`[[a,b,ax+by],[b,c,bx+cy],[ax+by,bx+cy,0]]=(b^2-ac)(ax^2+2bxy+cy^2)`

|{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|=(b^2-ac)(ax^2+2bxy+cy^2)

det[[ Prove that: ,b,ax+bya,c,bx+cyax+by,bx+cy,0]]=(b^(2)-ac)(ax^(2)+2bxy+cy^(2))

Prove that: |[a, b, ax+by],[ b, c, bx+cy], [ax+by, bx+cy,0]|=(b^2-a c)(a x^2+2b x y+c y^2)

Prove that: |(a,b, ax+by),(b,c,bx+cy), (ax+by, bx+cy,0)|=(b^2-a c)(a x^2+2b x y+c y^2) .

Prove that |(a,b,ax+by),(b,c,bx+cy),(ax+by, bx + cy, 0)| = (b^(2)-ac)(ax^(2) + 2bxy + cy^(2)) .

[[a, b, ax + byb, c, bx + cyax + by, bx + cy, 0]] = (b ^ (2) -ac) (ax ^ (2) + 2bxy + cy ^ (2))

If |{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|=0 and ax^2+2abxy+cy^2ne0," then "......

If a>0 and discriminant of ax^(2)+2bx+c is negative,then det[[a,b,ax+bb,c,bx+c]],+ve b.(ac-b)^(2)(ax^(2)+2bx+c) c.-ve d.0

If b^2 -aclt0 and alt 0, then thevalue of the determinant |(a,b,ax+by),(b,c,bx+cy),(ax + by,bx + cy,0)| is