Matrix Inverse Error

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

• 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.
– Ian
Commented Sep 23, 2019 at 1:14
• You have to switch the diagonal elements Commented Sep 23, 2019 at 1:15
• Thank you! You are 100% correct! Commented Sep 23, 2019 at 1:15

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).