Proven: Infinitely Many Solutions In A Consistent Matrix Equation Ax=B Leads To Ref With A Row Of Zeros

HW 3: If a consistent equation Ax = b, where A is a square matrix, has infinitely many solutions, then the REF of the augmented matrix has a row of zeros.

TRUE. If the equation has infinitely many solutions, then at least one of the variables is free, meaning that at least one of the columns does not have a pivot. Since there are the same number of rows as columns, this means that there will be a row of zeros as well.

To prove that if a consistent equation Ax=b, where A is a square matrix, has infinitely many solution, then the REF of the augmented matrix has a row of zeros, we can use the following steps:

Step 1: Assume that the equation Ax=b has infinitely many solutions.

Step 2: We can write the general solution for the system of equations as x = x_p + x_h, where x_p is a particular solution and x_h is the solution to the homogeneous system Ax = 0.

Step 3: If the equation has infinitely many solutions, then the homogeneous system Ax = 0 must have non-trivial solutions. This is because if the homogeneous system has only the trivial solution (x=0), then the general solution for the system Ax = b would be x = x_p + 0 = x_p, which is not infinitely many.

Step 4: Let H be the matrix whose columns are basis vectors for the homogeneous system Ax=0.

Step 5: Since H has linearly independent columns (by definition of linearly independent), we can apply the Rank-Nullity theorem to find that rank(H) + nullity(H) = n, where n is the number of columns in H.

Step 6: Since H has at least one non-trivial solution (from Step 3), nullity(H) is at least 1, which implies that rank(H) is at most n-1.

Step 7: Let A’ be the augmented matrix of the system Ax=b. By definition of the augmented matrix, A’ is obtained by concatenating A and b as columns.

Step 8: By performing row operations on A’ to reduce it to REF, we can show that the rank of A’ is the same as the rank of A.

Step 9: Since the system Ax=b is consistent (from our assumption in Step 1), rank(A’) is at least 1, which implies that rank(A) is at least 1.

Step 10: Combining Step 6 and Step 9, we have rank(H) < rank(A), which implies that there must be at least one row of zeros in the REF of A'. Step 11: Therefore, if a consistent equation Ax=b, where A is a square matrix, has infinitely many solutions, then the REF of the augmented matrix has a row of zeros.

More Answers: