Question

Find the value of the determinant.

$\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|$

Short answer

$\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|=-119$

Step-by-step solution

Step 1. Given information 

We have been given:

$\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|$

We have to find the value of this determinant.

Step 2. Use formula a11a12a13a21a22a23a31a32a33=a11a22a23a32a33-a12a21a23a31a33+a13a21a22a31a32.

$\left|\begin{array}{ccc}3& -9& 4\\ 1& 4& 0\\ 8& -3& 1\end{array}\right|=3\cdot \left|\begin{array}{cc}4& 0\\ -3& 1\end{array}\right|-(-9)\cdot \left|\begin{array}{cc}1& 0\\ 8& 1\end{array}\right|+4\cdot \left|\begin{array}{cc}1& 4\\ 8& -3\end{array}\right|$

Step 3. Simplify

$3\cdot \left|\begin{array}{cc}4& 0\\ -3& 1\end{array}\right|-(-9)\cdot \left|\begin{array}{cc}1& 0\\ 8& 1\end{array}\right|+4\cdot \left|\begin{array}{cc}1& 4\\ 8& -3\end{array}\right|$

$$\begin{gathered} =3\cdot [4\cdot 1-(-3)\cdot 0]+9\cdot [1\cdot 1-8\cdot 0]+4\cdot [1\cdot (-3)-8\cdot 4] \\ =3\cdot (4-0)+9\cdot (1-0)+4\cdot (-3-32) \\ =12+9+4\cdot (-35) \\ ==21-140 \\ =-119 \end{gathered}$$