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

non trivial solution

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no solution

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trivial solution

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infinite solution

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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 $D \neq 0$ , 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 $x = \frac{D_{1}}{D}, y = \frac{D_{2}}{D}, z = \frac{D_{3}}{D}$

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