Q.Find the value(s) of $x$ for which $\begin{vmatrix}x&3\\2&x-1\end{vmatrix}=0$.

Setting the determinant expression equal to zero and expanding gives a quadratic equation in $x$ to solve.

Expanding: $x(x-1)-3(2)=0 \Rightarrow x^2-x-6=0 \Rightarrow (x-3)(x+2)=0$, so $x=3$ or $x=-2$.

Expanding the determinant

Setting this equal to zero:

Solving by factorization

Check (independent recomputation): substituting each root back into the ORIGINAL determinant (not the expanded quadratic) — for $x=3$: $\begin{vmatrix}3&3\\2&2\end{vmatrix}=3(2)-3(2)=0$ ✓; for $x=-2$: $\begin{vmatrix}-2&3\\2&-3\end{vmatrix}=(-2)(-3)-3(2)=6-6=0$ ✓ — both roots genuinely make the determinant zero.

Stopping after finding the quadratic $x^2-x-6=0$ without solving it, or making a factorization error such as $(x-2)(x+3)$ instead of the correct $(x-3)(x+2)$ — always verify a factorization by expanding it back out.