## Precalculus (6th Edition) Blitzer

The value of $x$ is, $=3$
Consider, $\left| \begin{matrix} 2 & x & 1 \\ -3 & 1 & 0 \\ 2 & 1 & 4 \\ \end{matrix} \right|=39$ Evaluate the above determinant as follows: $\left| \begin{matrix} 2 & x & 1 \\ -3 & 1 & 0 \\ 2 & 1 & 4 \\ \end{matrix} \right|=39$ Evaluate the above determinant by expanding along the first row as follows: $2\left| \begin{matrix} 1 & 0 \\ 1 & 4 \\ \end{matrix} \right|-x\left| \begin{matrix} -3 & 0 \\ 2 & 4 \\ \end{matrix} \right|+1\left| \begin{matrix} -3 & 1 \\ 2 & 1 \\ \end{matrix} \right|=39$ Next, calculate the above determinant as follows: \begin{align} & 2\left( 4-0 \right)-x\left( -12-0 \right)+\left( -3-2 \right)=39 \\ & 8+12x-5=39 \\ & 12x+3=39 \end{align} Add $-3$ to both sides to get, $12x=36$ Divide by $12$. Now, $x=3$. Therefore, $x=3$