[Math] Why adjugate matrix 2×2 is different from 3×3 and others

linear algebramatrices

I've understand the simple way of calculating the adjugate matrix.

In short:

1). We need to calculate all cofactors:

Using the next formula: $A_{ij} = (-1)^{i+j} M_{ij}$

2). Transpose it.

It's also can be read from Wikipedia:
https://en.wikipedia.org/wiki/Adjugate_matrix

In linear algebra, the adjugate, classical adjoint, or adjunct of a
square matrix is the transpose of the cofactor matrix.

If to view examples, such short algorithm is correct for squared matrices 3×3 and larger…

But, for 2×2 is just a rule:

M = [ a b ]
    [ c d ]

adj( M ) = [  d -b ]
           [ -c  a ]

What do I dislike?

I dislike that, because it's some kind of prayer. And many forums in web are saying:

Just remember it? right?

Damn! People, it's a MATH, it's not some humanity science like history. It's A MATH, math consists of logic and explanation as any other technical stuff.

I don't want just to remember stupidly some formula, I want understand why is exactly using such adjugate calculation especially for matrix 2×2?

I don't understand the next… For the matrix 3×3 we calculate all cofactors and then transpose it, for e.g.:

Original:
1   2   3
2   5   4
5   2   3

Cofactor matrix:
  7  14 -21
  0 -12  8
 -7   2  1

Transposed cofactor matrix:
  7   0 -7
 14 -12  2
-21   8  1

As you can see the transpose process for 3×3 didn't exchange the value a11 and a33, but why there is such an exchange for 2×3 matrix? I don't understand… I want to get logical explanation, not just a strict prayer.

I want to hear explanation why? and such explanation must be logical.

Best Answer

Consider what happens when you're getting the matrix of minors for a $3\times 3$: You are getting the value for the $a_{ij}$ element by calculating the determinant of the non $i,j$ rows (I won't go through this since, judging by your question, you know how to do this already).

Now consider the same logic applied to a $2\times 2$ matrix: Do the same steps, except instead of having a $2\times 2$ matrix to calculate the determinant from, you have a $1\times 1$. In this case, the determinant is the single element in that matrix. From this, you can do the same steps as you would for a $3\times 3$.

I'll run through an example here (I'll compare with a $3\times 3$ matrix since that's the simplest matrix where these rules are first introduced):

Matrix $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$.

Matrix of minors $= \begin{bmatrix} d & c\\ b & a\end{bmatrix}$ - for element $(1,1)$ (i.e. $a$), "cover" row $1$ and column $1$ and you're left with $d$ so that equals the determinant for that element for the matrix of minors (as with a $3\times 3$).

Cofactor matrix, $C = \begin{bmatrix} d & -c\\ -b & a\end{bmatrix}$ - exact same as for a $3\times 3$.

$\operatorname{adj}A = C^T = \begin{bmatrix} d & -b\\ -c & a\end{bmatrix}$ - again, exact same as for a $3\times 3$.

This is where you get the "Just remember it" from. It follows the exact same steps as for any $n\times n$ matrix.

Then you calculate $A^{-1}$ as usual: $A^{-1} = \frac{1}{\det A}\operatorname{adj}A$.

Hope this helps.

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