Homework SourseFor the system AX B tell whether the system is consistent o
Solution
first find the basic theorem as below:
Consider any system of linear equation Ax = B, where A has dimension m x n.
Also consider the augmented matrix for this system [A|B].
Let p be the rank of A and q be the rank of [A|B]
Then the system has:
(a) no solution if p < q
(b) a unique solution if p = n, (i.e., if A has full rank, in which case p = q)
(c) infinitely many solutions if p = q but p < n
a)
as given we have a matrix A which has 3 rows and 4 columns.
we can say that the system has 3 linear equation with 4 variable.
and as given rank of the matrix is 3.
we know that the rank of the matrix is not greater than the number of non zero rows in a row echelon form of a specific matrix.
rank of a matrix = number of non zero rows in a row echelon form of a matrix.
we can say that augmented matrix [A|B] has 3 rows and 5 columns.
and as given rank of matrix A = 3 so it has 3 non zero rows.
we can say that the augmented matrix also has a 3 no zero rows.
hence rank of augmented matrix [A|B] is also equal to 3
we know that if rank of coefficient matrix A = rank of augmented matrix(B) then system is consistent.
we have rank(A) = rank([A|B]) = 3 hence we can say that system is consistent.
and as number of variable is greater than the number of linear equations we can say that the system has infinitely many solutions.
2)
A is 4x7 matrix , rank of A = 2 and rank [A|B] = 3
we know that,
if rank of coefficient matrix A = rank of augmented matrix(B) then system is consistent otherwise it is inconsistent.
we have,
rank of A = 2 and rank [A|B] = 3 so we have rank(A) is not equal to rank([A|B]
so we can say that system is incosistent.
and also rank of matrix A is less than rank of [A|B] we can say that the system has no solution.
