Question

Matrix problem:

Answer 1

110101110101011010101101010110101101011010101

Answer 2

Let A = show that A(adjA) = det(A)

Answer 3

Is that "A times the adjugate of A"?

Answer 4

I was kinda hoping to find out here what it is. lol. I don't know, but is it? Because I tried to study and I saw an "adjoint" thing.

Answer 5

Yes it is. Recall the formula for an inverse matrix:
\[A^{-1}=\frac{1}{|A|}\text{adj}(A)\]
You can rearrange this formula to get the equation you have to show.

Answer 6

Alright, so the adjugate matrix \(\text{adj}(A)\) is defined as the transpose of the cofactor matrix. Do you know what transpose and cofactor matrices are?

Answer 7

I am only familiar with the terms, sorry. But, no.

Answer 8

Is the det determinant?

Answer 9

Yeah, \(\det(A)\) is the determinant of \(A\), or \(|A|\).

Answer 10

So how do I prove they're equal?

Answer 11

Let's start with some examples to get your mind wrapped around transpose and cofactors.

Suppose \(A=\begin{pmatrix}1&2&1\\3&0&-1\\-1&-1&1\end{pmatrix}\).

The transpose of \(A\), denoted \(A^T\), is the matrix you get if you were to reflect all the elements across the left-to-right/top-to-bottom diagonal:
\[A^T=\begin{pmatrix}1&3&-1\\2&0&-1\\1&-1&1\end{pmatrix}\]
Got it?

Answer 12

Yes, yes I got it. :)

Answer 13

Okay. Cofactors are bit trickier to understand, but it's all a matter of computation.

The cofactor matrix \(C\) is the matrix of the cofactors of \(A\). A cofactor of the \(i,j\)-th element is the determinant of the submatrix of \(A\) that you get from deleting the \(i\)th row and \(j\)th column, multiplied by a power of -1.

That's a lot to take in at once, so here's an example. Let's find the cofactor of the (1,1) element (entry in first row, first column).
\[C_{(1,1)}=(-1)^{1+1}\begin{vmatrix}0&-1\\-1&1\end{vmatrix}=1(0\times1-(-1)\times(-1))=-1\]
As you should be able to see, the submatrix \(\begin{pmatrix}0&-1\\-1&1\end{pmatrix}\) is obtained by deleting the first row (1,3,-1) and first column (1,2,1). The power of (-1) depends on the entry's location in the original matrix; it's given by \((-1)^{i+j}\). In this case, \(i=1\) and \(j=1\), so \((-1)^{1+1}=1\).

Is any of this unclear?

Answer 14

Wait, I'll try to break it down and tell me if I'm wrong.

Answer 15

So is the submatrix always the ones you get after you deleted the first row and column?

Answer 16

This particular submatrix is the one you get for deleting the first row and column because it's the submatrix associated with the element in the first row, first column slot.

If we wanted the cofactor of the 2nd row, 2nd column entry (the 0 in the middle), we would delete the 2nd row (2,0,-1) and the 2nd column (2,0,-1), which gives the submatrix \(\begin{pmatrix}1&1\\-1&1\end{pmatrix}\).

Answer 17

Sorry, I meant "... 2nd column (3,0,-1) ..."

Answer 18

Hey man, sorry to ditch you after all this explanation but I gotta do something. This is really helpful though this would catch me up a lot in my math class. I'm gonna read on this later. Thanks! I'm a fan!

Answer 19

No problem. The actual work of computing \(A\cdot\text{adj}(A)=|A|\) is pretty tedious, but if you need help you can try asking me later.