Find x that satisfies the equality (matrix determinant):

Question

This is the exercise.

Find the value of $x\in\mathbb{R}$ that satisfies

$$ \begin{vmatrix} x & -1\\ 3 & 1-x \end{vmatrix} = \begin{vmatrix} 1 & 0 & -3 \\ 2 & x & -6 \\ 1 & 3 & x-5 \end{vmatrix} $$

This is what I've done.

\begin{align*} \det (A) &= \det (B)\\ \Rightarrow\det(x-x^{2}+3) &= \det(x^{2}-2x)\\ \Rightarrow x-x^{2}+3&=x^{2}-2x\\ \Rightarrow -2x^{2}+3x+3 &=0 \end{align*}

I've also tried to find $x$ with the quadratic formula but the above equation of determinants isn't equal when I plug in the $x$ I found. What's wrong? What step do I need to follow?

Answer 1

There's nothing wrong – just that $x=\frac{3\pm\sqrt{33}}4$, so simplifying the determinants after substituting $x$ in might take some work. The determinants on both sides should evaluate to $\frac{9\mp\sqrt{33}}8$.