1 votes 1 votes Given a system of equations: x+2y+2z=b1 5x+y+3z=b2 (a) the system will have infinitely many solutions for any given b1 and b2 (b) whether or not a solution exists depends on the given b1 and b2 Peach asked Dec 12, 2018 Peach 1.6k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply srestha commented Dec 12, 2018 reply Follow Share $a)$ will be correct 0 votes 0 votes Peach commented Dec 13, 2018 reply Follow Share But why? If I put b1=2, b2=3, rank of AB becomes 1 but rank of A is 2.so no solution. Hence, it depends on the given b1 and b2.?? 0 votes 0 votes akash.dinkar12 commented Dec 13, 2018 reply Follow Share Since rank of matrix A and augmented matrix A|B is same which is 2 and the number of variables is 3 So there will be infinitely many solutions.. 0 votes 0 votes Peach commented Dec 13, 2018 reply Follow Share What about the condition when b1=2 ,b2=3, then rank of AB is 1 na?? 0 votes 0 votes akash.dinkar12 commented Dec 13, 2018 reply Follow Share when the rank of A is itself 2 then the rank of A|B cannot be 1, will always greater than or equals to 2.it can never be less than that..... 0 votes 0 votes Please log in or register to add a comment.
