Q. 12 - Precalculus Enhanced with Graphing Utilities

Question

Find the value of the determinant.

$\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|$

Step-by-Step Solution

Answer

$\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|=-169$

Step 1. Given information

We have been given:

$\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|$

We have to find the value of this determinant. 

Step 2. Use formula a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = a 11 a 22 a 23 a 32 a 33 - a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 .

$\left|\begin{array}{ccc}1& 3& -2\\ 6& 1& -5\\ 8& 2& 3\end{array}\right|=1\cdot \left|\begin{array}{cc}1& -5\\ 2& 3\end{array}\right|-3\cdot \left|\begin{array}{cc}6& -5\\ 8& 3\end{array}\right|+(-2)\cdot \left|\begin{array}{cc}6& 1\\ 8& 2\end{array}\right|$

Step 3. Simplify

$$\begin{gathered} 1\cdot \left|\begin{array}{cc}1& -5\\ 2& 3\end{array}\right|-3\cdot \left|\begin{array}{cc}6& -5\\ 8& 3\end{array}\right|+(-2)\cdot \left|\begin{array}{cc}6& 1\\ 8& 2\end{array}\right| \\ =1\cdot [1\cdot 3-2\cdot (-5\left)\right]-3\cdot [6\cdot 3-8\cdot (-5\left)\right]-2\cdot [6\cdot 2-8\cdot 1] \\ =1\cdot (3+10)-3\cdot (18+40)-2\cdot (12-8) \\ =1\cdot 13-3\cdot 58-2\cdot 4 \\ =13-174-8 \\ =-169 \end{gathered}$$