# Math Help - Matrix Row Opterations problem

1. ## Matrix Row Opterations problem

Hi,

Im normally ok with these sort of questions but this one has me stumped!

I need to show using row operations that the following determinant = 0
i.e.:

| (y-z) (z-x) (x-y) |
| (z-x) (x-y) (y-z) | = 0
| (x-y) (y-z) (z-x) |

I tried doing it the usual way as you would if the matrix contained number instead of variables, but it ended up becoming very messy and I couldnt help but think there was a better way of doing it.

the horizontal lines :

|
|
|

are supposed be the determinant notation.

2. Each equation (or line of the matrix) turns out to be just zero. The determinant of a zero matrix is zero.

3. hi colby2152,

thanks for the reply. Do you think it will be enough to write that as the answer considering the question states to 'show using row operations'?

Thanks again.

4. Originally Posted by peterpan
hi colby2152,

thanks for the reply. Do you think it will be enough to write that as the answer considering the question states to 'show using row operations'?

Thanks again.
Multiply each row by one and simplify. You could also add up all the rows. Are those not row operations?

5. I guess they are I didnt realise that you could just simplify the rows in that way (multiplying by one and simplifying). Why is it that you can do that?

6. Originally Posted by peterpan
I guess they are I didnt realise that you could just simplify the rows in that way (multiplying by one and simplifying). Why is it that you can do that?
$1*((y-z)+(z-x)+(x-y) )= (y-z+z-x+x-y) \Rightarrow 0$

Is this true?

7. Yes that is certainly true and I understand the mechanics. What I am looking for is the intuition of why you can actually do that simplification to a row of a matrix.

8. Originally Posted by peterpan
Yes that is certainly true and I understand the mechanics. What I am looking for is the intuition of why you can actually do that simplification to a row of a matrix.
The row of a matrix is just an algebraic equation:

[x y z | 3]

is the same as:

$x+y+z=3$

Whatever you can do to algebraic equation, you can do to a matrix.

9. However in this case we do not have the last column (where you have 3 in the last example) i.e. its not an augmented matrix, the equations are not 'equal' to anything, so how do we justify the simplification?

Thanks for the help

10. Originally Posted by peterpan
However in this case we do not have the last column (where you have 3 in the last example) i.e. its not an augmented matrix, the equations are not 'equal' to anything, so how do we justify the simplification?

Thanks for the help
You are correct, yet the equations can be simplified to zero still. You have this given:

(y-z) (z-x) (x-y) |
| (z-x) (x-y) (y-z) |
| (x-y) (y-z) (z-x) |

In algebra, this looks like:

$(y-z)+(z-x)+(x-y)$
$(z-x)+(x-y)+(y-z)$
$(x-y)+(y-z)+(z-x)$

These equations all simplify to zero.