If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24 \), find the values of x.
Question 1 Report
If \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24 \), find the values of x.
Answer Details
Expand the determinant along the first row.
\[\Delta=(x-3)\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}-(-4)\begin{vmatrix}5&2\\2&6-x\end{vmatrix}+3\begin{vmatrix}5&2\\2&-4\end{vmatrix}\]
The \(2\times2\) minors:
\(\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}=2(6-x)+8=20-2x\).
\(\begin{vmatrix}5&2\\2&6-x\end{vmatrix}=5(6-x)-4=26-5x\).
\(\begin{vmatrix}5&2\\2&-4\end{vmatrix}=-20-4=-24\).
So
\[\Delta=(x-3)(20-2x)+4(26-5x)+3(-24)\]
\[=(-2x^{2}+26x-60)+(104-20x)-72=-2x^{2}+6x-28\]
Set \(\Delta=-24\):
\[-2x^{2}+6x-28=-24\Rightarrow-2x^{2}+6x-4=0\Rightarrow x^{2}-3x+2=0\]
\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2\]
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