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I need some help computing the inverse of the following problem. I am baffled as to why I cannot compute this. I'll list my steps please help me notice my error.

\begin{bmatrix} 1 & 2 \\ -1 & 2 \end{bmatrix}

To compute the inverse first find the Det = [($1 \times 2) - (-1 \times 2)] = 4 $.

As far as I am aware the only case in which a matrix will not have an inverse is if the determinant is zero which leads me to continue my process.

Now I divide all of the $ij$ components by the determinant and get.

\begin{bmatrix} 1/4 & 1/2 \\ -1/4 & 1/2 \end{bmatrix}

Taking into account the co-factor $C_{ij}$ = Minor $\times ({-1})^{i+j}$

I get the following matrix

\begin{bmatrix} 1/4 & -1/2 \\ 1/4 & 1/2 \end{bmatrix}

Now I transpose it.

\begin{bmatrix} 1/4 & 1/4 \\ -1/2 & 1/2 \end{bmatrix}

Now when I multiply this by my original matrix

\begin{bmatrix} 1 & 2 \\ -1 & 2 \end{bmatrix}

I should get the Identity matrix but keep utterly failing.

Edit: Thanks you guys for the answers my error was identifying my minor. Quick follow up. What is the interpretation of a minor? I have been struggling with that concept as every video I find simply tells me how to compute it rather than its purpose. Thanks

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    $\begingroup$ The inverse of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{D} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $D$ is the determinant, assuming $D \neq 0$. Your problem appears to be incorrectly identifying the minors; for example the $(1,1)$ minor here is the $(2,2)$ entry. $\endgroup$
    – Ian
    Commented Sep 23, 2019 at 1:14
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    $\begingroup$ You have to switch the diagonal elements $\endgroup$ Commented Sep 23, 2019 at 1:15
  • $\begingroup$ Thank you! You are 100% correct! $\endgroup$
    – Josue
    Commented Sep 23, 2019 at 1:15

1 Answer 1

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Copying from my comment:

The inverse of $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{D} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $D$ is the determinant, assuming $D \neq 0$. Your problem appears to be incorrectly identifying the minors; for example the $(1,1)$ minor here is the $(2,2)$ entry (which is why the positions of $a$ and $d$ get switched).

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  • $\begingroup$ Thank you very much, you are correct. I made a mistake with my minors. Can I ask you a quick fallow up question. What is the interpretation of a minor? I have been struggling with that concept as every video I find simply tells me how to compute it rather than its purpose. $\endgroup$
    – Josue
    Commented Sep 23, 2019 at 1:19
  • $\begingroup$ @Josue en.m.wikipedia.org/wiki/Minor_(linear_algebra) $\endgroup$
    – amd
    Commented Sep 23, 2019 at 6:29

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