Finding A solution to a matrix equation

Question

I have the following problem in a past exam that involves matrix equations:

Suppose that $X$is a $2 \times2$ matrix satisfying:

$X^{2} = 6X +I$

a) Find the values of $\alpha$ and $\beta$ such that $X^{3} = \alpha X + \beta I$

b) Hence find a solution of the matrix equation

$X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$

I need a hint for part (b) as I am unable to use the result for part (a). For part a the values for $\alpha $ and $\beta$ are 37 and 6. In particular, subbing in the values I have found in part (a), gives me the relation

$X^{3} = 37X + 6I$, however I fail to see how I can apply this to part b, as there is a matrix instead of a real number

I initially tried that, and got up to $37X = \begin{bmatrix} 36 &-1\ 2&35\end{bmatrix}X +\begin{bmatrix} 3 &8\ 2&4\end{bmatrix} $, however this is the part I got stuck with — user24376, May 10 '19 at 07:37
Are we just meant to find an expression for $X$ or actually find the value for it. — user24376, May 10 '19 at 07:40
It’s possible to get a specific matrix value for $X$. Remember that $37X=(37I)X$. — amd, May 10 '19 at 07:56
What I understand from this question is that -- you have to find the matrix $X$. If that is the case, then look here for a hint. — Vineet, May 10 '19 at 08:33
@VineetKumar I also went down that route but I don't think you can get a value for X since you have too many variables and not enough equations — user24376, May 10 '19 at 08:44
Why so, we have 4 equation and 4 variables matrix will be $a=1,b=3,c=2,d=5$ — Vineet, May 10 '19 at 09:04
@VineetKumar You should post your solution as an answer instead of linking to images in a comment. — amd, May 10 '19 at 18:53
@amd This question was marked as offtopic and wasn't taking answers. So, inline comment was the only way. However, now I have added an answer. :-) — Vineet, May 11 '19 at 06:44

score 1 · Accepted Answer · answered May 10 '19 at 18:49

1

Call the two matrices in the equation you are to solve $A$ and $B$, respectively. Substituting the result from the first part of the problem and rearranging, we have $37X-AX = B-6I$, which is where you got stuck. As I hinted in my comment, $37X=(37I)X$, so you can continue: $$(37I)X-AX=B-6I \\ (37I-A)X=B-6I \\ X=(37I-A)^{-1}(B-6I),$$ assuming that the inverse exists. I’ll leave it to you to compute the final value of $X$.

answered May 10 '19 at 18:49

amd

53,693

Thank you!! I didn't think of using the trick of $37X = 37IX$. – user24376 May 11 '19 at 14:05

score 1 · Answer 2 · answered May 11 '19 at 06:42

Suppose $X=\begin{bmatrix} a & b \\ c &d \end{bmatrix}$.

It is given that $X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$.

Therefore,$\begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} \times \begin{bmatrix} a & b \\ c &d \end{bmatrix}+ \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix} = \begin{bmatrix} 36a-c+9 & 36b-d+8 \\ 2a+35c+2 & 2b+35d+10 \end{bmatrix}$.

Since, $X^3=37X+6I$, we have $X^3= \begin{bmatrix} 37a+6 &37b\\ 37c&37d+6\end{bmatrix}$

Comparing both $X^3$'s. [4 variables and 4 distinct eqations]

We get, $37a+6-36a+c-9=0\Rightarrow \boxed{c+a=3}$, $37b-36b+d-8=0\Rightarrow \boxed{d+b=8}$, $37c-2a-35c-2=0\Rightarrow \boxed{c-a=1}$, and $37d+6-2b-35d-10=0\Rightarrow \boxed{d-b=2}$.

On solving these simultaneous equations we get $\bf{\boxed{a=1,b=3,c=2,d=5}}$.

Therefore, our matrix $X$ is $\begin{bmatrix} 1 & 3 \\ 2&5\end{bmatrix}$.

Finding A solution to a matrix equation

2 Answers2