Divide matrix on both sides of an equation

Question

Here is the question I have to solve:

Find value of $x$ if,
$$ \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\\ \end{bmatrix} \begin{bmatrix} 1 \\ 2\\ x\\ \end{bmatrix} =0 $$

My attempt:
I divided on both side of the equation by $\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\\ \end{bmatrix}$ to get: $$\frac{ \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\\ \end{bmatrix} \begin{bmatrix} 1 \\ 2\\ x\\ \end{bmatrix}}{\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\\ \end{bmatrix}} = \frac{0}{\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\\ \end{bmatrix}}$$ Which gives: $$ \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2\\ x\\ \end{bmatrix} = 0$$ Solving this: \begin{align} (1)(1)+(2)(x)+(1)(x) &= 0 \\ 1 + 2x + x &= 0 \\ 3x &= -1 \\ x &= \frac{-1}{3} \\ \end{align} ... which is the incorrect answer. Is dividing both sides of the equation by the same value invalid in this case or how else is this approach wrong?

Answer 1

The closest thing to dividing by a matrix is multiplying from the left by its inverse. It is true that...

$$ M^{-1}\begin{bmatrix} 1 & x & 1 \end{bmatrix} M \begin{bmatrix} 1 \\ 2\\ x\\ \end{bmatrix} = M^{-1}0=0$$ But it is not valid to change the order of multiplication and conclude that...

$$\begin{bmatrix} 1 & x & 1 \end{bmatrix} M^{-1} M \begin{bmatrix} 1 \\ 2\\ x\\ \end{bmatrix}=0$$

as you seem to have done

Answer 2

In linear algebra terminology, the expression you've having is known as quadratic forms. let $\boldsymbol{v}\in\mathbb{R}^n, A\in\mathbb{R}^{n\times n}$, the quadratic form is expressed as $$ \boldsymbol{v}^TA\boldsymbol{v}. $$

In your case, the expression would look like this $\boldsymbol{v}^TA\boldsymbol{u}$ where $\boldsymbol{u}\in\mathbb{R}^n$. As a result, your case only requires expanding the expression and equating it to zero as has been addressed in the comments by @Moo. Using Matlab, I get

$$ x^2 + 16x + 28 = 0 \implies x_{1,2} = -14,-2. $$

Matlab script

syms x real
v=[1;x;1];
k=[1;2;x];
A=[1 3 2;2 5 1;15 3 2];
expand(v'*A*k)