0
$\begingroup$

The problem is:

"The $2\times 2$ matrix A satisfies $A^2-4A-7I=0,$ where I is the $2\times 2$ identity matrix. Prove that A is invertible."

The hint given is:

"We are trying to a matrix that is the inverse of A."

I completed the square and was proceeding to take the square root of both sides when I realized the identity matrix has multiple square roots.

$\endgroup$
  • 1
    $\begingroup$ You're not following the hint. If there exists a matrix $B$ such that $AB=I_2$, then $A$ is invertible and $A^{-1}=B$. You're given that $A^2-4A=7I_2$. Finish. $\endgroup$ – Git Gud Oct 11 '14 at 23:27
2
$\begingroup$

Check that $$ AB=BA=I$$ where $B=7^{-1}(A-4I)$.

$\endgroup$
0
$\begingroup$

Here your question is "Prove that $A$ is invertible". For this you have to show that $det (A) \neq 0$.

Now since $A$ is a $2× 2$ matrix, satisfies the polynomial equation $x^{2} − 4x - 7 = 0$ (as $A^2-4A-7I=0$ is given) which contains non-zero constant term $(-7)$, so $det (A) =-7 \neq 0$. Therefore $A$ is invertible. [Q.E.D]

** Now if you have to find $A^{-1}$, then you can proceed as follows

$ A^{2}-4 A-7 I=0 $ .......(1)

Since A has an inverse, so multiplying both side of equation (1) by $A^{-1}$ we have

$A^{-1} ( A^{2}-4 A-7 I ) = A^{-1}. 0 \implies A^{-1} . A^{2} -4 A^{-1}. A - 7 A^{-1} . I=0 \implies A - 4 I - 7 A^{-1}=0 \implies A^{-1} = \frac{1}{7} (A - 4 I) $

$\endgroup$

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.