I'm currently studying some further maths because I will need to take these units alongside a future PHD in chemical engineering. One of the topics I'm looking at is matrices and I've been finding these to be relatively okay but just a couple of questions have stomped me. I've sorted one now and completely understand how the answer is derived.
The other one asks me to find possible values that x can take given that:
$A=\begin{bmatrix} x^2&3\\1&3x\end{bmatrix}$
$B = \begin{bmatrix} 3&6\\2&x\end{bmatrix}$
Also,
$AB = BA$
So I can figure out that the two are equal to each other by taking the determinant option. However because I have not seen a question of this standing before, I do not know where I am supposed to go with regards to x.
What would be the most logical step for me. Do I need to go down the divide by det route?
Thanks for any guidance on this issue!
Best Answer
Well, you know that $AB=BA$, so multiply the matrices together to get four equations:
$$AB=\begin{bmatrix}x^2&3\\1&3x\end{bmatrix}\begin{bmatrix}3&6\\2&x\end{bmatrix}=\begin{bmatrix}3x^2+6&6x^2+3x\\3+6x&6+3x^2\end{bmatrix}$$
Now, do the same for $BA$:
$$BA=\begin{bmatrix}3&6\\2&x\end{bmatrix}\begin{bmatrix}x^2&3\\1&3x\end{bmatrix}=\begin{bmatrix}3x^2+6&9+18x\\2x^2+x&6+3x^2\end{bmatrix}$$
Since $AB=BA$, you can compare the two termwise:
$$\begin{bmatrix}3x^2+6&6x^2+3x\\3+6x&6+3x^2\end{bmatrix}=\begin{bmatrix}3x^2+6&9+18x\\2x^2+x&6+3x^2\end{bmatrix}$$
So, there are 4 equations to compare here to solve for x.
$$ \begin{align} 3x^2+6&=3x^2+6\\ 6x^2+3x&=9+18x\\ 3+6x&=2x^2+x\\ 6+3x^2&=6+3x^2 \end{align} $$
Obviously the top and bottom ones cancel out – they're true regardless of the value of $x$. So, solve the middle two:
$$ \begin{align} 6x^2-15x-9&=0\\ 2x^2-5x-3&=0 \end{align} $$
Note that $3(2x^2-5x-3)=6x^2-15x-9$, so you actually only have one polynomial to solve.