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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

rank of matrix $=$ rank of augmented matrix $=$ no of unknown $=$ $3$
so unique solution..

Correct Answer: $B$
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Can case C arise? If Yes, how shall we determine?
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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
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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
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yes c option case can never arise
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@Angkit   rank(r)>n  this case will never arise

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

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in matrix $[A]_{3\times3},$ if $|A|_{3\times3}\neq0$ then rank should be $3$
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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

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