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I tried to prove the inverse of the matrix but I got a wrong formula. Here's the proof:

  1. Let $$aX + bY = m \quad , \quad cX + dY = n\\ \begin{align}X & = \frac{m - by}{a}\\ & = \frac{n - dy}{c}\end{align}\\ \therefore cm - cbY = an - adY\\ cm - an = Y(cb - ad)\\\therefore Y = \frac{cm - an}{cb - ad}$$ Similarly, $$\therefore X = \frac{dm - bn}{da - cb} $$

  2. If $B$ is inverse of matrix $A$, then $AB = BA = I$

Let $A = \begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix}$ and $B = \begin{pmatrix}b_1 & b_2 \\ b_3 & b_4\end{pmatrix}$

Multiplication forms a series of equations, like: $a_1b_1 + a_2b_3 = 1$ and $ a_1b_2 + a_2b_4 = 1$

But there is clearly something wrong here because according to 1, then $b_1 = b_2$ and $b_3 = b_4$ which is not always true. What's wrong with this proof ?

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  • $\begingroup$ $a_1b_2+a_2 b_4=0$ not $1$ $\endgroup$ – Ali Caglayan Jul 9 '16 at 21:54
  • $\begingroup$ @AliCaglayan how ? $\endgroup$ – Amr Ayman Jul 10 '16 at 14:36
  • $\begingroup$ Its matrix multiplication. You should get 4 expressions when you expand AB. The identity has zeros in the top right and bottom left corners. $\endgroup$ – Ali Caglayan Jul 11 '16 at 21:25
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Your second equation is wrong, hence the apparent contradiction.
You have the first row of $A$ multiplied by the second column of $B$. The result of that multiplication therefore gives the first row, second column entry of $AB=I$. The rest of the correction is trivial.

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