What is the value of k, if `|{:(k,b+c,b^(2)+c^(2)),(k,c+a,c^(2)+a^(2)),(k,a+b,a^(2)+b^(2)):}|=(a-b)(b-c)(c-a)`?

Options

A. 1
B. -1
C. 2
D. 0

Answer

A

Explanation

Let `|{:(k,b+c,b^(2)+c^(2)),(k,c+a,c^(2)+a^(2)),(k,a+b,a^(2)+b^(2)):}|=Delta`
`=k|{:(" "1," "1," "1),(b+c,c+a,a+b),(b^(2)+c^(2),c^(2)+a^(2),a^(2)+b^(2)):}|`
Applying `C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)`
`=k|{:(" "1," "0," "0),(b+c,a-b,a-c),(b^(2)+c^(2),a^(2)-b^(2),a^(2)-c^(2)):}|`
`=k|{:(" "1," "0," "0),(b+c," "a-b," "a-c),(b^(2)+c^(2),(a-b)(a+b),(a-c)(a+c)):}|`
`=k(a-b)(a-c)|{:(" "1," "1),(a+b,a+c):}|`
`=k(a-b)(a-c)(a+c-a-b)`
`=k(a-b)(b-c)(c-a)`
But given `Delta = (a-b)(b-c)(c-a)`
Thus, k = 1