Proving inverse of a 2 x 2 matrix is really an inverse

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The discussion centers on understanding the inverse of a 2x2 matrix and the relationship between the terms ad - bc and their role in the formula for the inverse. It clarifies that while ad - bc is not equal to 1, dividing by this determinant yields the correct inverse matrix. The formula for the inverse is highlighted as easy to memorize, involving swapping diagonal entries and negating off-diagonal entries. An example is provided to demonstrate how this method can be applied to solve linear equations effectively. The procedure is emphasized as a reliable algorithm for those who prefer structured approaches over intuitive methods.
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Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1682211587003.png

Dose someone please know how ##ad - bc## and ##-cb + da## are equal to 1?

Many thanks!
 
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They are not equal to 1. They are both divided by ad-bc, which gives 1.
 
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FactChecker said:
They are not equal to 1. They are both divided by ad-bc, which gives 1.
Oh thank you @FactChecker ! I see now
 
This formula
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\begin{pmatrix}d&-b\\-c&a\end{pmatrix} \cdot (ad-bc)^{-1}
$$

is so easy to memorize ...
  1. swap the diagonal entries ##a \leftrightarrow d##
  2. put a minus sign in front of the entries on the side diagonal ##b \rightarrow -b\, , \,c\rightarrow -c##
  3. and finally, divide by the determinant ##ad-bc.##
... that I use it whenever I have to solve a linear equation system in ##2## variables.

E.g. I wanted to write ##16 \cdot 24 ## and ##8\cdot 48## as ##(n-m)\cdot (n+m)## today. That goes:
\begin{align*}
\begin{pmatrix}1&-1\\1&1 \end{pmatrix}\cdot \begin{pmatrix}n\\m\end{pmatrix}=\begin{pmatrix}16\\24 \end{pmatrix}
\end{align*}
Then by doing the procedure as described I get
\begin{align*}
\begin{pmatrix}n\\m \end{pmatrix}=\begin{pmatrix}1&1\\-1&1 \end{pmatrix}\cdot \begin{pmatrix}16\\24 \end{pmatrix}\cdot \underbrace{(1\cdot 1- (-1\cdot 1))^{-1}}_{=1/2}=\begin{pmatrix}1&1\\-1&1 \end{pmatrix}\cdot \begin{pmatrix}8\\12 \end{pmatrix}=\begin{pmatrix}8+12\\ -8+12\end{pmatrix}=\begin{pmatrix}20\\4 \end{pmatrix}
\end{align*}
Same with the other product ##8\cdot 48.##

I know this was an easy example and could probably just "be seen". But I'm better with algorithms than "seeing" things. Once you get used to that procedure it is really easy to follow. Especially if the examples are a bit more complicated.
 
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