The system of linear equations x - y -0- x - y -0- z -0 will haveA- infinite solutionsB- no solutionsC- trivial solutionD- non trivial solution

The correct option is C trivial solution nbsp; Once you express the given equations in the form of matrices nbsp;determinant we get D = 2, D_1 = 0, D_2 = ...

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Standard XII

Mathematics

Cramer's Rule

The system of...

Question

The system of linear equations $x + y = 0, x - y = 0, z = 0$ 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 B
trivial solution

Once you express the given equations in the form of matrices determinant we get
$D = - 2, D_{1} = 0, D_{2} = 0, D_{3} = 0$ .
So $D \neq 0$ and $D_{1} = D_{2} = D_{3} = 0$ . So according to Cramer's rule there exists only trivial solutions and the solutions being $x = \frac{D_{1}}{D} = 0, y = \frac{D_{2}}{D} = 0, z = \frac{D_{3}}{D} = 0$

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The system of linear equations x+y=0,x–y=0,z=0 will have

The correct option is B trivial solution Once you express the given equations in the form of matrices determinant we get D=−2,D1=0,D2=0,D3=0. So D≠0 and D1=D2=D3=0. So according to Cramer's rule there exists only trivial solutions and the solutions being x=D1D=0,y=D2D=0,z=D3D=0

The system of linear equations $x + y = 0, x - y = 0, z = 0$ will have