Let $B$ be a matrix such that..

Question

$B=\begin{bmatrix}1&2x&-13\\1&2x+2&2x-13\\1&2x+6&7x+6\end{bmatrix}$ , $x$ is a parameter.
there exists a $y,z\in\mathbb{R}$ such that for this system theres a solution:
$B^{43}\begin{bmatrix}-285\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
$x=?$

I can't somehow get around how to solve this question, I know that it has something to do with inverse matrices and I know that if the matrix is invertible then $r(B)=3$, which means theres a unique answer for the equation $Bx=0$.

Any help or hints are really appreciated.
Thanks in advance.

Answer 1

Since $y$ and $z$ can be anything, the condition is equivalent to the fact that $B^{43}$ has nontrivial kernel, which means it is noninvertible. Now $B^{43}$ is noninvertible if and only if $B$ is noninvertible so you need precisely those $x \in \Bbb{R}$ such that $\det B = 0$.

We have $$0 = \det B = 2x+38 \implies x = -19.$$