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[" (iv) "x+y+z=lambda],[ax+bx+cz=mu],[a^...

[" (iv) "x+y+z=lambda],[ax+bx+cz=mu],[a^(2)x+by+c^(2)z=y]

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solve by cramer's rule : x+y+z=1 , ax+by+cz=k, a^2x+b^2y+c^2z=k^2[a ne b ne c] .

If a, b, c are all different, solve the system of equations x+y+z=1 ax+by+cz=lamda a^2x+b^2y+c^2z=lamda^2 using Cramer's rule.

If a, b, c are all different, solve the system of equations x+y+z=1 ax+by+cz=lamda a^2x+b^2y+c^2z=lamda^2 using Cramer's rule.

If a, b, c, x, y, z are real and a^(2)+b^(2) + c^(2)=25, x^(2)+y^(2)+z^(2)=36 and ax+by+cz=30 , then (a+b+c)/(x+y+z) is equal to :

If a, b, c, x, y, z are real and a^(2)+b^(2) + c^(2)=25, x^(2)+y^(2)+z^(2)=36 and ax+by+cz=30 , then (a+b+c)/(x+y+z) is equal to :

If a, b, c, x, y, z are real and a^(2)+b^(2) + c^(2)=25, x^(2)+y^(2)+z^(2)=36 and ax+by+cz=30 , then (a+b+c)/(x+y+z) is equal to :

If (x+y)/(3a-b) = (y+z)/(3b-c) = (z+x)/(3c-a) , then show that (x+y+z)/(a+b+c)= (ax + by+cz)/(a^(2)+b^(2)+c^(2))

If ax+cy+bz=X, cx+by+az=Y, bx+ay+cz=Z, show that (a^(2)+b^(2)+c^(2)-bc-ca-ab)(x^(2)+y^(2)+z^(2)-yz-zx-xy)=X^(2)+Y^(2)+Z^(2)-YZ-ZX-XY