19 votes 19 votes 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 no solution a unique solution more than one but a finite number of solutions an infinite number of solutions Linear Algebra gatecse-2005 linear-algebra system-of-equations normal + – gatecse asked Sep 21, 2014 gatecse 7.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 27 votes 27 votes rank of matrix $=$ rank of augmented matrix $=$ no of unknown $=$ $3$ so unique solution.. Correct Answer: $B$ Digvijay Pandey answered Apr 23, 2015 edited Apr 25, 2019 by Naveen Kumar 3 Digvijay Pandey comment Share Follow See all 8 Comments See all 8 8 Comments reply prasitamukherjee commented Sep 29, 2016 reply Follow Share Can case C arise? If Yes, how shall we determine? 0 votes 0 votes vijaycs commented Sep 29, 2016 reply Follow Share when rank of matrix = rank of augmented matrix < no of unknown 0 votes 0 votes prasitamukherjee commented Sep 30, 2016 reply Follow Share then it is infinite solutions. r < n, that is option D. I'm asking about option C 4 votes 4 votes Angkit commented Apr 26, 2017 reply Follow Share 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 3 votes 3 votes Kaluti commented Jul 23, 2017 reply Follow Share yes c option case can never arise 2 votes 2 votes Verma Ashish commented Feb 4, 2019 reply Follow Share @Angkit rank(r)>n this case will never arise 0 votes 0 votes Jhaiyam commented May 8, 2020 reply Follow Share How can we find determinant of augmented matrix ? 0 votes 0 votes abir_banerjee commented Oct 10, 2022 reply Follow Share @Jhaiyam then I must ask you how can you numbers of stars in the sky:) LOL determinant is only defined for square matrix and augmented matrix is not a square matrix. 0 votes 0 votes Please log in or register to add a comment.
9 votes 9 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 Rishi yadav answered Oct 14, 2017 Rishi yadav comment Share Follow See all 3 Comments See all 3 3 Comments reply Lakshman Bhaiya commented Jan 28, 2019 reply Follow Share in matrix $[A]_{3\times3},$ if $|A|_{3\times3}\neq0$ then rank should be $3$ 2 votes 2 votes Verma Ashish commented Feb 4, 2019 reply Follow Share 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 2 votes 2 votes sayan chowdhury commented Jan 31, 2022 i edited by sayan chowdhury Jan 31, 2022 reply Follow Share |A|=0 implies for a non-homogenous system implies that the solution may be inconsistent or infinitely many. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes (B) A unique solution . Rahul_kumar3 answered Nov 25, 2023 Rahul_kumar3 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Method-1:2 3 -1-1 2 53 5 1 In this matrix we can clearly see that three columns are Linearly independent and 3 L.I columns in R3 space So It will fill the space and Ax=b So there will be unique solution If the column space is filled and Ax=0 then there will be a trivial solutionMethod-2: Converting into echelon formAfter converting into echelon form we obtain augmented matrix as2 0 0-1 7 03 1 321 1 46We can clear see all the columns have pivot and there is no [00..00|b] form So there will be unique solutionMethod-3: By using rankrank[A] =3 and rank[A|b]=3 and number of columns =3Therefore unique solution is possibleAnswer is option-B SASIDHAR_1 answered Feb 25 SASIDHAR_1 comment Share Follow See all 0 reply Please log in or register to add a comment.