Consider the system of equations-a1 x-b1 y-c1 z-0a2 x-b2 y-c2 z-0a3 x-b3 y-c3 z-0if a1b1c1a2b2c2a3b3c3 - 0 - then the system hasa more than two solutionsb one trivial and one non-trivial solutionsc no solutiond only trivial solution 0-0-0

(a) #160;more #160;than #160;two #160;solutions #160;Here,A=0 #160;and #160;B=0 #160; #160; #160; #160; #160; #160; #160; #160; ...

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Cramer's Rule

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Question

Consider the system of equations:
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0,
if $|\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|$ = 0, then the system has
(a) more than two solutions
(b) one trivial and one non-trivial solutions
(c) no solution
(d) only trivial solution (0, 0, 0)

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a_{1} x + b_{1} y + c_{1} z = 0,

a_{2} x + b_{2} y + c_{2} z = 0,

a_{3} x + b_{3} y + c_{3} z = 0,

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

The system of linear equations $x + y = 0, x - y = 0, z = 0$ will have

A = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & a_{3} \end{matrix} ∣ ∣ ∣ ∣ \neq 0,

a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0

a_{3} x + b_{3} y + c_{3} z = 0

A = ⎡ ⎢ ⎣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ⎤ ⎥ ⎦

| A | \neq 0

a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0

a_{3} x + b_{3} y + c_{3} z = 0

b c + q r = c a + r p = a b + p q = - 1

∣ ∣ ∣ ∣ \begin{matrix} a p & a & p b q & b & q c r & c & r \end{matrix} ∣ ∣ ∣ ∣ = 0 (a b c, p q r \neq 0)

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0, a_{3} x + b_{3} y + c_{3} = 0

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

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