Having trouble solving using properties of determinants ....

In summary: 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...The 2x2 determinant would be (a2 - b2)(b - c) - (a - b)(b2 - c2).
  • #1
VoteSaxon
25
1

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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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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Likes VoteSaxon
  • #9
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
[tex] \left| \begin{array}{ccc}
a^2 - b^2 & b^2 - c^2 & c^2 \\
a - b & b - c & c \\
0 & 0 & 1
\end{array} \right|
[/tex]

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
[tex] \left| \begin{array}{ccc}
a^2 - b^2 & b^2 - c^2 & c^2 \\
a - b & b - c & c \\
0 & 0 & 1
\end{array} \right|
[/tex]

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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Likes VoteSaxon
  • #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.
 

FAQ: Having trouble solving using properties of determinants ....

What are determinants and why are they important for solving problems?

Determinants are mathematical values that can be calculated from matrices. They are important for solving problems because they can provide valuable information about the properties and behavior of the matrix, such as its invertibility and solutions to systems of linear equations.

What are some common properties of determinants that can be used to solve problems?

Some common properties of determinants include the fact that the determinant of a product of matrices is equal to the product of their determinants, and that the determinant of the transpose of a matrix is equal to the determinant of the original matrix. These properties can be useful in simplifying calculations and solving equations.

How can I use determinants to solve systems of linear equations?

Determinants can be used to solve systems of linear equations by setting up a matrix of coefficients and a matrix of constants, and then using the determinant to determine whether the system has a unique solution, no solution, or infinitely many solutions. The determinant can also be used to find the values of the variables in the solution.

What are some tips for solving problems using properties of determinants?

Some tips for solving problems using properties of determinants include understanding the basic properties and how they can be applied, breaking down complex matrices into smaller parts to make calculations easier, and practicing with various examples to become more familiar with the process.

Are there any limitations to using properties of determinants to solve problems?

While properties of determinants can be powerful tools for solving problems, there are some limitations to consider. For example, determinants can only be calculated for square matrices, and using determinants to solve systems of equations can become more complex with larger matrices. Additionally, determinants may not always provide a unique solution to a problem and may require additional methods to fully solve.

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