Prove that $\left|\begin{array}{lll}bc-a^{2} & ca-b^{2} & ab-c^{2} \\ ca-b^{2} & ab-c^{2} & bc-a^{2} \\ ab-c^{2} & bc-a^{2} & ca-b^{2}\end{array}\right|$ is divisible by $(a+b+c)$, and hence find the quotient.

Quotient $=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)^{2}$; the determinant equals $(a+b+c)^{2}\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)^{2}$

Determinants — CBSE Class 12 · 2016 Board Question with Solution

Prove that

b c - a^{2} c a - b^{2} ab - c^{2} c a - b^{2} ab - c^{2} b c - a^{2} ab - c^{2} b c - a^{2} c a - b^{2}

is divisible by

(a + b + c)

, and hence find the quotient.

Write

p = b c - a^{2}

q = c a - b^{2}

r = ab - c^{2}

, so the determinant is the reverse circulant

Δ = p q r q r p r p q .

Expanding along the first row collapses it to a symmetric cubic in

p, q, r

Δ = 3 pq r - (p^{3} + q^{3} + r^{3}) = - (p + q + r) (p^{2} + q^{2} + r^{2} - pq - q r - r p) .

The first bracket is already symmetric in

a, b, c

p + q + r = (b c - a^{2}) + (c a - b^{2}) + (ab - c^{2}) = - (a^{2} + b^{2} + c^{2} - ab - b c - c a) .

For the second bracket note that each difference carries a factor

(a + b + c)

p - q = (b - a) (a + b + c), q - r = (c - b) (a + b + c), r - p = (a - c) (a + b + c),

so, using

p^{2} + q^{2} + r^{2} - pq - q r - r p = \frac{1}{2} [(p - q)^{2} + (q - r)^{2} + (r - p)^{2}]

p^{2} + q^{2} + r^{2} - pq - q r - r p = (a + b + c)^{2} (a^{2} + b^{2} + c^{2} - ab - b c - c a) .

Multiplying the two brackets, the two minus signs cancel:

Δ = (a + b + c)^{2} (a^{2} + b^{2} + c^{2} - ab - b c - c a)^{2} .

(a + b + c)

divides

Δ

, and the quotient is

\frac{Δ}{a + b + c} = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)^{2}

Check with

a = 1, b = 2, c = 3

p = 5, q = - 1, r = - 7

give

Δ = 324 = 6^{2} \cdot 3^{2}

, and the quotient is

54 = 6 \cdot 3^{2}

CBSE Class 12 Board · 2016 · Subjective