Q. 56 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 51–56, solve for x, 

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

Step-by-Step Solution

Answer

The value of x is $0 \text{or} \frac{1}{3}$.

Step 1. Given Information

We are given, 

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

We have to find the value of x.

Step 2. Solving the determinant

Solving the determinant, we get 

$$\begin{gathered} \left|\begin{array}{lll}x& 1& 2\\ 1& x& 3\\ 0& 1& 2\end{array}\right|=-4x \\ (-1{)}^{1+1}\times x\times \left|\begin{array}{ll}x& 3\\ 1& 2\end{array}\right|+(-1{)}^{2+1}\times 1\times \left|\begin{array}{ll}1& 2\\ 1& 2\end{array}\right|+(-1{)}^{3+1}\times 0\times \left|\begin{array}{ll}1& 2\\ x& 3\end{array}\right|=-4x \\ x(3 x-3)-1(0)+0=-4x \\ 3x^2-3x+4x=0 \\ 3x^2+x=0 \\ x(3x-1)=0 \\ x=0 \end{gathered}$$

or

$$\begin{gathered} 3x-1=0 \\ x=\frac{1}{3} \end{gathered}$$