Invese of a 2x2 matrix

soroban

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In case you didn't know . . .


Find the inverse of: .\(\displaystyle A \;=\;\begin{bmatrix}a&b\\c&d\end{bmatrix}\)


[1] Switch the two on the main diagonal \(\displaystyle (a\text{ and }d): \;\begin{bmatrix}d & . \\ . & a\end{bmatrix}\)

[2] Change the signs of the other two: .\(\displaystyle \begin{bmatrix}. & -b \\ -c & . \end{bmatrix}\)

[3] Divide everything by the determinant: .\(\displaystyle \begin{vmatrix}a&b\\c&d\end{vmatrix} \:=\:ad - bc\)


\(\displaystyle \text{Therefore: }\:A^{-1} \;=\;\begin{bmatrix}\dfrac{d}{ad-bc} & \dfrac{-b}{ad-bc} \\ \dfrac{-c}{ad-bc} & \dfrac{a}{ad-bc}\end{bmatrix}\)
 
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Is the "invese" of a matrix in any way related to the "inverse"?:twisted:


From a different thread and subforum:

I don't understand what you are doing!
You say that the problem is to find the derivative but you seem to
be trying to do an . > > itegral. , < < <

First, you understand that it is not easy to turn a computer monitor on it side!
But I think your function is \(\displaystyle 4x+ \int_{x^2}^{x^3}e^{t^2}dt\).
That integral cannot be done in terms of elementary functions,
and this problem does not require you to > > > itegrate < < < it.

Is the "itegral" of a calculus problem in any way related to the "integral?"

Is "itegrate" of a calculus problem in any way related to "integrate?"
 
This is oe of those thigs that I "kow" but always have to look up whe I wat to "emembe".

Hee's aothe example: Aea of tiagle i tems of the thee vetex-coodiates.

;)
 
From a different thread and subforum:



Is the "itegral" of a calculus problem in any way related to the "integral?"

Is "itegrate" of a calculus problem in any way related to "integrate?"
What! You dare accuse me of muking a mystgreake?
 
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