This shouldn't be too dfficult

 

I need to prove (e.g.) ;

 

[math]

b_1A_1 + b_2A_2 + b_3A_3 = 0

 

[/math]

 

Ive managed to get;

 

[math]

 

a_1a_3b_2 + a_1a_2b_3 - a_2a_3b_1 + a_2b_3a_1 - a_3a_1b_2 +a_3a_2b_1

 

[/math]

 

which nicely cancels out with the exception of 2 terms; giving;

 

[math]

 

2a_1a_2b_3

 

[/math]

 

Can anybody see where I've gone wrong?

What are the small and capital letters with the indices?

With a 3*3 matrix you have;

[math]

 

\left( \begin{array}{ccc}

a_{1} & b_{1} & c_{1} \\

a_{2} & b_{2} & c_{2} \\

a_{3} & b_{3} & c_{3}

\end{array} \right)

 

 

[/math]

 

then

 

[math]

A_1 =\left( \begin{array}{ccc}

b_{2} & c_{2} \\

b_{3} & c_{3} \\

\end{array} \right)

[/math]

 

i.e with a_1 row and column removed.

 

Remember you need to introduce the signs;

 

[math]

\left( \begin{array}{ccc}

+ & - & +\\

- & + & - \\

+ & - & +

\end{array} \right)

 

[/math]

Let's assume not everyone uses exactly the same notation as you do in school: There is no chance to figure out what you are trying to do and say. Slow it down, try explaining what you are talking about from the very start assuming none of the notations is obvious to the reader.

 

As a total shot in the dark: If you are trying to prove that the determinant of a matrix is zero, then this is not the case in general.

Ok sorry.

 

So what I'm doing is multiplying the elements of the second column by the co-factors of the first column- this the same as doing a cyclic interchange of the columns which we know from the rules of matricies = 0.

 

So what I've done is multiply the second column (b's) by the cofactor of the first column (A) yielding the second line of my original post.

 

However we need to make sure that we attatch the correct signs to each minor (i.e. what is shown in the middle line of my last post) to yield the correct cofactor which is what I have shown in the bottom of my last post.

 

OK I a cyclic interchange leaves the determinant unchanged so ignore that bit. And also A_1 should be the determinant of the matrix I've shown not just the matrix.

A_1 should be the determinant of the matrix I've shown not just the matrix.

That already clears up quite a bit.

 

OK I a cyclic interchange leaves the determinant unchanged so ignore that bit.

Is that what you want to show? That a cyclic interchange leaves the determinant unchanged?

 

 

So what I'm doing is multiplying the elements of the second column by the co-factors of the first column- this the same as doing a cyclic interchange of the columns which we know from the rules of matricies = 0.

What =0 ? Surely not the matrix.

The first line of working should = 0.

 

This has something to do with cyclic interchanges but I dont think that is relevent and probaby just confused things by mentioning it.