Given that aα$$2+2b$$α$$+c$$≠$$0and$$ that the system of equations $$(a$$α$$+b)x+ay+bz=0$$,$$(b$$α$$+c)x+by+cz=0$$,$$(a$$α$$+b)y+(b$$α$$+c)z=0has$$ a non trivial solution, then a,b,care in

Question

Given that aα$$2+2b$$α$$+c$$≠$$0and$$ that the system of equations $$(a$$α$$+b)x+ay+bz=0$$,$$(b$$α$$+c)x+by+cz=0$$,$$(a$$α$$+b)y+(b$$α$$+c)z=0has$$ a non trivial solution, then a,b,care in

Infinity Learn · Accepted Answer

The condition for a homogeneous system of equations to have a $$non-trivial$$ solution is Δ$$=0$$⇒Δ$$=a$$α$$+babb$$α$$+cbc0a$$α$$+bb$$α$$+c=0$$                $$R3$$→$$R3$$−$$($$α$$R1+R2)$$⇒aα$$+babb$$α$$+cbc-a$$α$$2+2b$$α$$+c00=0$$ ⇒$$-a$$α$$2+2b$$α$$+cac-b2=0$$ ⇒$$ac-b2=0$$    ∵aα$$2+2b$$α$$+c$$≠$$0$$ ⇒$$b2=ac$$ ⇒a,b,c are in G.P.

Given that $a α^{2} + 2 b α + c \neq 0$ and that the system of equations $(a α + b) x + a y + b z = 0,$ $(b α + c) x + b y + c z = 0$ , $(a α + b) y + (b α + c) z = 0$ has a non trivial solution, then $a, b, c$ are in

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Given that aα2+2bα+c≠0and that the system of equations (aα+b)x+ay+bz=0,(bα+c)x+by+cz=0,(aα+b)y+(bα+c)z=0has a non trivial solution, then a,b,care in

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Given that $a α^{2} + 2 b α + c \neq 0$ and that the system of equations $(a α + b) x + a y + b z = 0,$ $(b α + c) x + b y + c z = 0$ , $(a α + b) y + (b α + c) z = 0$ has a non trivial solution, then $a, b, c$ are in