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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
Starship - On Strike
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Eric mansen
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2 Answers2

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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?

ConMan
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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

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

Daniel Cunha
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