How to intuit dividing the inverse by the determinant of the matrix?

Question

I am learning Determinants and inverses but cannot comprehend this:

But the interesting thing is that this determinant there, this now that it scales space. If we then take this matrix when we do the flipping around we haven't changed its scaling of space we need to undo that scaling and bring it back down to a scale of one. So the determinant here is what we need to divide the inverse matrix by in order for it to probably be an inverse.

I mean, what does it mean to undo that scaling and bring it back down to a scale of one?

How to geometrically interpret undoing that scaling of the matrix?

This needs more context. Your link is to a course. Surely you are not expecting us to enroll in the course and go through it until we get to this passage, right? — John Douma, Aug 31 '22 at 16:50
@JohnDouma I have updated the link, and now it links to YouTube. — Lerner Zhang, Sep 01 '22 at 12:53
A YouTube video isn't good neither though. Regarding your question: think of the determinant as a volume, maybe this will help? — Anthony, Sep 01 '22 at 12:56
See this post and here . The only right answer is through exterior algebra — Clemens Bartholdy, Sep 01 '22 at 12:57
@AnthonySaint-Criq Why not scale it twice, one for A and the other for $A^{-1}$? — Lerner Zhang, Sep 01 '22 at 12:58
@LernerZhang Good thought. Notice for a real number $c$ and $n$-dimensional matrix $A$, $\det(cA) = c^n \det(A)$. In this example with $n=2$, dividing the second matrix by $ad-bc$ does reduce the volume change factor by $(ad-bc)^2$. — aschepler, Sep 01 '22 at 13:13
I'd suggest you actually check out exterior algebra, I am pretty sure you would reach the same conclusion @FShrike — Clemens Bartholdy, Sep 01 '22 at 15:36

Constantinos Pisimisis · Answer 1 · 2022-09-01T14:40:47.603

Generic formula for inverse matrix

$\text{Let A be a 2$\times$2 matrix.To be more specific consider this matrix:}$ $\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$
$\text{Generic formula for inverse matrix: } A^{-1}= \displaystyle \frac 1 {det(A)} \cdot Adj(A)$ $\text{Explanation of symobols: detA is the determinant of matrix A.}$ $\text{Adj(A) is the adjoint matrix.}$

Generic Formula for Adjoint matrix

$\text{Given a square matrix $B=b_{ij}$, the generic formula for adjoint matrix is: }$
$ \displaystyle Adj(B)=({b_{ij}}) \text{ where: }b_{ij}=(-1)^{i+j}\cdot det(b_{ij})$ $\text{In other words the element ${b_{ij}}$ is refering to the value of the determinant}$
$\text{obtained by deleting the $i^{th}$ row and $j^{th}$ column of the B matrix.}$

What about 2$\times$2 matrix?

Given the previous formulas the inverse of matrix A is: $$ A^{-1}=\frac 1 {det(A)} \cdot \begin{pmatrix}d&-b\\\ -c&a\end{pmatrix} \Rightarrow A^{-1}=\frac 1 {ad-cd} \cdot \begin{pmatrix}d&-b\\\ -c&a\end{pmatrix} $$ $\text{Where: } \begin{pmatrix}d&-b\\\ -c&a\end{pmatrix}=adj(A) $

Conclusion

The lack of knowledge when it comes to basic formulas of inverse matrix its easy to create some misunderstandings.The determinant "changes" the factor that its multiplied.And not only as a number but also "changes" the space.Since adjoint is calculated by determinant we have to divide by the determinant to undo that "change".I strongly recommend to take a look on this site:inverse and adjoint matrix and this:Determinant as scaling factor.

How to intuit dividing the inverse by the determinant of the matrix?

1 Answers1