# Proof with matrices

I gave some thoughts about ortogonal matrices and the inverse matrix. I found a interesting point:

Let $A \in \mathbb{R}^2$ with $A=\begin{pmatrix} a & b \\ -c & d \end{pmatrix}$ and $det(A)=1$. The inverse matrix has the form $A^{-1}=\begin{pmatrix} a & -b \\ c & d \end{pmatrix}$. Is it always true, that the entries are the same, except $\pm$?

I hope my question is understandable.

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In general, if $$A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$$ then $$A^{-1}=\frac{1}{\det(A)}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix},$$ provided that $\det(A)\neq0$.

If the determinant equals $1$, just change the position of the element on the main diagonal and replace the other two elements by their negatives.

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Thanks, I got it. Quite simple, if I would use the definition... – ulead86 Jan 30 '14 at 21:07
@ulead86 what do you mean with "only provides $det(A)\neq0$? This formula holds for every invertible $2\times 2$-matrix. If $det(A)=0$ the matrix is not invertible and the question what the entries of $A^{-1}$ look like is meaningless – user127.0.0.1 Jan 30 '14 at 21:12
If the determinant equals $1$, just change the position of the element on the main diagonal and replace the other two elements by their negatives. – Michael Hoppe Jan 30 '14 at 21:12

Yeah just plonk $2x2$ matrices together:

$\left ( \begin{array}{cc} a & b \\ c & d \end{array} \right )\left ( \begin{array}{cc} a' & b' \\ c' & d' \end{array} \right )= \left ( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right )$

Which gives that:

$aa'+bc'=1$

$ab'+bd'=0$

$ca'+dc'=0$

$cb'+dd'=0$

Then just solve these to get the inverse!

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