The system of linear equations $x + y + z = 6, x + 2 y + 3 z = 10,$
$x + 2 y + 2 z = 4$ will have

non trivial solution

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trivial solution

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infinite solutions

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no solutions

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Solution

The correct option is A
non trivial solution

Once you express the given equations in the form of matrices then the determinant
$D = - 1, D_{1} = - 8, D_{2} = 8, D_{3} = - 6$ .
Here $D \neq 0$ , also none of $D_{1}, D_{2}, D_{3}$ are zeros.
So according to Cramer's rule there exists non trivial solutions and the solutions being given by
$x = D_{1} / D, y = D_{2} / D, z = D_{3} / D$ .

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