Find values of $x$ so that the matrix is invertible

Question

Find values of $x$ so that the matrix is invertible $$A=\begin{pmatrix} x & 0 & x \\ x & 2 & 1 \\ 2x & 0 & 2x \\ \end{pmatrix}$$

I know that a matrix is invertible if determinant is not $0$, but I don't know how to find the $x$ values. I feel is a tricky question and this matrix will not be invertible no matter which value $x$ takes, but I don't know how to prove that either.

Answer 1

You just have to calculate it determinant:

$$ \det(A) = 4x^2 -4x^2 $$

Since it is always $0$ it is never invertibile.

Answer 2

$\det(A)=\begin{vmatrix} x & 0 & x \\ x & 2 & 1 \\ 2x & 0 & 2x \\ \end{vmatrix}$ $=\begin{vmatrix} x & 0 & x \\ x & 2 & 1 \\ 0 & 0 & 0 \\ \end{vmatrix}=0$

The first step equals second step by row operations.

Answer 3

If $C_1$, $C_2$, and $C_3$ are the three columns of $A$, then $C_1-\frac x2C_2=C_3-\frac12C_2$. Therefore, the columns are not linearly independent and so the matrix is not invertible (whatever $x$ is).

Answer 4

Note that $\forall x, R_3=2R_1\implies Rank(A)<3\implies \det(A)=0$.