$A$ root of the equation $\left| \begin{array}{ccc} 3 - x & -6 & 3 \\ -6 & 3 - x & 3 \\ 3 & 3 & -6 - x \end{array} \right| = 0$ is

The values of $x$ in the following determinant equation,$\left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$ are

The characteristic equation of the matrix $A = \begin{bmatrix} 2 & 3 & 0 \\ 1 & 2 & 5 \\ 3 & -1 & 2 \end{bmatrix}$ is:

Let $A = \begin{bmatrix} x+2 & 3x \\ 3 & x+2 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 5 & x+2 \end{bmatrix}$. Then all solutions of the equation $\det(AB) = 0$ are:

If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is: