If matrix \(A=\left[\begin{array}{lll}x-1 & 2 &2\\ 3& x-1 & 2\\ 3& 3& x-1\end{array}\right]\), then the number of real values of \(x\) satisfying the equation \(\frac{d}{dx}\left|A\right|=0\), is

\(2\)

\(A=\left[\begin{array}{lll}x-1 & 2 &2\\ 3& x-1 & 2\\ 3& 3& x-1\end{array}\right]\)

$\begin{bmatrix}x+3&2&2\\x+4&x-1&2\\x+5&3&x-1\end{bmatrix}$   [$C_1→C_1+C_2$]

$\begin{bmatrix}x+3&2&2\\1&x-3&0\\2&1&-3\end{bmatrix}$   [$R_2→R_2-R_1,\,R_3→R_3-R_1$]

$=(x-3)\begin{bmatrix}(x-3)&(x+3)&-2\end{bmatrix}+2(1-2(x-3))$

$|A|=x^3-3x^2-15x+47$

so $\frac{d|A|}{dx}=3x^2-6x-15=0$

$⇒x^2-2x-5=0$

$x=\frac{2±\sqrt{24}}{2}=1±\sqrt{6}$

Two real solutions