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For ex,

$$A^{-1}\cdot A=\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]\cdot\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$

and

$$A\cdot A^{-1}=\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]\cdot\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$

so we can make two systems:

\begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \end{cases} \end{equation*}

and

\begin{equation*} \begin{cases} x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}

after line addition we get the system:

\begin{equation*} \begin{cases} 2x_2 + 5x_4 = 1, \\ 2x_1 + 5x_3 = 1, \\ 3 x_1 + 4 x_2 = 1, \\ 3 x_3 + 4 x_4 = 1 \end{cases} \end{equation*}

We solve the system and find the roots: $(x_1 = -2, x_2 = -2, x_3 = 1, x_4 = 1)$. This is not the inverse matrix. What is obvious. The inverse matrix found by the standard algorithm is equal to

$$A^{-1}=\left[\begin{matrix}3 & -2\\-1 & 1\end{matrix}\right]$$

But where are my arguments wrong?

Edited correct roots of the last system are: $x_1 - \frac{10x_4}{3} = \frac{-1}{3}, x_2 + 2.5x_4 = 0.5, x_3 + \frac{4x_4}{3} = \frac{1}{3}$

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  • $\begingroup$ "line addition": what on earth is that? $\endgroup$ – Lord Shark the Unknown Jan 20 at 5:08
  • $\begingroup$ @LordSharktheUnknown I just added the lines with the same variables in the systems ... $\endgroup$ – Just do it Jan 20 at 5:09
  • $\begingroup$ What is the rank of your new system? $\endgroup$ – Lord Shark the Unknown Jan 20 at 5:10
  • $\begingroup$ The coefficients of the actual inverse do satisfy your new system, so what's the problem? $\endgroup$ – Lord Shark the Unknown Jan 20 at 5:15
  • $\begingroup$ @LordSharktheUnknown I think 3 because the last line is 0 $\endgroup$ – Just do it Jan 20 at 5:15
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It is even easier that what you have done. Consider $$A^{-1}\cdot A=\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]\cdot\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$

You get your system: \begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \end{cases} \end{equation*}

Solve it, it is easy to see: $(x_1,x_2,x_3,x_4)=(3,-2,-1,1)$.

You can do the same in the other way. Consider $$A\cdot A^{-1}=\left[\begin{matrix}1 & 2\\1 & 3\end{matrix}\right]\cdot\left[\begin{matrix}x_1 & x_2\\x_3 & x_4\end{matrix}\right]=\left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$$

You get your system: \begin{equation*} \begin{cases} x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}

Solve it, it is easy to see: $(x_1,x_2,x_3,x_4)=(3,-2,-1,1)$.

Hence $$A^{-1}= \left[\begin{matrix}3 & -2\\-1 & 1\end{matrix}\right]$$

Your problem comes when adding these two systems. If you have your new system:

\begin{equation*} \begin{cases} x_1 + 2x_3 = 1, \\ x_2 + 2x_4 = 0, \\ x_1 + 3 x_3 = 0, \\ x_2 + 3x_4 = 1 \\ x_1 + x_2 = 1, \\ 2x_1 + 3x_2 = 0, \\ x_3 + x_4 = 0, \\ 2 x_3 + 3x_4 = 1 \end{cases} \end{equation*}

I think that you have done some bad addition between these equations. Because it should lead the same answer as you have seen above.

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  • $\begingroup$ Yes, I have already figured out, thanks) $\endgroup$ – Just do it Jan 20 at 11:41

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