Finding the linear factors of a matrix

i have the following question...

express the following determinant as the product of four linear factors:

| 1 a a^2 |
| 1 b b^2 |
| 1 c c^2 |

can anyone talk me through this?.. im not sure of how to go about doing it!! (should have paid attention in class!! :s-smilie:

)

hiroki

Reply 1

Juwel

18

hiroki

i have the following question...

express the following determinant as the product of four linear factors:

| 1 a a^2 |
| 1 b b^2 |
| 1 c c^2 |

can anyone talk me through this?.. im not sure of how to go about doing it!! (should have paid attention in class!! :s-smilie:

)

hiroki

The determinant can be written as:
|M| =
|0 (a-b) (a²-b²)|
|0 (b-c) (b²-c²)|
|1 c c² |

And therefore |M| = (a-b)(b²-c²) - (b-c)(a²-b²)
= (a-b)(b-c)(b+c) - (b-c)(a-b)(a+b) = (a-b)(b-c)[(b+c)-(a+b)]
= (a-b)(b-c)(c-a)

I get three linear factors. I did this question before, I'll go find my answer.

Reply 2

hiroki

OP

oh god wait sorry...

switch that with...

| 1 a a^3 |
| 1 b b^3 |
| 1 c c^3 |

sorry!!

hiroki

Reply 3

Juwel

18

hiroki

oh god wait sorry...

switch that with...

| 1 a a^3 |
| 1 b b^3 |
| 1 c c^3 |

sorry!!

hiroki

OK, |M| =
|0 (a-b) (a³-b³)|
|0 (b-c) (b³-c³)|
|1 c c³ |

|M| = (a-b)(b³-c³) - (b-c)(a³-b³)

The difference of two cubes, a³-b³ = (a-b)(a²+ab+b²)
>>>
|M| = (a-b)(b-c)(b²+bc+c²) - (b-c)(a-b)(a²+ab+b²)
= (a-b)(b-c)[b²+bc+c²-a²-ab-b²]
= (a-b)(b-c)[c²-a²+bc-ab]
= (a-b)(b-c)[(c-a)(c+a)+b(c-a)]
= (a-b)(b-c)(c-a)[a+b+c]

Phew, four linear factors.