2
$\begingroup$

Determine all diagonal matrices $X$ of order $3$ that verify the equation $X^2 - X - 21 = 0$.

I came up to this problem, but I don't know how to solve it. Nay thoughts please.

$\endgroup$
  • 5
    $\begingroup$ Algebraic operations on diagonal matrices are the same as applying those operations directly on the elements of the diagonal. So, that $X^2-X-21=0$ means that for each entry $d$ of the diagonal one must have $d^2-d-21=0$. So, the condition is equivalent to each entry of the diagonal is one of the two roots of the polynomial $x^2-x-21=0$. $\endgroup$ – MoonLightSyzygy Jan 7 at 19:26
  • 1
    $\begingroup$ Can you please give me an example on how to do this? Thank you $\endgroup$ – JOJO Jan 7 at 19:40
  • $\begingroup$ Why are you subtracting a scalar from a matrix? $\endgroup$ – Rodrigo de Azevedo Jan 8 at 7:26
1
$\begingroup$

This is a matrix equation, do not mix matrices and numbers.
I assume the equation to solve is $$X^2-X-2I={\bf{0}},$$ where $I$ states for the matrix $I=\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}$ and the RHS is the matrix $\begin{pmatrix} 0&0&0\\0&0&0\\0&0&0\end{pmatrix}.$

As said in a comment, we need to solve the quadratic equation $$r^2-r-2=0,$$ or equivalently $$(r+1)(r-2)=0.$$ The solutions are $-1$ and $2.$

EDIT

The convenient diagonal matrices have $-1$ or $2$ as diagonal entries.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I believe that there are eight solutions, not just two. $\endgroup$ – egreg Jan 7 at 22:23
  • $\begingroup$ Right. Thank you. $\endgroup$ – user376343 Jan 7 at 23:10
  • $\begingroup$ The first line just criticises what is a common notational convenience. $\endgroup$ – ancientmathematician Jan 8 at 7:34
  • $\begingroup$ @ancientmathematician from my side it was not criticism. In the first, well accepted comment, the author uses 21 which doesn't give "nice" results. In a school exercise, it is probable that the right quadratic equation is that from my solution. Moreover, the first comment was not clear to OP. Thus I explained also the possible "I" instead of 1. $\endgroup$ – user376343 Jan 8 at 10:12
1
$\begingroup$

Follows a fully worked example of "how to do this" as requested by our OP JOJO in her/his comment to the question itself:

$X$ being a $3 \times 3$ diagonal matrix it may be written

$X = \begin{bmatrix} x_1 & 0 & 0 \\ 0 & x_2 & 0 \\ 0 & 0 & x_3 \end{bmatrix}; \tag 1$

then

$X^2 = \begin{bmatrix} x_1^2 & 0 & 0 \\ 0 & x_2^2 & 0 \\ 0 & 0 & x_3^2 \end{bmatrix}; \tag 2$

thus

$\begin{bmatrix} x_1^2 - x_1 - 21 & 0 & 0 \\ 0 & x_2^2 - x_2 - 21 & 0 \\ 0 & 0 & x_3^2 - x_3 - 21 \end{bmatrix} = X^2 - X - 21I = 0; \tag 3$

it is now easily seen that

$ x_i^2 - x_i - 21 = 0, \; 1 \le i \le 3; \tag 4$

we may now deploy the quadratic formula:

$x_i = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(-21)}}{2} = \dfrac{1 \pm \sqrt{85}}{2}, \; 1 \le i \le 3; \tag 5$

returning to (1), since each of the $x_i$ may take on either of the two values (5), we see there are precisely $2^3 = 8$ possible matrices $X$ satisfying (2).

Nota Bene: I have taken the equation satisfied by $X$ to be

$X^2 - X - 21I = 0, \tag 6$

rather than

$X^2 - X - 2I = 0, \tag 7$

as was done by user376343 in her/his answer; but since either equation has $2$ real roots, there are $8$ possible matrices $X$ in either case. End of Note.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.