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The $2 \times 2$ matrix $A$ satisfies

$$A^2 - 4A -7I = 0,$$

where $I$ is the identity matrix. Prove that $A$ is invertible.

I'm not sure how to do this. Help would be appreciated.


marked as duplicate by user1551 matrices May 24 '15 at 23:09

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  • $\begingroup$ Hint: $A(A-4I) = 7I$. $\endgroup$ – Anurag A May 24 '15 at 22:28
  • 6
    $\begingroup$ Not in characteristic $7$, though. $\endgroup$ – Pedro Tamaroff May 24 '15 at 22:28

Observe that $$ \left(\frac17(A-4I)\right)A=I $$


Hint: write $I = AA^{-1}$, so that: $$A^2-4A - 7AA^{-1} = 0.$$Can you solve for $A^{-1}$?

  • $\begingroup$ I'm still a little new to the concept of matrices. How exactly would I start to solve for A^-1 $\endgroup$ – idk May 24 '15 at 22:31
  • $\begingroup$ It's similar to real numbers, you only have to remember that not always matrices commute, and not always you can "divide" (i.e., multiply everything by the inverse). Here we would have: $$7AA^{-1} = A(A-4I) \implies AA^{-1} = \frac{1}{7}A(A-4I).$$ Since we assume that $A^{-1}$ exists, you can multiply both sides by $A^{-1}$ by the left, so we get only $A^{-1}$ on the left side. I'll let you finish it now. $\endgroup$ – Ivo Terek May 24 '15 at 22:34
  • $\begingroup$ ohh ok it makes sense now. Thanks :) $\endgroup$ – idk May 24 '15 at 22:38

you can rewrite $A^2 - 4A - 7I = 0$ as $$A\left(\frac17A-\frac47I\right) = I.$$ therefore $A$ is invertible and $$A^{-1} = \frac17A-\frac47I. $$

  • 2
    $\begingroup$ Based on the level of the question, I think it's fair to assume the OP is working over the reals. Considerations of characteristics would be a needlessly high level of sophistication for the question, just as we don't worry about fields of characteristic 2 when teaching the quadratic equation. $\endgroup$ – Zach Effman May 24 '15 at 22:35
  • $\begingroup$ @ZachEffman, i agree. anyone who bothers with characteristic $7$ and the like is unlikely ask this question in the first place. $\endgroup$ – abel May 24 '15 at 22:38

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