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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.

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

4 Answers 4

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You just have to calculate it determinant:

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

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

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

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

The first step equals second step by row operations.

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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).

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