Having trouble solving using properties of determinants ....

AI Thread Summary
The discussion revolves around solving a determinant problem where the user initially struggles with manipulating the matrix. They express confusion about the usefulness of a row of '1's and the resulting expressions from their calculations. Participants suggest using properties of determinants, such as column operations and recognizing when rows are multiples of each other, to simplify the problem. The user eventually realizes the importance of factoring expressions like a² - b² and b² - c² to find a solution. Ultimately, they express gratitude for the guidance received, indicating they have made progress in understanding the problem.
VoteSaxon
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Homework Statement


dY0tvZ6.png

I'm a bit at a loss - I thought the last row with '1's would be useful, but it just gave me:
(b2c - bc2) - (a2c - ac2) + (a2b - ab2)
and
bc(b - c) - ac(a - c) + ab(a - b)

But then it is a dead end. I am probably doing something stupid again ...

Any help appreciated.
 
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I will give you a couple hints. First, what are the properties of determinants that you might possibly choose?

For example: What will you get if a = b? In particular, notice what happens to the columns when a = b. What does that tell you about the form of the determinant?

For example: What do you get if you swap the first and second columns? That is, what do you get if you swap a for b? What does that tell you about the form of the determinant?

For example: What do you get if you multiply each element of the matrix by a constant K? What does that tell you about the form of the determinant?
 
Remember you can add or subtract a column from an other one. With that, you can make all but one element in a row equal to zero.
 
VoteSaxon said:

Homework Statement


dY0tvZ6.png

I'm a bit at a loss - I thought the last row with '1's would be useful, but it just gave me:
(b2c - bc2) - (a2c - ac2) + (a2b - ab2)
and
bc(b - c) - ac(a - c) + ab(a - b)

But then it is a dead end. I am probably doing something stupid again ...

Any help appreciated.

In addition to the hint in #3, remember that you do not change the value of the determinant when you add a multiple of one row onto another row, or a multiple of one column onto another column.
 
Well thanks guys, I have managed to get rid of the bottom row with your methods. Again, not too sure where to go from there. As the first row is kind of a multiple of the second row (R1 = R22) I thought I could go from there, but it didn't seem to pan out. Am I on the right track, or have I completely misunderstood?

Again, thanks a ton.
 
VoteSaxon said:
Well thanks guys, I have managed to get rid of the bottom row with your methods.
Then you have probably made a mistake. If a row or column of a square matrix consists only of 0 entries, its determinant is 0. From the problem statement, the determinant is zero only under certain conditions. It would be helpful if you showed what you did to get rid of the bottom row.
VoteSaxon said:
Again, not too sure where to go from there. As the first row is kind of a multiple of the second row (R1 = R22) I thought I could go from there, but it didn't seem to pan out. Am I on the right track, or have I completely misunderstood?

Again, thanks a ton.
 
Mark44 said:
Then you have probably made a mistake. If a row or column of a square matrix consists only of 0 entries, its determinant is 0. From the problem statement, the determinant is zero only under certain conditions. It would be helpful if you showed what you did to get rid of the bottom row.
I basically subtracted the columns from each other until the row of 1s disappeared.
z9sAg02.jpg

sigh... this is really wrong isn't it?
 
VoteSaxon said:
sigh... this is really wrong isn't it?
The first two determinants look OK, but not the third. If you subtract C1 from C3, which is what your notation says, you don't get 0 in the lower right corner.
 
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VoteSaxon said:
I basically subtracted the columns from each other until the row of 1s disappeared.
z9sAg02.jpg

sigh... this is really wrong isn't it?

You went too far: you could stop at the second line and evaluate the determiant
\left| \begin{array}{ccc}<br /> a^2 - b^2 &amp; b^2 - c^2 &amp; c^2 \\<br /> a - b &amp; b - c &amp; c \\<br /> 0 &amp; 0 &amp; 1 <br /> \end{array} \right|<br />

If you "expand" this along the 3rd column, you just end up needing to evaluate a ##2 \times 2## determinant, which is pretty easy.
 
  • #10
Ray Vickson said:
You went too far: you could stop at the second line and evaluate the determiant
\left| \begin{array}{ccc}<br /> a^2 - b^2 &amp; b^2 - c^2 &amp; c^2 \\<br /> a - b &amp; b - c &amp; c \\<br /> 0 &amp; 0 &amp; 1<br /> \end{array} \right|<br />

If you "expand" this along the 3rd column, you just end up needing to evaluate a ##2 \times 2## determinant, which is pretty easy.

Wouldn't the 2x2 determinant be (a2 - b2)(b - c) - (a - b)(b2 - c2)? Because I do not think this gives the correct answer - it gives me (a2 + bc)(b - c). Maybe I could manipulate this somehow to give me what the answer is looking for ...
 
  • #11
VoteSaxon said:
Wouldn't the 2x2 determinant be (a2 - b2)(b - c) - (a - b)(b2 - c2)? Because I do not think this gives the correct answer - it gives me (a2 + bc)(b - c). Maybe I could manipulate this somehow to give me what the answer is looking for ...

Sure: just remember that ##a^2-b^2 = (a-b)(a+b)## and ##b^2 - c^2 = (b-c)(b+c)##, and pull out common factors.
 
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  • #12
Ray Vickson said:
Sure: just remember that ##a^2-b^2 = (a-b)(a+b)## and ##b^2 - c^2 = (b-c)(b+c)##, and pull out common factors.
YES!
Thank you! I think I got it!
bxRfFhO.jpg


Really appreciate your patience and cleverness guys! Keep up the good work.
 
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