for what values of the parameter is the matrix diagonazible?

Question

for what values of the real parameter t the matrix diagonizable?
$$\begin{pmatrix} 2t & 1 & 0 \\ 3t & 0 & 0 \\ t & 0 & -t \end{pmatrix}$$

I can't find a solution any help?

Answer 1

Hint

$$\left|\begin{pmatrix} 2t & 1 & 0 \\ 3t & 0 & 0 \\ t & 0 & -t \end{pmatrix}-\lambda I\right|= \left|\begin{pmatrix} 2t-\lambda & 1 & 0 \\ 3t & -\lambda & 0 \\ t & 0 & -t-\lambda \end{pmatrix}\right| =-(t+\lambda)(\lambda(\lambda-2t)-3t)=0 $$ The roots are $\lambda =-t$, $\lambda=t\pm\sqrt{t^2+3 t}$

so you need $t>0$ and $t^2+3 t>0$, i.e. $t < -3$ or $t > 0$

Answer 2

The characteristic polynomial of your matrix is \begin{equation} p(x) = {x}^{3}-t{x}^{2}- \left( 2\,{t}^{2}+3\,t \right) x-3\,{t}^{2} \end{equation}

Which has roots: \begin{equation} \begin{array}[ccc] \\ x_1 = -t,& x_2 = t+\sqrt{t^2+3t},& x_3 = t-\sqrt{t^2+3t} \end{array} \end{equation}

The values for which $x_2,x_3$ exists are such that $t^2 + 3t \geq 0$

I'll let you fill in the details.