Inverse of a generic Matrix A such that A^2+xA+yI_n = 0, x,y real

[diogomgf]

Hello again,

Been wrapping my head around yet another linear algebra problem:

For the generic matrix A_nxn such that:

A²+xA+yI_n = 0 (matrix 0), with x and y belonging to the real numbers and y can't be 0...

I need to know what is the inverse of A.

 

[Dr.Peterson]

For the generic matrix A_nxn such that:

A²+xA+yI_n = 0 (matrix 0), with x and y belonging to the real numbers and y can't be 0...

I need to know what is the inverse of A.

See if you can rewrite A²+xA+yI_n = 0 in the form A(...) = I_n. Then the expression in parentheses will be A^-1. This will, of course, be an expression in A, x, and y.

Then pick a simple 2x2 matrix A and see if you can check this by finding values for x and y, and see that your inverse is correct. It's an interesting exercise.

 

[topsquark]

My two cents: I'm assuming that you know A.

Suppose that \(\displaystyle Det(A) \neq 0\) so that an inverse to A exists.

Then
\(\displaystyle A^2 + x A + y I_{n} = 0\)

\(\displaystyle (A^2 + x A + y I_{n} ) A^{-1} = 0 \cdot A^{-1}\)

\(\displaystyle A^2 \cdot A^{-1} + x A \cdot A^{-1} + y I_{n} A^{-1} = 0\)

\(\displaystyle A + x I_{n} + y A^{-1} = 0\)

Thus
\(\displaystyle A^{-1} = - \dfrac{1}{y} (A + x I_{n})\)

-Dan

 

[Steven G]

My two cents: I'm assuming that you know A.

Suppose that \(\displaystyle Det(A) \neq 0\) so that an inverse to A exists.

Then
\(\displaystyle A^2 + x A + y I_{n} = 0\)

\(\displaystyle (A^2 + x A + y I_{n} ) A^{-1} = 0 \cdot A^{-1}\)

\(\displaystyle A^2 \cdot A^{-1} + x A \cdot A^{-1} + y I_{n} A^{-1} = 0\)

\(\displaystyle A + x I_{n} + y A^{-1} = 0\)

Thus
\(\displaystyle A^{-1} = - \dfrac{1}{y} (A + x I_{n})\)

-Dan

Nicely done. I would have solved to xA, then solve for A and take inverse of both sides. But your 2 cents method is certainly better.

 

[Dr.Peterson]

My two cents: I'm assuming that you know A.

Suppose that \(\displaystyle Det(A) \neq 0\) so that an inverse to A exists.

Then
\(\displaystyle A^2 + x A + y I_{n} = 0\)

\(\displaystyle (A^2 + x A + y I_{n} ) A^{-1} = 0 \cdot A^{-1}\)

\(\displaystyle A^2 \cdot A^{-1} + x A \cdot A^{-1} + y I_{n} A^{-1} = 0\)

\(\displaystyle A + x I_{n} + y A^{-1} = 0\)

Thus
\(\displaystyle A^{-1} = - \dfrac{1}{y} (A + x I_{n})\)

-Dan

My suggestion was equivalent to this, but factoring rather than multiplying by the inverse (which requires assuming the inverse exists), just as we prefer solving equations by factoring rather than by dividing by the variable:

\(\displaystyle A^2 + x A + y I_{n} = 0\)

\(\displaystyle A(A + xI_{n}) = -yI_{n}\)

\(\displaystyle A(A + xI_{n})\cdot\dfrac{1}{-y} = I_{n}\)

Therefore \(\displaystyle A^{-1} = -\dfrac{1}{y}(A + xI_{n})\)

 

[diogomgf]

My two cents: I'm assuming that you know A.

Suppose that \(\displaystyle Det(A) \neq 0\) so that an inverse to A exists.

Then
\(\displaystyle A^2 + x A + y I_{n} = 0\)

\(\displaystyle (A^2 + x A + y I_{n} ) A^{-1} = 0 \cdot A^{-1}\)

\(\displaystyle A^2 \cdot A^{-1} + x A \cdot A^{-1} + y I_{n} A^{-1} = 0\)

\(\displaystyle A + x I_{n} + y A^{-1} = 0\)

Thus
\(\displaystyle A^{-1} = - \dfrac{1}{y} (A + x I_{n})\)

-Dan

Thanks alot for the replies every one.

Problem solved =)