Variable in a matrix

Question

I'm currently stuck on this problem. You have the following formules:

6αx1 + 4x2 = 5
 9x1 + 2αx2 = -2

For which values of α does this system have a unique solution?

I currently got this but dont know if it's correct.
So the matrix must be 1 0
                      0 1
for it to have an unique solution.
This is only possible when 9-6α = 0
and 4-2α = 0

so α would be 2 and 1,5 but this is not possible

Answer 1

Your system is

$\begin{bmatrix} 6\,\alpha & 4 \\ 9 & 2\,\alpha \end{bmatrix} \left\{\begin{array}{cc} x_1\\x_2\end{array}\right\} = \left\{\begin{array}{cc} 5\\-2\end{array}\right\}$

It will have a unique solution iff the determinant of the matrix is different than zero

$det\left(\begin{bmatrix} 6\,\alpha & 4 \\ 9 & 2\,\alpha \end{bmatrix}\right) = 12\,\alpha^2 - 36 \neq0 \Rightarrow \alpha^2\neq3\Rightarrow\boxed{\alpha\neq\pm\sqrt{3}}$

Answer 2

A set of $n$ linear equations in $n$ unknowns has a unique solution when the matrix formed by the coefficients is invertible. In this case, the matrix is $\begin{bmatrix} 6\alpha & 4 \\ 9 & 2\alpha\end{bmatrix}$ (assuming I've read your notation right).

There are a couple of ways to text the invertibility of a matrix, but the main one would be to test the determinant. Are you familiar with calculating the determinant, and how you can use it to check whether the matrix is invertible?