Question

Without expanding evaluate the determinant:

Answer 1

\[\LARGE \left|\begin{matrix}a^2 &a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\end{matrix}\right|\]

Answer 2

Ya you told me trick , it slipped of my mind :( ,

Answer 3

the*

Answer 4

Also, here's some extra info in case you were wondering. \[\Large a \ne b \ne c\] if any of them are equal the whole determinant is obviously 0 haha.

Answer 5

first observation is
\[

\large \left|\begin{matrix}a^2 &a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\end{matrix}\right| = F(a,b,c)

\]

3rd degree function

Answer 6

like this ?
\(det \left ( \begin{bmatrix}
a^2 & a & 1\\
b^2 & b& 1\\
c^2& c & 1
\end{bmatrix} \right ) = det \left ( a^2\begin{bmatrix}
b& 1\\
c&1
\end{bmatrix}-a \begin{bmatrix}
b^2& 1\\
c^2&1
\end{bmatrix}+ \begin{bmatrix}
b^2& b\\
c^2&c
\end{bmatrix} \right ) \)

then what ?

Answer 7

Yeah you're not allowed to do that @ikram002p heh =P

Answer 8

ohh lol without expantion hehe i thought with expantion and said wth xD

Answer 9

well
in case
c= n (a)
b=m (a)


this might be weird xD

Answer 10

\(a=b\) makes \(F(a,b,c) = 0\)
so \((a-b)\) is a factor of \(F(a,b,c)\) :

\[F(a,b,c) = (a-b)(something)\]

Answer 11

That's on the right path I think ganeshie! I got to that result by a different method though so I am really surprised, very cool!

Answer 12

wait :o
how this matrix relatied to this
\(\begin{bmatrix}
1& 1 &1 \\
a&b &c \\
a^2& b^2 & c^2
\end{bmatrix} \)
?

Answer 13

I dunno @ikram002p lol. That's like the backwards transpose (is that a thing?) idk lol

Answer 14

lol i dont know as well :P

Answer 15

Kainui, il conclude :
similar reasoning yields (b-c) and (c-a) as other facors,
we stop at this point cuz it is clear that F is a 3rd degree function

So F(a,b,c) = k(a-b)(b-c)(c-a)

Answer 16

k has to be 1 for obvious reasons.
cant wait to see the another method of solving this xD

Answer 17

You almost got it, but I think you're wrong by a negative sign!

What I did was subtract the second row from the first and factor out a (a-b)

d=k(a-b)(b-c)(c-a)

now we know that the diagonal is one of the elements of the determinant, a^2b so if we look in our determinant we have -ka^2b so then k=-1.

Answer 18

:o

Answer 19

\[\LARGE \left|\begin{matrix}a^2-b^2 & a-b & 0\\ b^2 & b &1 \\ c^2 & c &1\end{matrix}\right|\]\[\LARGE (a-b)\left|\begin{matrix}a+b & 1 & 0\\ b^2 & b &1 \\ c^2 & c &1\end{matrix}\right|\]

Answer 20

Step by step answer
http://www.mathskey.com/question2answer/16731/without-expanding-evaluate-the-determinant?show=16733

Answer 21

@bradely you killed the fun my friend lol
following this guys how they were approaching the problem was interesting

Answer 22

ill sighn in in that site :P