There is a short-cut for finding the inverse of 2x2 matrices
Can someone please explain why it works ? because I can't find any proof ?
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3Did you try multiplying the claimed result by $A$? – Simply Beautiful Art Dec 13 '17 at 01:18
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I know that it works, I want to know why it works – Random person Dec 13 '17 at 01:19
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1Well, it works because, it works. There are no hidden reasons here. It is just a simple calculation. – eranreches Dec 13 '17 at 01:20
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I'm not sure I understand your question. "why it works" isn't a very good question, especially if you swat away suggestions such as my first comment. – Simply Beautiful Art Dec 13 '17 at 01:21
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The formula can be easily derived with Gauss Jordan – Riley Dec 13 '17 at 01:22
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Oh man, mathmeeting.com, free homework help and solutions. Hm... just to let you know, our site policies are probably different than mathmeeting's. – Simply Beautiful Art Dec 13 '17 at 01:26
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I'm not asking for the answer to a homework question simply a reason why a short-cut I was taught works – Random person Dec 13 '17 at 01:28
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Yes, I know. But since this seems to be your first question, I wanted to point you towards our help center. Note that your question may be downvoted and closed due to a lack of effort/context; this site tends to encourage questions that show thought has been put into them. – Simply Beautiful Art Dec 13 '17 at 01:30
4 Answers
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If $A$ is an invertible $2 \times 2$ matrix, then it satisfies:
$$A A^{-1} = I$$
Let $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$. Then you need to solve the augmented system:
$$ \left[ \begin{array}{cc|cc} a&b&1&0\\ c&d&0&1\\ \end{array} \right] $$
Now row-reduce this system until the left side becomes $I$. The right side will then give you the desired $A^{-1}$
infinitylord
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The r.h.s. is the adjugate matrix of $A$, i.e. the transpose of the matrix of cofactors of $A$, which is easy to calculate in the case of $2\times2$ matrices.
One proves, independently of the size of the matrix, that $$A\cdot\operatorname{adj}A=(\det A)\,I.$$ This formula is a rewriting of Laplace's formula, and is valid for matrices over any commutative ring.
Bernard
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If an inverse for A exists, it is unique.
Simply multiply out the two matrices; the off-diagonal elements in the result are zero ($-ab+ab$, $cd-cd$). The diagonal elements are both $ad-bc$. Now, recall the determinant is simply $ad-bc$, so when you divide by that the result is the identity matrix.
There really isn't any more to prove here.
eSurfsnake
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This comes from the classical adjoint matrix.
If we have $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$, then $A^{-1}=\dfrac{1}{\det(A)}*adj(A)$
Now the $adj(A)= \begin{bmatrix} (-1)^{1+1}d & (-1)^{1+2}c\\ (-1)^{2+1}b & (-1)^{2+2}a \end{bmatrix}^T=\begin{bmatrix} d & -c\\ -b & a \end{bmatrix}^T = \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$
Now $\det(A)$ is simply $ad-cb$, and so we have that $$A^{-1}=\dfrac{1}{ad-cb}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$$
If you don't know the adjoint method, please look here:
K Split X
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