# Using properties of determinant prove that `|[1, a, a^2],[1, b, b^2],[1, c, c^2]| = (a-b)(b-c)(c-a)`

justaguide | College Teacher | (Level 2) Distinguished Educator

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The determinant of a matrix ` [[a, b, c],[d,e,f],[g,h,i]]` is given by:

`|[a, b, c],[d,e,f],[g,h,i]|`= `a*(e*i - f*g) - b*(d*i - g*f) + c*(d*h -e*g)`

The determinant that has to be found `|[1, a, a^2], [1, b, b^2],[1, c, c^2]|`

= `1*(b*c^2 - b^2*c) - a*(c^2 - b^2) + a^2*(c - b)`

=> `b*c^2 - b^2*c + a*b^2 - a*c^2 + a^2*c - a^2*b`

=> `bc(c - b) + a(b - c)(b + c) + a^2(c - b)`

=> `(b - c)(ab + ac - bc - a^2)`

=> `(b - c)(a(c - a) -b(c - a))`

=> `(a - b)(b - c)(c - a)`

This proves that `|[1, a, a^2], [1, b, b^2],[1, c, c^2]|` = `(a - b)(b - c)(c - a)`

Sources:

wimudu | Student, Undergraduate | (Level 1) eNoter

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1 a a^2

1 b b^2

1 c c^2      appyling the R1= (R1-R2) and R2=(R2-R3)

then,    0  (a-b)  (a^2-b^2)

0  (b-c)  (b^2-c^2)

1    c         c^2

taking{ (a-b) (b-c) }common from

(a-b)(b-c) , 0 1 (a+b)

0 1 (b+c)

1  c  c^2

expanding by C1

(a-b)(b-c)(a-c)