Finding the determinant of a matrix using determinant properties

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Yatayyat
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Hi! I have been having some trouble in finding the determinant of matrix A in this Q

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Which relevant determinant property should I make use of to help me find the determinant of matrix A and maybe matrix B also

This is what I have tried for matrix A (using addition property) so far but it's not much help really

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Any help would be great! Thanks
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Muttley79
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(Original post by Yatayyat)
Hi! I have been having some trouble in finding the determinant of matrix A in this Q

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Views: 13
Size:  66.4 KB

Which relevant determinant property should I make use of to help me find the determinant of matrix A and maybe matrix B also

This is what I have tried for matrix A (using addition property) so far but it's not much help really


Any help would be great! Thanks
Do you need that for the first one? Just do it 'normally' the algebra is simple.
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ghostwalker
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(Original post by Yatayyat)
Hi! I have been having some trouble in finding the determinant of matrix A in this Q

Name:  Screenshot 2020-09-25 at 14.24.12.png
Views: 13
Size:  66.4 KB

Which relevant determinant property should I make use of to help me find the determinant of matrix A and maybe matrix B also

This is what I have tried for matrix A (using addition property) so far but it's not much help really

Name:  IMG_5289.jpg
Views: 12
Size:  482.6 KB

Any help would be great! Thanks

Not the best method to use to start, IMO, but it does lead to an elegant solution. Try flipping the rows/columns of the 2nd determinant so that it looks like the first.
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DFranklin
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The key rule is that adding a multiple of a row to a different row doesn't change the determinant.

A corollary is that if two rows are a multiple of each other the determinant is 0. (because you could use the first rule to reduce one row to all zeroes, and then expand along that row to evaluate the determinant).

This sorts the first matrix out very quickly. 2nd one looks a bit more tricky.
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DFranklin
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(Original post by ghostwalker)
Not the best method to use to start, IMO, but it does lead to an elegant solution. Try flipping the rows/columns of the 2nd determinant so that it looks like the first.
Any luck with B? Blasting the algebra gives a "nice" expression, but a nice expression that feels like it would be hard to get by fiddling with row-column ops.
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ghostwalker
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(Original post by DFranklin)
Any luck with B? Blasting the algebra gives a "nice" expression, but a nice expression that feels like it would be hard to get by fiddling with row-column ops.
Add rows 3 and 2 to 1, take out the factor, then blast.
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Yatayyat
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(Original post by ghostwalker)
Not the best method to use to start, IMO, but it does lead to an elegant solution. Try flipping the rows/columns of the 2nd determinant so that it looks like the first.
How does making the 2nd determinant look like the 1st determinant (via swapping rows/columns around) help if we don't know how to work out the determinant of the 1st?
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MarkFromWales
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(Original post by Yatayyat)
How does making the 2nd determinant look like the 1st determinant (via swapping rows/columns around) help if we don't know how to work out the determinant of the 1st?
We established that the first determinant is 0.
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Yatayyat
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(Original post by MarkFromWales)
We established that the first determinant is 0.
Im confused how did you figure it was 0? by which property?
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ghostwalker
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(Original post by Yatayyat)
How does making the 2nd determinant look like the 1st determinant (via swapping rows/columns around) help if we don't know how to work out the determinant of the 1st?
If you've done it correctly you don't need to work out any determinants.
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DFranklin
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(Original post by ghostwalker)
Add rows 3 and 2 to 1, take out the factor, then blast.
PRSOM. Feels it's fairly spoiler free to say that:

It does factor as (x+y)(x+\omega y)(x+ \omega^2y) for the typical value of omega, so I suspect there's some "pure" method if you're sufficiently sneaky.
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MarkFromWales
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(Original post by Yatayyat)
Im confused how did you figure it was 0? by which property?
See post #4.
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Yatayyat
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(Original post by ghostwalker)
If you've done it correctly you don't need to work out any determinants.
DFranklin suggested that I somehow add a multiple of a row to a different row. From that do I need to somehow form 2 equivalent rows which then means the determinant is zero?
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MarkFromWales
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(Original post by Yatayyat)
DFranklin suggested that I somehow add a multiple of a row to a different row. From that do I need to somehow form 2 equivalent rows which then means the determinant is zero?
Here's how I'd do it.
Add row 2 to row 1.
Multiply row 3 by x+y+z.
So now rows 1 and 3 are identical.
Subtract row 3 from row 1 to get a zero row.
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ghostwalker
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(Original post by DFranklin)
PRSOM. Feels it's fairly spoiler free to say that:

It does factor as (x+y)(x+\omega y)(x+ \omega^2y) for the typical value of omega, so I suspect there's some "pure" method if you're sufficiently sneaky.
Considered as a polynomial is x or y perhaps, though it would be difficult to spot (for me at least).

Adding to my prev. suggestion:
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we can then reduce a row to 1 0 0 , and no blasting required.
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DFranklin
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(Original post by ghostwalker)
Considered as a polynomial is x or y perhaps, though it would be difficult to spot (for me at least).

Adding to my prev. suggestion:
Spoiler:
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we can then reduce a row to 1 0 0 , and no blasting required.
I found "just expand the original matrix" the easiest approach by quite a long way; to be honest I gave up on reducing a row to 1 0 0, although that's somewhat limitations of trying to solve based solely on what I can type quickly into the TSR reply window. As for the factorizing, I'd seen it somewhere before I think.
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Yatayyat
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(Original post by MarkFromWales)
Here's how I'd do it.
Add row 2 to row 1.
Multiply row 3 by x+y+z.
So now rows 1 and 3 are identical.
Subtract row 3 from row 1 to get a zero row.
Thank you this makes sense to me from what I followed from your steps

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I could even skip the last step because it's a known fact that 2 equal rows would give a zero determinant from what I saw in my notes, but making a row filled with zero elements would also give a zero determinant

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MarkFromWales
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(Original post by Yatayyat)
Thank you this makes sense to me from what I followed from your steps

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I could even skip the last step because it's a known fact that 2 equal rows would give a zero determinant from what I saw in my notes, but making a row filled with zero elements would also give a zero determinant

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Yes, that's correct.
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ghostwalker
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(Original post by DFranklin)
I found "just expand the original matrix" the easiest approach by quite a long way; to be honest I gave up on reducing a row to 1 0 0, although that's somewhat limitations of trying to solve based solely on what I can type quickly into the TSR reply window. As for the factorizing, I'd seen it somewhere before I think.
Unless I'm mistaken, this whole thing reduces to
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2(x^3+y^3)



I worked from my reduction, rather than yours.
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DFranklin
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(Original post by ghostwalker)
Unless I'm mistaken, this whole thing reduces to
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2(x^3+y^3)


I worked from my reduction, rather than yours.
Agreed. (What I posted is off by a factor of 2, but that's because I was thinking about "how to factor the bit that's difficult to factor" when I posted it, not because I got a different determinant).
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