Determinant Prove

The yz at the end should be inside the brackets.

Then I'd expand inside; there will be four terms, so factoring by grouping might be a good idea.

Another way you could have done this is just to fully expand the determinant as a sum of products, and then do the same to the right-hand side of what you are to prove. If the results are equal, you're done.

But your way is good factoring practice!
 
Dr.Peterson said:
The yz at the end should be inside the brackets.

Then I'd expand inside; there will be four terms, so factoring by grouping might be a good idea.

Another way you could have done this is just to fully expand the determinant as a sum of products, and then do the same to the right-hand side of what you are to prove. If the results are equal, you're done.

But your way is good factoring practice!
Click to expand...

Thank you. I will try this tomorrow morning after the museum. Going to work now.
 
Dr.Peterson said:
The yz at the end should be inside the brackets.

Then I'd expand inside; there will be four terms, so factoring by grouping might be a good idea.

Another way you could have done this is just to fully expand the determinant as a sum of products, and then do the same to the right-hand side of what you are to prove. If the results are equal, you're done.

But your way is good factoring practice!
Click to expand...

I really tried to complete this prove on paper but to no avail. Can you please show me the complete prove?
 
harpazo said:
I really tried to complete this prove on paper but to no avail. Can you please show me the complete prove?
Click to expand...
If I were to do this, I would expand along the third column.
Then the LHS becomes \(\displaystyle zy^2-xy^2+xz^2-yz^2+yx^2-zx^2\)
Expand the RHS: \(\displaystyle x^2y-xy^2-xyz+y^2z-x^2z+xyz+xz^2-yz^2\)
By inserting \(\displaystyle \pm xyz\) into the LHS & rearranging you should see the equality.
 
pka said:
If I were to do this, I would expand along the third column.
Then the LHS becomes \(\displaystyle zy^2-xy^2+xz^2-yz^2+yx^2-zx^2\)
Expand the RHS: \(\displaystyle x^2y-xy^2-xyz+y^2z-x^2z+xyz+xz^2-yz^2\)
By inserting \(\displaystyle \pm xyz\) into the LHS & rearranging you should see the equality.
Click to expand...

I'll try it again.
 
If you know a bit about determinant properties, there is a nice way using them. Consider the determinant:

[MATH] \large P(x) = \left | \begin{array}{ccc} x^2 & x & 1\\ a^2 & a & 1\\ b^2 & b & 1 \end{array} \right | [/MATH]
where I have used [MATH]a[/MATH] and [MATH]b[/MATH] instead of [MATH]y[/MATH] and [MATH]z[/MATH] and called it [MATH]P(x)[/MATH]. The first two entries in column 1 are [MATH]x^2[/MATH] and [MATH]a^2[/MATH] but they are displaying incorrectly on my screen. I don't know why and maybe one of the mods can correct it. Anyway, expanding by the first row we know [MATH]P(x)[/MATH] is a polynomial of degree [MATH]2[/MATH] with leading coefficient the cofactor of [MATH]x^2[/MATH]. Also note that [MATH]P(a) = P(b) = 0[/MATH]. Now use the factor theorem on [MATH]P(x)[/MATH].
 
Dr.Peterson said:
The yz at the end should be inside the brackets.

Then I'd expand inside; there will be four terms, so factoring by grouping might be a good idea.

Another way you could have done this is just to fully expand the determinant as a sum of products, and then do the same to the right-hand side of what you are to prove. If the results are equal, you're done.

But your way is good factoring practice!
Click to expand...


D = (y - z)[x^2 - x(y + z) + yz]

D = (y - z)[x^2 - xy - xz + yz]

D = (y - z)[x(x - y) - z(x + y)]

D = (y - z)(x - z)(x - y)(x + z)

I got 4 terms after completing the square.

What did I do wrong?
 
Yes, since the original has degree 3, the factored form can't have 4 factors.

It looks like you tried to factor x^2 - xy - xz + yz by grouping, but fell into a common error. When you get x(x - y) - z(x + y), in which the two binomials are not equal, you can't continue; your next line's (x - z)(x - y)(x + z), is not equal to this. Either the grouping has to be changed, or you made a mistake somewhere, or it just can't be factored by this method.

So look carefully at your work, checking that the signs are right from each line to the next. You'll find that the third line should have had x(x - y) - z(x - y), so there are like binomials and the next line will have (x - z)(x - y), without (x + z).
 
Dr.Peterson said:
Yes, since the original has degree 3, the factored form can't have 4 factors.

It looks like you tried to factor x^2 - xy - xz + yz by grouping, but fell into a common error. When you get x(x - y) - z(x + y), in which the two binomials are not equal, you can't continue; your next line's (x - z)(x - y)(x + z), is not equal to this. Either the grouping has to be changed, or you made a mistake somewhere, or it just can't be factored by this method.

So look carefully at your work, checking that the signs are right from each line to the next. You'll find that the third line should have had x(x - y) - z(x - y), so there are like binomials and the next line will have (x - z)(x - y), without (x + z).
Click to expand...

Ok. I will go back to the start. I never give up.
 
Dr.Peterson said:
Yes, since the original has degree 3, the factored form can't have 4 factors.

It looks like you tried to factor x^2 - xy - xz + yz by grouping, but fell into a common error. When you get x(x - y) - z(x + y), in which the two binomials are not equal, you can't continue; your next line's (x - z)(x - y)(x + z), is not equal to this. Either the grouping has to be changed, or you made a mistake somewhere, or it just can't be factored by this method.

So look carefully at your work, checking that the signs are right from each line to the next. You'll find that the third line should have had x(x - y) - z(x - y), so there are like binomials and the next line will have (x - z)(x - y), without (x + z).
Click to expand...

See picture for solution.

20190928_120554.jpg
 
The next to last line is wrong. The common factor of the two terms within the brackets is (x-y), so you factor that out once.

Then in the last line, you dropped the (x-z), rather than the extra (x-y).

I'm hoping that you just didn't write what you meant.
 
Dr.Peterson said:
The next to last line is wrong. The common factor of the two terms within the brackets is (x-y), so you factor that out once.

Then in the last line, you dropped the (x-z), rather than the extra (x-y).

I'm hoping that you just didn't write what you meant.
Click to expand...

In that case, can you complete the prove for me?
 
I also think you need a little rest - may be a little shut-eye - then press-on!!
 
More rest and sleep is needed. I will leave this for my next day off.
 
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