How do I solve for $x$, when given a $1 \times 3$, $3 \times 3$, and $3 \times 1$ matrix, all multiplied together set equal to zero with an $x$ variable in two of the matrices?
The question asks to determine values of $x$ such that:
$$ \begin{bmatrix} x && 2 && 1 \end{bmatrix} \begin{bmatrix} 2 && 0 && -2 \\ 0 && 0 && 0 \\ -2 && -4 && -1 \end{bmatrix} \begin{bmatrix} x \\ -1 \\ 4 \end{bmatrix} = 0 $$
Lets call the matrices $A$, $B$, and $C$ respectively as they are ordered. I started by multiplying matrix $A$ and $B$, and resulted in a $1 \times 3$ matrix $AB$:
$$ \begin{bmatrix} 2x - 2 && -4 && -2x + 1 \end{bmatrix} $$
I then multiplied our new matrix with matrix $C$. The resultant $1 \times 1$ matrix was:
$$ \begin{bmatrix} 2x^2 - 10x + 8 \end{bmatrix} $$
I then factored out a $2$ to get:
$$ 2 \begin{bmatrix} x^2 - 5x + 4 \end{bmatrix} $$
I then solved for values of $x$ that made it $0$, since the three matrices multiplied were initially set to equal $0$:
$$ 2 \begin{bmatrix} (x - 1)(x - 4) \end{bmatrix} $$
So it seems the problem is looking for the values $x = 1$ and $x = 4$, but upon submission it told me, the answers were wrong, so I feel I messed up somewhere or I’m missing other values. But I’ve gone through the problem a couple times and reached the same result. Can anyone point out my error or show me what I’m doing wrong?
It is the correct method, but note that the AB should be
$$[2x-2,-4,-2x-1]$$
and thus
$$ABC=(2x-2)x-1(-4)+4(-2x-1)=0 \implies 2x^2-10x=0$$