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Question: \(x_{1} + 2x_{2} + 3x_{3} = a2x_{1} + 3x_{2} + x_{3} =\) \(b3x_{1} + x_{2} + 2x_{3} = c\) this syst...

x1+2x2+3x3=a2x1+3x2+x3=x_{1} + 2x_{2} + 3x_{3} = a2x_{1} + 3x_{2} + x_{3} = b3x1+x2+2x3=cb3x_{1} + x_{2} + 2x_{3} = c this

system of equations has.

A

Infinite solution

B

No solution

C

Unique solution

D

None of these

Answer

Unique solution

Explanation

Solution

We have, x1+2x2+3x3=cx_{1} + 2x_{2} + 3x_{3} = c

2ax1+3x2+x3=c2ax_{1} + 3x_{2} + x_{3} = c

3bx1+x2+2x3=c3bx_{1} + x_{2} + 2x_{3} = c

Let a=b=c=1a = b = c = 1.

Then D=∣123231312∣D = \left| \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix} \right| = 1(5)−2(1)+3(−7)=−18≠01(5) - 2(1) + 3( - 7) = - 18 \neq 0

1 & 2 & 3 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \end{matrix} \right| = - 3$$ Similarly $D_{y} = D_{z} = - 3$. Now,$\Delta = (2 + i)\left| \begin{matrix} 1 & 1 & i \\ 1 & 1 + 2i & 1 + i \\ 1 & 2 & 1 - i \end{matrix} \right|$ $y = z = \frac{1}{6}$ Hence $D \neq 0$, $x = y = z$, i.e., unique solution.