Q. 11 - Precalculus Enhanced with Graphing Utilities

Question

Find the value of the determinant.

$\left|\begin{array}{ccc}3& 4& 2\\ 1& -1& 5\\ 1& 2& -2\end{array}\right|$

Step-by-Step Solution

Answer

$\left|\begin{array}{ccc}3& 4& 2\\ 1& -1& 5\\ 1& 2& -2\end{array}\right|=10$

Step 1. Given information

We have been given:

$\left|\begin{array}{ccc}3& 4& 2\\ 1& -1& 5\\ 1& 2& -2\end{array}\right|$

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 .

$\left|\begin{array}{ccc}3& 4& 2\\ 1& -1& 5\\ 1& 2& -2\end{array}\right|=3\left|\begin{array}{cc}-1& 5\\ 2& -2\end{array}\right|-4\left|\begin{array}{cc}1& 5\\ 1& -2\end{array}\right|+2\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|$

Step 3. Simplify

$$\begin{gathered} 3\left|\begin{array}{cc}-1& 5\\ 2& -2\end{array}\right|-4\left|\begin{array}{cc}1& 5\\ 1& -2\end{array}\right|+2\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right| \\ =3\cdot \left[\right(-1)\cdot (-2)-2\cdot 5]-4\cdot [1\cdot (-2)-1\cdot 5]+2\cdot [1\cdot 2-1\cdot (-1\left)\right] \\ =3\cdot (2-10)-4\cdot (-2-5)+2\cdot (2+1) \\ =3\cdot (-8)-4\cdot (-7)+2\cdot 3 \\ =-24+28+6 \\ =10 \end{gathered}$$