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Solve the system of equations

                        x + y + z = 6

                        x + 2y + 3z = 14

                        x + 4y + 7z = 30 has

Option: 1

no solution


Option: 2

unique solution


Option: 3

infinite solutions


Option: 4

none of these


Answers (1)

best_answer

As we have learned

Non-homogeneous system of linear equation -

b\neq 0

- wherein

Given system of equation is

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30 

\Delta=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right|=1(14-12)-1(7-3)+1(4-2)=0

\Delta_1=\left|\begin{array}{lll} 6 & 1 & 1 \\ 14 & 2 & 3 \\ 30 & 4 & 7 \end{array}\right|=0

\Delta_2=\left|\begin{array}{lll} 1 & 6 & 1 \\ 1 & 14 & 3 \\ 1 & 30 & 7 \end{array}\right|=0

\Delta_3=\left|\begin{array}{lll} 1 & 1 & 6 \\ 1 & 2 & 14 \\ 1 & 4 & 30 \end{array}\right|=0

Aso, 

x + y + z = 6 ……(1) 

y + 2z = 8 ….(2)

x = 6 – y – z = 6 – (8 – 2z) – z = z – 2 

Taking z = k, we get x = k – 2, y = 8 – 2k; k ∈ R 

Putting k = 1, we have one solution as x = – 1, y = 6, z = 1. 

Thus by giving different values for k we get different solutions.

Hence the given system has an infinite number of solutions.

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