Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - determinant problem

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    26

    determinant problem

    Assuming that det (AB)= det(A)det(B) for 3 x 3 matrices, prove that

    | 3 a+b+c a^3+b^3+c^3 |
    | a+b+c a^2+b^2+c^2 a^4+b^4+c^4 |
    | a^2+b^2+c^2 a^3+b^3+c^3 a^5+b^5+c^5 |


    is equal to (a+b+c) (a-b)^2 (a-c)^2 (b-c)^2

    I do this and get 0 which doesnt help me at all
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    A = \left[ {\begin{array}{rrr}<br />
   1 & 1 & 1  \\<br />
   a & b & c  \\<br />
   {a^2 } & {b^2 } & {c^2 }  \\<br />
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}<br />
   1 & a & {a^3 }  \\<br />
   1 & b & {b^3 }  \\<br />
   1 & c & {c^3 }  \\<br />
\end{array}} \right]

    Your matrix is the product of AB.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by Plato View Post
    A = \left[ {\begin{array}{rrr}<br />
   1 & 1 & 1  \\<br />
   a & b & c  \\<br />
   {a^2 } & {b^2 } & {c^2 }  \\<br />
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}<br />
   1 & a & {a^3 }  \\<br />
   1 & b & {b^3 }  \\<br />
   1 & c & {c^3 }  \\<br />
\end{array}} \right]

    Your matrix is the product of AB.
    How on Earth did you factor that??

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    Quote Originally Posted by Plato View Post
    A = \left[ {\begin{array}{rrr}<br />
   1 & 1 & 1  \\<br />
   a & b & c  \\<br />
   {a^2 } & {b^2 } & {c^2 }  \\<br />
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}<br />
   1 & a & {a^3 }  \\<br />
   1 & b & {b^3 }  \\<br />
   1 & c & {c^3 }  \\<br />
\end{array}} \right]

    Your matrix is the product of AB.
    but how does that help me prove that its equal to (a+b+c) (a-b)^2 (a-c)^2 (b-c)^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    Find det(A). Then find det(B).
    Then find det(A)det(B).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    yeah i do that and just get lines and lines of algebra and cant find anyway of cancelling it down to get (a+b+c) (a-b)^2 (a-c)^2 (b-c)^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    Maybe if you worked ‘backwards’?
    That is, multiply \left( {a + b + c} \right)\left( {a - b} \right)^2 \left( {a - c} \right)^2 \left( {b - c} \right)^2 out.
    Be careful to see how the product works.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    I still cant get them to be the same theres just so many terms all over the place i cant find anyway of putting them together.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744
    Hello, feage7!

    Who assigned this problem . . . Professor deSade?


    Assuming that \text{det }(AB) \:= \:\text{det}(A)\cdot\text{det}(B)prove that

    \begin{vmatrix}3 &  a+b+c  & a^3+b^3+c^3 \\ a+b+c & a^2+b^2+c^2 & a^4+b^4+c^4 \\ a^2+b^2+c^2 & a^3+b^3+c^3 & a^5+b^5+c^5\end{vmatrix} \;= \;(a+b+c) (a-b)^2 (a-c)^2 (b-c)^2

    Thanks to Plato's incredible factoring, the rest is "just algebra" . . . Ha!
    It still takes Olympic-level gymnastics to simplify the determinants.

    It involves at lot of "factor by grouping".
    I won't bother explaining every step; I hope you can follow anyway.

    We have: / \text{det}(A) \;=\;\begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix} \;=\;(bc^2 - b^2c) - (ac^2 - a^2c) + (ab^2 - a^2b)

    . . = \;bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \;=\;a^2c - a^2b - ac^2 + ab^2 + bc^2 - b^2c

    . . = \;a^2(c-b) - a(c^2-b^2) + bc(c - b) \;=\;a^2\underbrace{(c-b)} - a\underbrace{(c-b)}(c+b) + bc\underbrace{(c-b)}

    . . = \;(c-b)\left[a^2 - a(c+b) + bc\right] \;= \;(c-b)(a^2 - ac - ab + bc)

    . . = \;(b-c)\left[a\underbrace{(a-b)} - c\underbrace{(a - b)}\right] \;=\;(c-b)(a-b)(a-c) \;= \;(a-b)(b-c)(c-a)

    Hence: . \boxed{\text{det}(A) \;=\;(a-b)(b-c)(c-a)}


    We have: . \text{det}(B)\;=\;\begin{vmatrix}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{vmatrix} \;=\;(bc^3-b^3c) - (ac^3 -a^3c) + (ab^3-a^3b)

