# Thread: determinant problem

1. ## 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

2. $A = \left[ {\begin{array}{rrr}
1 & 1 & 1 \\
a & b & c \\
{a^2 } & {b^2 } & {c^2 } \\
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}
1 & a & {a^3 } \\
1 & b & {b^3 } \\
1 & c & {c^3 } \\
\end{array}} \right]$

Your matrix is the product of AB.

3. Originally Posted by Plato
$A = \left[ {\begin{array}{rrr}
1 & 1 & 1 \\
a & b & c \\
{a^2 } & {b^2 } & {c^2 } \\
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}
1 & a & {a^3 } \\
1 & b & {b^3 } \\
1 & c & {c^3 } \\
\end{array}} \right]$

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

-Dan

4. Originally Posted by Plato
$A = \left[ {\begin{array}{rrr}
1 & 1 & 1 \\
a & b & c \\
{a^2 } & {b^2 } & {c^2 } \\
\end{array}} \right]\quad \& \quad B = \left[ {\begin{array}{rcr}
1 & a & {a^3 } \\
1 & b & {b^3 } \\
1 & c & {c^3 } \\
\end{array}} \right]$

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

5. Find det(A). Then find det(B).
Then find det(A)det(B).

6. 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

7. 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.

8. 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.

9. 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 . . .

10. 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?

11. 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:

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

12. 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.

13. Originally Posted by feage7
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

14. 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.

15. Originally Posted by feage7
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

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