For what values of x is the matrix A invertible? $A = \begin{bmatrix}-3x & 2\\4 & 4\end{bmatrix}$
How would I go about solving for what values of $x$ is the matrix $A$ invertible? I know that if the determinant is 0 the matrix is not invertible. I've looked at many answers on here but all of them seem very complicated and on matrices $3\times 3$ which I haven't learned yet.
 A: First approach:
The determinant of $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is given by $ad-bc$. In this case the determinant will be $-12x-8$. If this happens to be zero then matrix is NOT invertible.
Second approach:
A matrix is invertible if and only if the rows(or columns) are independent. The second row is $[4 \,\, 4]=4[1 \,\, 1]$ so for the first row to be dependent, it should also be a multiple of $[1 \,\, 1]$. Said differently, the components in the second row must be equal. Thus $x=-2/3$.
A: You know that the columns of $A$ are linearly independent (i.e. $A$ is invertible) if and only if 
$$\alpha \begin{pmatrix}-3x \\ 4\end{pmatrix}+\beta\begin{pmatrix}2 \\ 4\end{pmatrix}=0 \qquad \implies \qquad \alpha=\beta =0$$
We can rewrite it as 
$$\begin{cases} -3x\alpha+2\beta = 0 \\ 4(\alpha+\beta)=0\end{cases}\qquad \implies \qquad \alpha=\beta =0$$
From the second equation $\alpha=-\beta$, plugging that in the first equation, we get
$$ 0=3x\beta+2\beta=(2+3x)\beta \implies \beta =0$$
We note that this relation is true whenever $2+3x\neq 0$. It follows that $A$ is invertible if $x\neq-2/3$. Noting that the columns of $A$ are identical when $x=-2/3$ shows the equivalence.
