Given that a square matrix A satisfies the equation A^2 -7A -I=0 Show that A is invertible and find its inverse.

- chrisplusian

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Empty

is A a 2x2 matrix?

- chrisplusian

It doesn't say that

- chrisplusian

I typed the problem verbatim

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- chrisplusian

\[A^2-7A-I=0\]

- Empty

If it's not, I don't think this is enough to imply that it's invertible, since every matrix satisfies its own characteristic equation this is essentially the characteristic equation of a 2x2 matrix, which means it has two distinct eigenvalues. But I think if it's say a 3x3 this isn't really enough.

- Empty

Maybe I'm over thinking this though and there's like some easier way to go about this haha

- chrisplusian

I don't know what eigenvalues are, and I have asked my professor about the eigenvalues before when someone on here helped me and started by trying to explain them. He said that until he explains it to us the solutions to our problems won't require comprehending eigenvalues.

- Empty

Ok, I think I found something better then for you, we don't need to go so deep, sorry about that!

- chrisplusian

What we have covered are matrix algebra, matrix inverses, vectors, transpose of a matrix, lines and planes in this linear algebra course

- Empty

Ok what we have to do is factor it and rearrange this equation, that will give you the inverse:
\[A^2-7A-I=0\]
here I add the identity to both sides and rewrite \(A=AI\) which is fine since a matrix times the identity is just itself.
\[A^2-7AI=I\]
Now I can factor out one of the A's
\[A(A-7I)=I\]
Oh hey, a matrix times another matrix equal to the identity matrix means that it must be its inverse!
\[AA^{-1}=I\]
So here you go:
\[A^{-1}=A-7I\]

- chrisplusian

Could it be that simple? I must be overthinking it too. I was thinking they wanted specific matrix components...... I was wondering how I was even supposed to decide what size matrix it was. Thanks

- chrisplusian

ONe thing....

- Empty

Haha yeah, if there's any step in particular that you seem doubtful about I could help explain more, yeah sure anything.

- chrisplusian

How does this show that it is invertible?

- chrisplusian

Because at one point it is equal to the identity matrix?

- Empty

Ok, so suppose I show you this matrix equation,
\[AB=I\]
Is B the inverse of A?

- chrisplusian

Or would I need to show that multiplication both ways results in the identity matrix?

- chrisplusian

AB=I, and BA=I is how you prove it is invertible right?

- Empty

Yeah you could show that if you like, but once you know \[AB=I\] then you know \[BA=I\] since without even knowing anything we can prove it.
In this example if you want to show that, you can do that if you like as well, just do what I've done but factor out A on the right instead.
I'll give the proof that \[AB=I\] implies that \[BA=I\] in my next post.

- chrisplusian

I actually looked back at the work I did before I posted this question and I have \[(A-7)A=I\] does that seem like it is ok without the identity attached to the 7?

- Empty

Not quite, because what does it mean to add a scalar to a matrix?

- Empty

However it is common to just represent a scalar multiplied by an identity matrix as a so-called 'scalar matrix' so as long as you understand that the 7 you have written there represents a matrix with 7's down the diagonal you're fine.

- chrisplusian

I know this sounds silly but what does it mean to add a scalar to a matrix? You can't do that right?

- Empty

Yeah you can't! That's why you have to have the identity matrix there, so really that 7 you've written is shorthand:
\[7 \implies 7I\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.