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

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.

• Your feeling is correct (compare the top and bottom rows) Commented Jul 13, 2019 at 13:55
• Even if you don't notice the first and the third row, you should be able to reduce this to $2\times2$ determinant: $\det A= 2\det\begin{pmatrix} x&x\\2x&2x\end{pmatrix}$. Commented Jul 14, 2019 at 6:37

You just have to calculate it determinant:

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

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

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

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

The first step equals second step by row operations.

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).