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- infinite solutionsC- no solutionsD- non trivial solution

The correct option is C no solutions nbsp; By Cramer's rule x =D_1D, y = D_2D, z =D_3D. Given D=0 and at least one of nbsp; D_1, D_2, D_3 is not 0. Therefo ...

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Standard X

Mathematics

Classification of Pair of Linear Equations

If D =0 and a...

Question

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

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 D
no solutions

By Cramer's rule $x = \frac{D_{1}}{D}, y = \frac{D_{2}}{D}, z = \frac{D_{3}}{D}$ . Given $D = 0$ and at least one of $D_{1}, D_{2}, D_{3}$ is not 0. Therefore by Cramer's rule we have no possible solutions for this set of equations..

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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 D no solutions By Cramer's rule x=D1D,y=D2D,z=D3D. Given D=0 and at least one of D1,D2,D3 is not 0. Therefore by Cramer's rule we have no possible solutions for this set of equations..

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

The correct option is D
no solutions

By Cramer's rule $x = \frac{D_{1}}{D}, y = \frac{D_{2}}{D}, z = \frac{D_{3}}{D}$ . Given $D = 0$ and at least one of $D_{1}, D_{2}, D_{3}$ is not 0. Therefore by Cramer's rule we have no possible solutions for this set of equations..