22 votes 22 votes Consider the following set of equations $x+2y=5$ $4x+8y=12$ $3x+6y+3z=15$ This set has unique solution has no solution has finite number of solutions has infinite number of solutions Linear Algebra gate1998 linear-algebra system-of-equations easy + – Kathleen asked Sep 25, 2014 Kathleen 7.3k views answer comment Share Follow See all 12 Comments See all 12 12 Comments reply Show 9 previous comments Veenit commented Jan 22, 2020 reply Follow Share @rohith1001 Yes I too got the same option. 1 votes 1 votes rishabhsharma commented Aug 26, 2020 reply Follow Share Constant matrix also linearly dependent with same factor as coefficient matrix →Yes (Then Inconsistent System of Equation- No Solution) If →No (Then consistent System of Equation- Infinite Solutions) 0 votes 0 votes Shreed commented Oct 29, 2023 reply Follow Share In row echelon form we get [0 0 0 | b ] form where b!=0. Therefore system of linear eq is inconsistent and has no solution. 0 votes 0 votes Please log in or register to add a comment.
Best answer 42 votes 42 votes There are no solutions. If we multiply $1^{\text{st}}$ equation by $4,$ we get $4x + 8y = 20$ But $2^{\text{nd}}$ equation says $4x + 8y = 12$ Clearly, there can not be any pair of $(x,y),$ which satisfies both equations. Correct Answer: B. Happy Mittal answered Sep 28, 2014 • edited May 31, 2021 by Lakshman Bhaiya Happy Mittal comment Share Follow See all 2 Comments See all 2 2 Comments reply radha gogia commented Jan 27, 2016 reply Follow Share I am getting rank of augmented matrix as 3 as well as rank of A =3 so according to this condition there must be a unique solution , although it shouldn't exist but then how can it contradict ? 0 votes 0 votes Happy Mittal commented Jan 29, 2016 reply Follow Share Rank of A is not 3. Recheck your method. 3 votes 3 votes Please log in or register to add a comment.
14 votes 14 votes First transformed the matrix into echelon form and rest can seen from the pic. Tuhin Dutta answered Dec 22, 2017 Tuhin Dutta comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes (B) Thus, The R1 is Dependent Upon R2 And as per calculation |A| = 0 So, It is linearly dependent then it is Having No solution Rahul_kumar3 answered Nov 25, 2023 Rahul_kumar3 comment Share Follow See 1 comment See all 1 1 comment reply SASIDHAR_1 commented Feb 25 i edited by SASIDHAR_1 Feb 25 reply Follow Share Before solving this question we can roughly identify what kind of solutions can be possible. For that we have to write the matrix A first.1 4 32 8 60 0 3From this we can clearly see that C1 and C2 are Linearly dependent. That means it do not fill the column space.So solution may or may not exist .The possible solutions can be infinite or 0Method-1:Step-1: Find rank[A] and rank[A|b] i) if rank[A] = rank[A|b] then there exist a solutionii)if rank[A] != rank[A|b] then solution does not exist In this case rank[A] =2 and rank[A|b] = 3. Therefore the system has No solution.Method-2: Convert the augmented matrix [A|b] into echelon form After solving it comes out to be this1 0 02 0 00 3 05 0 -8So clearly we can see there exist [00….0|b] form Which is a condition for No solution.so option-B is correct answer 0 votes 0 votes Please log in or register to add a comment.