Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 51–56, solve for x, 

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

Step-by-Step Solution

Answer

The value of x is $\frac{13}{11}$.

Step 1. Given Information

We are given, 

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

We have to find the value of x.

Step 2. Solving the determinant

Solving the determinant, we get 

$$\begin{gathered} x\left|\begin{array}{ll}3& 2\\ 2& 5\end{array}\right|-1\left|\begin{array}{cc}4& 2\\ -1& 5\end{array}\right|+1\left|\begin{array}{cc}4& 3\\ -1& 2\end{array}\right|=2 \\ x(3·5-2·2)-1(4·5-2(-1\left)\right)+1(4·2-3(-1\left)\right)=2 \\ x(15-4)-1(20+2)+1(8+3)=2 \\ 11 x-11=2 \\ 11x=13 \end{gathered}$$

Dividing by $11$, we get

$x=\frac{13}{11}$