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asked in Linear Algebra by Boss (7.3k points) | 212 views
No..no of nonzero rows =3..so rank 3
i thnk it should be B
Okk...made a mistake... actually if I move 3rd and 4th row up and then exchange first column with 2nd...there is a 3x3 matrix of non-zero determinant.

 

What is the answer...A?
Ans C
When det A = 0 and equations of form AX= 0 and Rank of A < N (order or matrix or no of unknowns) then no of linearly independent solutions are N-rank of matrix. Here its 4-3 =1 no of linearly independent solutions as per my calculation.I dont know how answer C is coming, I feel its wrong.It should be A here.

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The rank of the matrix=3 by the method of Row-reduced Echelon form.

 

Now, if there are 'm' linearly independent vectors(or rank of matrix) and 'n' is number of unknows, then 

#linearly independent solutions = n - m   (m<=n)

 

So,here, #linearly independent solutions = 4 - 3 = 1

So, A should be the correct answer.

 

Ref: http://math.stackexchange.com/questions/1235474/the-number-of-linearly-independent-solution-of-the-homogeneous-system-of-linear

answered by Veteran (18k points)
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