# Inverting matrix with unknown.

• Sep 6th 2009, 09:11 PM
Peleus
Inverting matrix with unknown.
Hi all,

Quick question.

I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

This is the matrix however I have to do it on -

$\displaystyle \begin{array}{cc}-2&1\\1&x\end{array}$

Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

Cheers.
• Sep 6th 2009, 09:32 PM
Rapha
Hi Peleus!

Quote:

Originally Posted by Peleus
Hi all,

Quick question.

I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

This is the matrix however I have to do it on -

$\displaystyle \begin{array}{cc}-2&1\\1&x\end{array}$

Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

Cheers.

This is a 2x2 matrix and there is a formula for this problem

Invertible matrix - Wikipedia, the free encyclopedia
(title: Inversion of 2×2 matrices )

But you don't want to use this, right?

Yours
Rapha
• Sep 6th 2009, 09:40 PM
Peleus
I've only got one question on it, so it will get me by for this, but as always the saying goes give a man a fish and eat for a day, teach a man.....

Is there any other way of doing it?
• Sep 7th 2009, 03:36 AM
mr fantastic
Quote:

Originally Posted by Peleus
Hi all,

Quick question.

I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

This is the matrix however I have to do it on -

$\displaystyle \begin{array}{cc}-2&1\\1&x\end{array}$

Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

Cheers.

Swap the rows:

$\displaystyle \left( \begin{array}{cc} 1 & x \\ -2 & 1 \end{array}\right)$

$\displaystyle R2 \rightarrow R2 + 2 R1$:

$\displaystyle \left( \begin{array}{cc} 1 & x \\ 0 & 1 + 2x\end{array}\right)$

$\displaystyle R2 \rightarrow \frac{1}{1 + 2x} R2 \left( \text{provided } 1 + 2x \neq 0 \Rightarrow x \neq - \frac{1}{2} \right)$:

$\displaystyle \left( \begin{array}{cc} 1 & x \\ 0 & 1\end{array}\right)$

$\displaystyle R1 \rightarrow R1 - x R2$:

$\displaystyle \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$

Note that $\displaystyle x \neq - \frac{1}{2}$ is consistent with the requirement that the determinant is not equal to zero.