If the system of linear equations ax $$+$$ ky $$+$$ $$2az$$ $$=$$ $$0$$ ; bx $$+$$ ky $$+$$ $$3bz$$ $$=$$ $$0$$ ; cx $$+$$ ky $$+$$ $$4cz$$ $$=$$ $$0where$$ k,a,b,c ∈ Rhas $$non-$$ zero solution then

Question

If the system of linear equations ax $$+$$ ky $$+$$ $$2az$$ $$=$$ $$0$$ ; bx $$+$$ ky $$+$$ $$3bz$$ $$=$$ $$0$$ ; cx $$+$$ ky $$+$$ $$4cz$$ $$=$$ $$0where$$ k,a,b,c ∈ Rhas $$non-$$ zero solution then

Infinity Learn · Accepted Answer

System has non – zero solution ⇒Δ$$=0$$ ⇒ $$ak2abk3bck4c=0$$⇒$$ka12ab13bc14c=0$$ ⇒$$a12ab13bc14c=0$$ Expanding along column $$C2$$, we get        $$-14bc-3bc+14ac-2ac-13ab-2ab=0$$ ⇒$$2ac=bc+ab$$ Dividing throughout by abc, we get        $$2b=1a+1c$$ ⇒a,b,c are in H.P.

If the system of linear equations $ax + ky + 2az = 0 ; bx + ky + 3bz = 0 ; cx + ky + 4cz = 0$ where $k,a,b,c \in R$ has non- zero solution then

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If the system of linear equations ax + ky + 2az = 0 ; bx + ky + 3bz = 0 ; cx + ky + 4cz = 0where k,a,b,c ∈ Rhas non- zero solution then

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If the system of linear equations $ax + ky + 2az = 0 ; bx + ky + 3bz = 0 ; cx + ky + 4cz = 0$ where $k,a,b,c \in R$ has non- zero solution then