If $D \neq 0$ and at least one of $D_{1}, D_{2}, D_{3}$ is not 0 then according to Cramer's rule the system of linear equations 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

By Cramer's rule
$x = \frac{D 1}{D}, y = \frac{D 2}{D}, z = \frac{D 3}{D}$ . Given $D$ not equal to 0 and at least one of $D_{1}, D_{2}, D_{3}$ is not 0.
Therefore we have a non trivial solution.

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Similar questions

Q. If

D \neq 0

and at least one of

D_{1}, D_{2}, D_{3}

is not 0 then according to Cramer's rule the system of linear equations will have

If $D = 0$ and at least one of $D_{1}, D_{2}, D_{3}$ is not 0 then according to Cramer's rule the system of linear equations will have

Q. If D = 0 and at least one of

D_{1}, D_{2}, D_{3}

is not 0, then according to Cramer's rule, the system of linear equations will have

Which of the following cases will not lead to non trivial solutions in case of system of linear equations according to Cramer's rule Convention given D != 0 ?

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If D≠0 and at least one of D1,D2,D3 is not 0 then according to Cramer's rule the system of linear equations will have

The correct option is A non trivial solution By Cramer's rule x=D1D,y=D2D,z=D3D. Given D not equal to 0 and at least one of D1,D2,D3 is not 0. Therefore we have a non trivial solution.

If $D \neq 0$ and at least one of $D_{1}, D_{2}, D_{3}$ is not 0 then according to Cramer's rule the system of linear equations will have

The correct option is A
non trivial solution

By Cramer's rule
$x = \frac{D 1}{D}, y = \frac{D 2}{D}, z = \frac{D 3}{D}$ . Given $D$ not equal to 0 and at least one of $D_{1}, D_{2}, D_{3}$ is not 0.
Therefore we have a non trivial solution.