Row Echelon Form: What A Row Of Zeros In The Matrix Means For Solving Systems Of Equations

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

FALSE. If the matrix A has more rows than columns, then all the variables can be basic and there will be a row of zeros (see the same example as in part a)).

If the REF (row echelon form) of the augmented matrix of a consistent equation Ax = b has a row of zeros, it means that the corresponding equation in the system has fewer variables than the other equations.

For example, consider the following system of equations:

2x + 3y – 5z = 10
-4x – 6y + 10z = -20
6x + 9y – 15z = 30

The augmented matrix for this system is:

\begin{bmatrix}
2 & 3 & -5 & 10 \\
-4 & -6 & 10 & -20 \\
6 & 9 & -15 & 30
\end{bmatrix}

We can use row operations to put this matrix into row echelon form:

\begin{bmatrix}
2 & 3 & -5 & 10 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}

In this case, the second and third rows contain all zeros after the row reduction. This means that there are fewer variables than equations, and therefore there are infinitely many solutions to the system of equations.

To see why this is the case, we can rewrite the equations in terms of the variables that are not free variables. In this case, the first column (corresponding to x) contains a leading coefficient (the 2 in the first row), so we can solve for x in terms of y and z:

2x + 3y – 5z = 10
2x = -3y + 5z + 10
x = -3/2 y + 5/2 z + 5

Now we can substitute this expression for x into the other two equations:

-4x – 6y + 10z = -20
6x + 9y – 15z = 30

Replacing x with -3/2 y + 5/2 z + 5, we get:

-4(-3/2 y + 5/2 z + 5) – 6y + 10z = -20
6(-3/2 y + 5/2 z + 5) + 9y – 15z = 30

Simplifying these equations, we get:

9y – 15z = 30
9y – 15z = 30

Since both equations are identical, any values of y and z that satisfy one of them will also satisfy the other. Therefore, there are infinitely many solutions to the system of equations.

In general, if the REF of the augmented matrix of a consistent equation Ax=b has one or more rows of zeros, then the system of equations has fewer equations than variables and therefore has infinitely many solutions. The number of free variables will be equal to the number of rows with zeros in the REF.

More Answers: