{2} & 2 & 1
\end{array} \right]$

$\displaystyle \det (A) = \left[ \begin{array}{ccc}
-2 & \displaystyle \frac{-3}{2} & \displaystyle \frac{1}{2} \\
2 & 4 & 0 \\
\displaystyle \frac{1}{2} & 2 & 1
\end{array} \right] = 2 \left| \begin{array}{cc}
4 & 0 \\
2 & 1
\end{array} \right| - \left( \frac{-3}{2} \right) \left| \begin{array}{cc}
2 & 0 \\
\displaystyle \frac{1}{2} & 1
\end{array} \right| + \frac{1}{2} \left| \begin{array}{cc}
2 & 4 \\
\displaystyle \frac{1}{2} & 2
\end{array} \right| = -2 (4 \cdot 1 - 0 \cdot 2) - \left( \frac{-3}{2} \right) \left(2 \cdot 1 - 0 \cdot \frac{1}{2} \right) + \frac{1}{2} \left(2 \cdot 2 - 4 \cdot \frac{1}{2} \right)$

$\det (A) = -8 + 3 + 1$

$\det (A) = -4$

The given matrix has an inverse.