# Thread: Inverting matrix with unknown.

1. ## 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 -

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

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.

2. Hi Peleus!

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 -

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

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

3. 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?

4. 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 -

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

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:

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

$R2 \rightarrow R2 + 2 R1$:

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

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

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

$R1 \rightarrow R1 - x R2$:

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

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