Prove that: |[a, b, ax+by],[ b, c, bx+cy...

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)`

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To prove that \[ \left| \begin{array}{ccc} a & b & ax + by \\ b & c & bx + cy \\ ax + by & bx + cy & 0 \end{array} \right| = (b^2 - ac)(ax^2 + 2bxy + cy^2), ...

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If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

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Prove that the lines ax+by + c = 0, bx+ cy + a = 0 and cx+ay+b=0 are concurrent if a+b+c = 0 or a+b omega + c omega^(2) + c omega = 0 where omega is a complex cube root of unity .

STATEMENT-1 : If a, b, c are distinct and x, y, z are not all zero and ax + by + cz = 0, bx + cy + az = 0, cx + ay + bz = 0 , then a + b + c = 0 and STATEMENT-2 : a^2 + b^2 + c^2 > ab + bc + ca , if a, b, c are distinct.

Let a,b,c be real numbers with a^2 + b^2 + c^2 =1 . Show that the equation |[ax-by-c,bx+ay,cx+a],[bx+ay,-ax+by-c,cy+b],[cx+a,cy+b,-ax-by+c]|=0 represents a straight line.

If the system of equations ax + by + c = 0,bx + cy +a = 0, cx + ay + b = 0 has infinitely many solutions then the system of equations (b + c) x +(c + a)y + (a + b) z = 0 (c + a) x + (a+b) y + (b + c) z = 0 (a + b) x + (b + c) y +(c + a) z = 0 has

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