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If matrix $A$ is given with dimensions $2 \times2 $ then,

A is invertible if, and only, if $ad - bc \neq 0$:

$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$

How can this be derived?

I just need hints, I am not good at properties etc.. of matrices, please help!

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    $\begingroup$ Have you heard of $\;\text{Adj}\,A\;$ = the classical adjoint or adjugate of a matrix? $\endgroup$ – Timbuc Jan 5 '15 at 13:55
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Notice that

$$\begin{align}\begin{bmatrix}a & b\ |\ 1 & 0 \\c & d \ |\ 0 & 1\end{bmatrix} &\Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & \frac{ad - bc}{a} \ &|\ -\frac{c}{a} & 1\end{bmatrix} \Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & 1 \ &|\ -\frac{c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\\&\Leftrightarrow \begin{bmatrix}a & 0\ &|\ \frac{ad}{ad-bc} & \frac{-ba}{ad-bc} \\0 & 1 \ &|\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix} \Leftrightarrow \begin{bmatrix}1 & 0\ &|\ \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\0 & 1 \ &|\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\end{align}$$

Then

$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$

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You may use the Cayley-Hamilton theorem for $2\times 2$-matrices, $$ A^2-(a+d)A+(ad-bc)I_2=0. $$ This can be computed easily. Multiplying with $A^{-1}$ we obtain $A-(a+d)I_2=-(ad-bc)A^{-1}$, hence the formula for $A^{-1}$.

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The proof that your expression really is the inverse of $\;A\;$ is pretty easy. How it is derived can be done as follows without deep knowledge in matrix theory:

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x&y\\z&w\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\iff$$

$$\begin{align}&ax&&+bz&&=1\\ &&ay&&+bw&=0\\ &cx&&+dz&&=0\\ &&cy&&+dw&=1\end{align}$$

Multiply second equation by $\;-c\;$ and fourth one by $\;a\;$ and add them to get:

$$(-bc+ad)w=a\implies w=\frac1{ad-bc}a$$

and etc. with the rest of unknowns.

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  • $\begingroup$ Do you have to prove the other way around because the definition is $AB = BA = I$? I am also new to linear algebra $\endgroup$ – GalaxyVintage Jan 31 '16 at 0:41
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2*2 Matrices inverse proof

As $A \times A^{-1}=I$

$\begin{bmatrix} x_{11} &x_{12} \\ x_{21}&x_{22} \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

$ax_{11}+cx_{12} = 1$
$ax_{21}+cx_{22} = 0$
$bx_{11}+dx_{12} = 0$
$bx_{21}+dx_{22} = 1$

$b(ax_{11}+cx_{12}) = abx_{11}+bcx_{12}=b$
$a(bx_{11}+dx_{12}) = abx_{11}+adx_{12}=0$
$(abx_{11}+adx_{12})-(abx_{11}+bcx_{12})=-b$
$x_{12}(ad-bc)=-b$
Row 1 Column 2 : $x_{12} =$ $-b\over{ad-bc}$

$b(ax_{21}+cx_{22}) = abx_{21}+bcx_{22}=0$
$a(bx_{21}+dx_{22}) = abx_{21}+adx_{22}=a$
$(abx_{21}+adx_{22})-(abx_{21}+bcx_{22})=a$
$x_{22}(ad-bc)=a$
Row 2 Column 2 : $x_{22} = $$a \over ad-bc$

$d(ax_{11}+cx_{12}) = adx_{11}+cdx_{12}=d$
$c(bx_{11}+dx_{12}) = bcx_{11}+cdx_{12}=0$
$(adx_{11}+cdx_{12})-(bcx_{11}+cdx_{12})=d$
$x_{11}(ad-bc)=d$
Row 1 Column 1 : $x_{11} = $$d\over ad-bc$

$d(ax_{21}+cx_{22}) = adx_{21}+cdx_{22}=0$
$c(bx_{21}+dx_{22}) = bcx_{21}+cdx_{22}=c$
$(adx_{21}+cdx_{22})-(bcx_{21}+cdx_{22})=-c$
$x_{21}(ad-bc)=-c$
Row 2 Column 1 : $x_{21} = $$-c \over ad-bc$

As $x_{11}=$$d \over ad-bc$, $x_{12}=$$-b \over ad-bc$, $x_{21}$$=-c \over ad-bc $and $x_{22}=$$a \over ad-bc$
$$ A^{-1}= \begin{bmatrix} \frac{d}{detA} & \frac{-b}{detA}\\ \frac{-c}{detA}& \frac{a}{detA} \end{bmatrix} = \frac{1}{detA}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix} $$

My first reply was unreadable due to lack of formatting, hope this new revision could help!

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  • $\begingroup$ Please use MathJax, and it's not scary. Your answer is hard to read. $\endgroup$ – Krish Nov 26 '17 at 3:53

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