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 haveA- trivial solutionB- non trivial solutionC- infinite solutionsD- no solutions

The correct option is B non trivial solution nbsp; By Cramer's rule x =D1D, y =D2D, z =D3D. Given D not equal to 0 and at least one of D_1, D_2, D_3 is not 0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Solution of a Linear Equation in Two Variables

If D =0 and a...

Question

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

trivial solution

No worries! We‘ve got your back. Try BYJU‘S free classes today!

infinite solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

no solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

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

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

Q. The system of linear equations x + y = 0, x – y = 0, z = 0 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 ?

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

Join BYJU'S Learning Program

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.