    . . =\;bc^3 -b^3c - ac^3 + a^3c + ab^3 - a^3b \;= \;a^3c - a^3b - ac^3 + ab^3 + bc^3 - b^3c

    . . = \;a^3(c-b) - a(c^3-b^3) + bc(c^2-b^2) \;= \; a^3\underbrace{(c-b)} - a\underbrace{(c-b)}(c^2+bc + b^2) + bc\underbrace{(c-b)}(c+b)

    . . = \;(c-b)\left[a^3 - a(c^2+bc+b^2) + bc(c+b)\right] \;= \; (c-b)\left[a^3 - ac^2 - abc - ab^2 + bc^2 + b^2c\right]

    . . = \;(c-b)\left[a^3-ab^2-ac^2+bc^2-abc+b^2c\right] \;= \; (c-b)\left[a(a^2-b^2) - c^2(a-b) - bc(a-b)\right]

    . . = \;(c-b)\left[a\underbrace{(a-b)}(a+b) - c^2\underbrace{(a-b)} - bc\underbrace{(a-b)}\right] \;= \; (c-b)(a-b)\left[a(a+b) - c^2 - bc\right]

    . . = \;(c-b)(a-b)\left[a^2-c^2 + ab - bc\right] \;= \; (c-b)(a-b)\left[\underbrace{(a-c)}(a+c) + b\underbrace{(a-c)}\right]

    . . = \;(c-b)(a-b)(a-c)[a+c+b] \;= \;(a-b)(b-c)(c-a)(a + b + c)

    Hence: . \boxed{\text{det}(B) \;= \;(a-b)(b-c)(c-a)(a+b+c)}



    Therefore: . \text{det}(AB) \;= \;\text{det}(A)\cdot\text{det}(B)

    . . . . . . . . . . . . . . . = \;[(a-b)(b-c)(c-a)]\cdot[(a-b)(b-c)(c-a)(a+b+c)]

    . . . . . . . . . . . . . . . = \;(a+b+c)(a-b)^2(b-c)^2(c-a)^2 . . . ta-DAA!

    I need a nap . . .

    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    cheers for that just want to ask how you can change (c-b) to (b-c) on the last line of working out det B?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744


    You might notice that I "switched" two pairs.

    . . (c - b)(a - b)(a - c)(a + c + b) = (a - b)(b - c)(c - a)(a + b + c)


    The baby steps go something like this:

    . . (c-b)(a-c) \;= \;[\text{-}(b-c)][\text{-}(c-a)] \;=\;(b-c)(c-a)

    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    yeah but the questions asks for the bract (a-c) not (c-a) so then the answer isnt correct. I am just presuming but not checked it out yet at the moment that if u just swap all the b and c terms around in matrice A and B it would work.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by feage7 View Post
    yeah but the questions asks for the bract (a-c) not (c-a) so then the answer isnt correct. I am just presuming but not checked it out yet at the moment that if u just swap all the b and c terms around in matrice A and B it would work.
    a - c = -(c - a)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Nov 2006
    Posts
    26
    yeah i know that i mean the question asks for (a+b+c)(a-b)^2(a-c)^2(b-c)^2 not (a+b+c)(a-b)^2(c-a)^2(b-c)^2 which is why im saying u swap (a-c) and (c-b) to get the (b-c) term correct but then (a-c) becomes (c-a) in the final answer which means by making one term right ur making another wrong.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by feage7 View Post
    yeah i know that i mean the question asks for (a+b+c)(a-b)^2(a-c)^2(b-c)^2 not (a+b+c)(a-b)^2(c-a)^2(b-c)^2 which is why im saying u swap (a-c) and (c-b) to get the (b-c) term correct but then (a-c) becomes (c-a) in the final answer which means by making one term right ur making another wrong.
    You want (a+b+c)(a-b)^2(a-c)^2(b-c)^2 and have (a+b+c)(a-b)^2(c-a)^2(b-c)^2. The only difference is between (a-c)^2 and (c-a)^2. But:
    (c-a)^2 = (-[a-c])^2 = (-1)^2(a-c)^2 = (a-c)^2

    so there is no difference.

    With respect to each determinant, Both of Soroban's Det(A) and Det(B) had a single factor of (c-a) so switching to (a-c) in both gives a negative sign for each. But since we are multiplying the two determinants, we get two negative signs, which cancel.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Determinant problem
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 4th 2012, 04:39 AM
  2. determinant problem
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 30th 2011, 08:12 AM
  3. Determinant problem
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: February 18th 2010, 02:46 PM
  4. determinant problem
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: September 3rd 2009, 10:15 AM
  5. Determinant problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 13th 2008, 05:59 PM

Search Tags


/mathhelpforum @mathhelpforum