Question

Find the value of the determinant.

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

Short answer

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

Step-by-step solution

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 a11a12a13a21a22a23a31a32a33=a11a22a23a32a33-a12a21a23a31a33+a13a21a22a31a32.

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

Step 3. Simplify

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