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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?

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It is the correct method, but note that the AB should be

$$[2x-2,-4,-2x-1]$$

and thus

$$ABC=(2x-2)x-1(-4)+4(-2x-1)=0 \implies 2x^2-10x=0$$

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  • $\begingroup$ Thank you! I made the same mistake of having 2x+1 instead of 2x-1 all 3 times I did the problem. I found x=0 and x=5 to be the solutions and those were accepted as correct. $\endgroup$ Feb 15 '18 at 22:39
  • $\begingroup$ @DrewPesall Well done! You are welcome, Bye. $\endgroup$
    – user
    Feb 15 '18 at 22:40

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