I have a question where $\alpha$ and $\beta$ are elements of $\Bbb R$, and I have to show that a system of linear equations will have infinitely many solutions whenever $\beta$ is an element of $\bigl[ -\sqrt 2, \sqrt 2\bigr]$.
There are three variables that have values like $\sin\alpha,\cos\alpha, \beta x$, etc. Now I know how to solve regular multi-variable linear equations to check if a linear system has infinite solutions or not, but this one is really tricky.... I don't even know where to start.
PS. I am really bad at math, I don't even know how to read notations.
Here's the actual question:
Let $\alpha$, $\beta$ $\in$ R; then show that the following system of linear equations, will have infinite many solutions whenever $\beta$ $\in$ [- $\sqrt 2$, $\sqrt 2$ ];
$\ - x + (\sin \alpha )y - (\cos \alpha )z = 0\\ \beta x + (\sin \alpha )y + (cos\alpha )z = 0\\ x + (cos\alpha )y + (sin\alpha )z = 0 $
Taking user36196’s suggestion, compute the determinant of $$\begin{bmatrix}-1&\sin\alpha&-\cos\alpha \\ \beta & \sin\alpha & \cos\alpha \\ 1&\cos\alpha&\sin\alpha \end{bmatrix}.$$ For there to be an infinite number of solutions to the system, this determinant must vanish. After a bit of simplification, this condition results in the equation $$\beta = \cos2\alpha+\sin2\alpha.$$ Now, the right-hand side can be rewritten as $\sqrt2 \sin\left(2\alpha+\frac\pi4\right),$ so clearly $\beta\in\left[-\sqrt2,\sqrt2\right]$ for this to hold, but that condition by itself is not sufficent for there to be an infinite number of solutions to the system.