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Definition of a Determinant

Consider the ...

Question

Consider the system of linear equations: $x_{1} + 2 x_{2} + x_{3} = 3$ $, 2 x_{1} + 3 x_{2} + x_{3} = 3$ and $3 x_{1} + 5 x_{2} + 2 x_{3} = 1$ The system has

exactly 3 solutions

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a unique solution

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no solution

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infinite number of solutions

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Solution

The correct option is C no solution
$D = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 1 2 & 3 & 1 3 & 5 & 2 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Since, $D = 0$ , given system does not have any solution.
Hence, option 'C' is correct.

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Similar questions

x_{1} + 2 x_{2} + x_{3} = 3

2 x_{1} + 3 x_{2} + x_{3} = 3

3 x_{1} + 5 x_{2} + 2 x_{3} = 1

Consider the system of equations
$x_{1} + 2 x_{2} + 3 x_{3} = 1$
$x_{2} + 2 x_{3} + 3 x_{1} = 2$
$x_{3} + 2 x_{1} + 3 x_{2} = 3$
then

A. $x_{1} = \frac{2}{3}$

B. $x_{1} + x_{2} = \frac{4}{3}$

C. $x_{3} = - \frac{1}{3}$

D. $x_{1} - x_{2} = \frac{1}{3}$

λ

2 x_{1} - 2 x_{2} + x_{3} = λ x_{1}

2 x_{1} - 3 x_{2} + 2 x_{3} = λ x_{2}

- x_{1} + 2 x_{2} = λ x_{3}

λ

2 x_{1} - 2 x_{2} + x_{3} = λ x_{1}, 2 x_{1} - 3 x_{2} + 2 x_{3} = λ x_{2}

- x_{1} + 2 x_{2} = λ x_{3}

x_{1} + 2 x_{2} + 3 x_{3} = 6

x_{1} + 3 x_{2} + 5 x_{3} = 9

2 x_{1} + 5 x_{2} + a x_{3} = b

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