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Consider the following system of linear equations : $$2x_1 - x_2 + 3x_3 = 1$$ $$3x_1 + 2x_2 + 5x_3 = 2$$ $$-x_1+4x_2+x_3 = 3$$ The system of equations has

  1. no solution
  2. a unique solution
  3. more than one but a finite number of solutions
  4. an infinite number of solutions
asked in Linear Algebra by Boss (18.2k points) | 1.1k views

2 Answers

+16 votes
Best answer
rank of matrix $=$ rank of augmented matrix $=$ no of unknown $=$ $3$
so unique solution..

Correct Answer: $B$
answered by Veteran (59.7k points)
edited by
0
Can case C arise? If Yes, how shall we determine?
0

when rank of matrix = rank of augmented matrixno of unknown 

+3
then it is infinite solutions. r < n, that is option D. I'm asking about option C
+1
i think more than one but a finite number of solutions will never arise

as we have only 3 cases r=n,r<n and r>n
+1
yes c option case can never arise
0

@Angkit   rank(r)>n  this case will never arise

+5 votes

Determinant of matrix =14 which is non zero

If The determinant of the coefficient matrix is non zero then definitely the system of given equation has a unique solution 

 so option B

answered by Boss (11k points)
+1
in matrix $[A]_{3\times3},$ if $|A|_{3\times3}\neq0$ then rank should be $3$
0
if we get |A|=0 then we have to check for either infinite solⁿ or no solution ...so we have to go with our fundamental method..

Then i think finding determinant is not fruitful
Answer:

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