0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks, I got it. Quite simple, if I would use the definition... $\endgroup$ – ulead86 Jan 30 '14 at 21:07
  • 1
    $\begingroup$ @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 $\endgroup$ – user127.0.0.1 Jan 30 '14 at 21:12
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hoppe Jan 30 '14 at 21:12
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.