Q. 54 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 51–56, solve for x, 

$\left|\begin{array}{rrr}3& 2& 4\\ 1& x& 5\\ 0& 1& -2\end{array}\right|=0$

Step-by-Step Solution

Answer

The value of x is $-\frac{3}{2}$.

Step 1. Given Information

We are given, 

$\left|\begin{array}{rrr}3& 2& 4\\ 1& x& 5\\ 0& 1& -2\end{array}\right|=0$

We have to find the value of x.

Step 2. Solving the determinant

Solving the determinant, we get 

$$\begin{gathered} \left|\begin{array}{ccc}3& 2& 4\\ 1& x& 5\\ 0& 1& -2\end{array}\right|=0 \\ (-1{)}^{1+2}\times 2\times \left|\begin{array}{cc}1& 5\\ 0& -2\end{array}\right|+(-1{)}^{2+2}\times x\times \left|\begin{array}{cc}3& 4\\ 0& -2\end{array}\right|+(-1{)}^{2+3}\times 1\times \left|\begin{array}{lc}3& 4\\ 1& 5\end{array}\right|=0 \\ 2-6x-11=0 \\ 6x=-9 \end{gathered}$$

Dividing by $6$, we get

$$\begin{gathered} x=\frac{-9}{6} \\ x=\frac{-3}{2} \end{gathered}$$