Question

\(A*adj(A) = |A|I\)

I am finding it hard to see how the elements not in diagonal become 0... can somebody explain please

Answer 1

@sidsiddhartha

Answer 2

\[
\left[
\begin{array}{ccc}

A_{11} &A_{12}&\cdots \\
A_{21} &A_{22}&\cdots \\
\cdots \\

\end{array}
\right]
\left[
\begin{array}{ccc}

C_{11} &C_{21}&\cdots \\
C_{12} &C_{22}&\cdots \\
\cdots \\

\end{array}
\right]
=\left[
\begin{array}{ccc}
|A| &0&\cdots \\
0 &|A|&\cdots \\
\cdots \\

\end{array}
\right]
\]

Answer 3

\(\large A_{11}C_{11} + A_{12}C_{12} + \cdots +A_{1n}C_{1n} = |A|\)

\(\large A_{21}C_{11} + A_{22}C_{12} + \cdots +A_{2n}C_{1n} = 0\)

Answer 4

all right
take
\[A*adj(A) =B_{i,j}\]
so B(i,j) is the product of the i-th row of A by the j-th column of adj(A)
ok?

Answer 5

ok if \(i \ne j\), then it will be 0 <<<< this is the confusing part

Answer 6

yes
if i equals j then this sum is the cofactor expansion of A along the i-th row, so it is equal to det(A)
and
B(i,j)=0 (i and j are different numbers from 1 to n).

Answer 7

\(B_{ij} = |A|\) if \(i = j\)

this part is clear

Answer 8

I have lil bit difficulty in seeing why below is true :
\(B_{ij} = 0 \) if \(i \ne j\)

Answer 9

why is the dot product of a "row" with the "cofactors of any other row" is 0

Answer 10

Try multiplying a sample matrix by its adjugate.
\[A=\begin{pmatrix}a&b\\c&d\end{pmatrix}~~\iff~~\text{adj}(A)=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\]
What do you get for
\[A\cdot \text{adj}(A)~~?\]

Answer 11

ad-bc -ab+ab
cd-cd -bc+ad

Answer 12

|A| 0
0 |A|


XD is there any hidden/special reason for "-ab+ab" and "cd-cd" showing up in non-diagonal places ?

Answer 13

Maybe not so hidden, but the top row of \(A\) and the right column of the adjugate is negative of each other (meaning the terms you add together).

Answer 14

For a higher dimension matrix it might not be so easy to see, but the cancellation stems from the \((-1)^{i+j}\) factor of the cofactor matrices, I believe.

Answer 15

looking at the 2x2 matrix, it is somewhat convincing... but 3x3 will have a messy products in cofactor matrix instead of single elements and it becomes hard to see

Answer 16

You can try working through an example and convincing yourself that way, then try extending it to a general matrix.

Answer 17

I did that already, it helped a bit... but didn't give 100% clarity

Answer 18

Yeah, examples tend to have that about them... I suppose you can use variables in place of actual numbers. Some pattern will start to emerge as you increase the dimension of the matrix, but the work to get there is tedious.

Answer 19

I have used variables A_ij and C_ij,
Noticed that adding all 6 terms in the expansion of a row with cofactors of another row produced 0. it was really some joy to see the calculations working out nicely and producing 0, but thats the end of it. It was not giving more insight into why the dot product becomes 0

Answer 20

I was looking at a proof yesterday... let me pull it up, one sec..

Answer 21

http://www.math.vanderbilt.edu/~msapir/msapir/jan29.html#cofactorexpansion

this page has a proof, but its very confusing and short and doesn't explain much

Answer 22

It's the second corollary right?

Answer 23

yes thats the one :)

Answer 24

`Proof. Indeed, let us replace the i-th row of A by the j-th row of A and call the resulting matrix B.`

how is it legal to replace a row by another row ?

Answer 25

\(A = \left|
\begin{array}{ccc}

A_{11} &A_{12}&\cdots \\
A_{21} &A_{22}&\cdots \\
\cdots \\

\end{array}
\right|

\)


\(B = \left|
\begin{array}{ccc}

A_{21} &A_{22}&\cdots \\
A_{21} &A_{22}&\cdots \\
\cdots \\

\end{array}
\right|

\)

Answer 26

It's not so much replacing as it is considering a different row altogether - the wording is a bit confusing. I think what they mean is "replace \(i\) with \(j\) in the equation". We know that
\[\large\sum_{k=1}^nA(i,k)C_{j,k}=|A|\]
by the second theorem. If we consider any row other than the \(i\)-th row, say the \(j\)-th row (meaning we set \(i=j\)), then
\[\large\sum_{k=1}^nB(j,k)C_{j,k}=0\]
because this is the cofactor expansion of a matrix with equal rows.

Answer 27

At least that's how I interpreted it...

Answer 28

I'll give this another glance some other time. I have to get to work in the morning.

Answer 29

I don't see yet how A and B are related.... I can wait... thanks a lot, really appreciate your help :)

Answer 30

Something to consider: for an \(n\times n\) matrix \(A\) with nonzero determinant,
\[A^{-1}=\frac{1}{|A|}\text{adj}(A)\]
If you can accept this formula (or prove it), it might provide some insight on the other proof.