# Quadratic Equation with Matrix [Prove Invertible]

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.

• 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. – Git Gud Oct 11 '14 at 23:27

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

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

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