The reduced row-echelon forms of the augmented matrices of four systems are give

Question

The reduced row-echelon forms of the augmented matrices of four systems are given below. How many solutions does each system have?

1.

A. Unique solution
B. No solutions
C. Infinitely many solutions
D. None of the above

2.

A. Unique solution
B. No solutions
C. Infinitely many solutions
D. None of the above

3.

A. No solutions
B. Infinitely many solutions
C. Unique solution
D. None of the above

4.

A. Infinitely many solutions
B. No solutions
C. Unique solution
D. None of the above

Explanation / Answer

The general rules to follow (as long as the number of rows (equations) is equal to the number of columns in the coefficient matrix (the variables)):

In the reduced row-echelon form, if there is no row of all 0's, on the coefficient side, then the system is consistent with a unique solution. If there is a row of all zeroes, then the system is consistent if the augmented column in that row is also 0, as when you multiply 0 by all the variables in the system you should get 0! In that case, the system will have infinitely many solutions, as the zero row means an equation has cancelled out, leaving the system underdetermined. On the other hand, if the augmented column corresponding to a row of all 0's is not 0, then the system is inconsistent (no solution), because you are taking a linear combination of all 0's and getting something that isn't 0, which is impossible.

If the number of rows is less than the number of columns in the coefficient matrix, there are less equations than there are variables. The system is underdetermined. This is equivalent to having rows of all zeros appended at the bottom until the number of equations is equal to the number of variables -- you can always add more equations that just equal themselves.

If the number of rows is greater than the number of columns in the coefficient matrix, there are more equations than there are variables, and the system is overdetermined. There will be at least one row of all 0's in reduced row echelon form and from that you can determine if the system is consistent or not according to whether the augmented column is zero or nonzero. However, even with a row of all 0's that equals a 0 in place, you will only have infinitely many solutions if there are less nonzero rows than the number of variables (i.e., the number of equations that didn't cancel out is less than the number of variables); if the number of nonzero rows equals the number of columns in the coefficient matrix and the system is consistent, there will be exactly one solution.

Bearing all that in mind:

1) B - no solution, because the third row has all zeros equalling a 1.

2) B - no solution, because the third row has all zeros equalling a 1.

3) B - infinitely many solutions; the system is underdetermined, with only 2 rows and 3 variable coefficients. You can think of adding a third row of all zeroes to make the coefficient matrix 3x3. Either way of looking at it, the system is dependent.

4) C - unique solution -- two equations, two variables.

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