(New page: '''''Statement: I am going to derive through a series of statements that transposing a matrix does NOT change its determinant.''''' First we will start with a 2x2 matrix as follows: '''...) |
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c & d \end{bmatrix}</math> | c & d \end{bmatrix}</math> | ||
− | So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the | + | So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the transpose). |
A^T= | A^T= | ||
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Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough... | Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough... | ||
+ | |||
+ | Now, we will use the power of induction to make some powerful assumptions, which will be proven in a bit. | ||
+ | |||
+ | Lets assume this is true for all cases of nxn... | ||
+ | |||
+ | So now assume we have a nxn matrix called B: | ||
+ | |||
+ | Then we can say that det(B)=det(B^T) | ||
+ | |||
+ | |||
+ | '''''I now propose a question that is food for thought for the rest of this derivation. | ||
+ | |||
+ | Question: Is it true for all (n+1)x(n+1) matrices as well??''''' | ||
+ | |||
+ | We shall now see... | ||
+ | |||
+ | Let a (n+1)x(n+1) matrix called A= | ||
+ | |||
+ | <math>\begin{bmatrix} | ||
+ | a11 & a12 & a13 & a1m \\ | ||
+ | a21 & a22 & a23 & a2m \\ | ||
+ | a31 & a32 & a33 & a3m \\ | ||
+ | am1 & am2 & am3 & amm \\ | ||
+ | |||
+ | \end{bmatrix}</math> |
Revision as of 11:00, 7 December 2010
Statement: I am going to derive through a series of statements that transposing a matrix does NOT change its determinant.
First we will start with a 2x2 matrix as follows:
Let the 2x2 matrix A=
$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $
So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the transpose).
A^T=
$ \begin{bmatrix} a & c \\ b & d \end{bmatrix} $
So, det(A^T)=ad-cb.
Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough...
Now, we will use the power of induction to make some powerful assumptions, which will be proven in a bit.
Lets assume this is true for all cases of nxn...
So now assume we have a nxn matrix called B:
Then we can say that det(B)=det(B^T)
I now propose a question that is food for thought for the rest of this derivation.
Question: Is it true for all (n+1)x(n+1) matrices as well??
We shall now see...
Let a (n+1)x(n+1) matrix called A=
$ \begin{bmatrix} a11 & a12 & a13 & a1m \\ a21 & a22 & a23 & a2m \\ a31 & a32 & a33 & a3m \\ am1 & am2 & am3 & amm \\ \end{bmatrix} $