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Difficult matrix question (3*3) fp3

Show that, for all real values of x, the matrix:

(2 -2 4)
(3 x -2)
(-1 3 x)

is non-singular.

Great from here I get:

2x^2 +10x +44.

What to do from here,
Reply 1
Avatar for math42
math42
20
Original post by Mihael_Keehl
Show that, for all real values of x, the matrix:

(2 -2 4)
(3 x -2)
(-1 3 x)

is non-singular.

Great from here I get:

2x^2 +10x +44.

What to do from here,


If a quadratic has no real roots then it never equals zero
Reply 2
Avatar for joostan
joostan
13
Original post by Mihael_Keehl
Show that, for all real values of x, the matrix:

(2 -2 4)
(3 x -2)
(-1 3 x)

is non-singular.

Great from here I get:

2x^2 +10x +44.

What to do from here,


I've not checked your computation of the determinant, but assuming it's correct, and that your matrix is real, you can compute the discriminant of the quadratic.
Original post by 1 8 13 20 42
If a quadratic has no real roots then it never equals zero


Original post by joostan
I've not checked your computation of the determinant, but assuming it's correct, and that your matrix is real, you can compute the discriminant of the quadratic.


Do you guys mean make the quadratic a term that yield positive answer i.e. the value being squared and so it will be always greater than zero?

Does finding the discriminant actually help here, I did it but it got me confused whehter it was the right thing to do ;P
Reply 4
Avatar for Zacken
Zacken
22
Original post by Mihael_Keehl
Do you guys mean make the quadratic a term that yield positive answer i.e. the value being squared and so it will be always greater than zero?

Does finding the discriminant actually help here, I did it but it got me confused whehter it was the right thing to do ;P


If you show that the discriminant is Δ<0\Delta < 0 then you're done. Because that means the quadratic then always lies above the xx-axis.

You could also just work off 2(x+52)2−252+442(x + \frac{5}{2})^2 - \frac{25}{2} + 44, as you said.
(edited 8 years ago)
Original post by Zacken
If you show that the discriminant is Δ<0\Delta < 0 then you're done. Because that means the quadratic then always lies above the xx-axis.

You could also just work off 2(x−5x2)2−252+442(x - \frac{5x}{2})^2 - \frac{25}{2} + 44, as you said.


Thanks for the reply.

What does that triangle mean?!

Also isn't it:

2(x+5x2)2−3122(x + \frac{5x}{2})^2 - \frac{31}{2} ?

What do I do from here, going afk though will respond tomorrow :biggrin:
Reply 6
Avatar for Kummer
Kummer
7
Original post by mihael_keehl
show that, for all real values of x, the matrix:

(2 -2 4)
(3 x -2)
(-1 3 x)

is non-singular.

Great from here i get:

2x^2 +10x +44.

What to do from here,
Row reduction gives the identity matrix, therefore it has a rank of 3, thus non-singular.
Reply 7
Avatar for Zacken
Zacken
22
Original post by Mihael_Keehl
Thanks for the reply.

What does that triangle mean?!

Also isn't it:

2(x+5x2)2−3122(x + \frac{5x}{2})^2 - \frac{31}{2} ?

What do I do from here, going afk though will respond tomorrow :biggrin:


Discriminant.

And, it's not exactly what you wrote, but close enough - see my edited post for the correction .

If you use the determinant, showing that it is less than 0 means you are done.

If you use the other method, then you can state: hence the quadratic is always positive and you are done.
Original post by Zacken
Discriminant.

And, it's not exactly what you wrote, but close enough - see my edited post for the correction .

If you use the determinant, showing that it is less than 0 means you are done.

If you use the other method, then you can state: hence the quadratic is always positive and you are done.


Thank you have solved it all now, so you would say that either method is acceptable.

Which would you say is your preferred method?
Reply 9
Avatar for Zacken
Zacken
22
Original post by Mihael_Keehl
Thank you have solved it all now, so you would say that either method is acceptable.

Which would you say is your preferred method?


I'd go with the determinant method.
Original post by Zacken
I'd go with the determinant method.


Thank you.

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