MATLAB Tutor, Part 11

MATLAB Tutor

Part 11: Row Operations on Matrices

In the last step of Part 10 of this Tutorial, we tried to solve the linear system of equations

       x + 4y + 3z = 10
     2x +   y –   z = -1
     3x –   y – 4z = 11

We found that the system had no solutions. In this part, we want to investigate why not. We will use matrix row operations to find out.

First we need to save the coefficients of the system in a matrix, and the right-hand side vector in another matrix, and then form the matrix of the augmented system. Enter

     A = [1 4 3
             2 1 -1
             3 -1 -4]
     b = [10; -1; 11]
     M = [A, b]          %Augmented matrix

The augmented matrix is an equivalent representation of the system of equations. When we multiply an equation by a constant and add it to another equation, then the solution set of the new system is the same as the old. This is what we are doing when we use row operations on the augmented matrix.

We multiply a row of the augmented matrix by a constant and add it to another row to get a simpler matrix. The solution set of the simplified system corresponding to the new augmented matrix is the same as the old. In the steps below, we will row reduce the augmented matrix to a very simple form, step by step. Later we will see that it can all be done with a single command.

First we will try to get all zeros in the first column beneath the 1 in the first row. To do that, we multiply the first row by -2 and add it to the second row. Then, we multiply the first row by -3 and add it to the third row. Recall that the ith row is just M(i, :) in MATLAB. Enter

M(2, :) = -2*M(1, :) + M(2, :)
M(3, :) = -3*M(1, :) + M(3, :)

Now, we need to get a 1 into the (2,2) element of the new augmented matrix. We can do that by multipling the second row by (-1/7). Enter

M(2, :) = (-1/7)*M(2, :)

Add multiples of the second row to the first and third rows to "zero out" the second column, except for the 1 in the (2,2) position.

Normally, the next step would be to divide the third row by a constant to get a 1 in the (3,3) position and then add multiples of the third row to the first two rows to get zeros above the 1 in the third column. That procedure will not work for this augmented matrix. Do you see now why the corresponding linear system has no solution?

The row reduction we have just accomplished can be done with a single MATLAB command, "rref", which stands for row-reduced echelon form. Enter

M = [A, b]
rref(M)

Note that this reduction goes a step futher than ours. If we had divided our row three by 20 and used the resulting 1 in the (3,4) place to eliminate the ones in the fourth column above it, then we would have obtained the "rref" result.

If a linear system has a solution, it is easy to just read it off from the row-reduced echelon form. Form the augmented matrix and use rref to solve each of the following systems from Part 10 of this Tutorial:
-         x - 2y +    z = 0
                2y - 8z = 8
       -4x + 5y + 9z = -9
-         x - 2y +    z + 2w = 0
                2y -   8z +   w   = 8
       -4x + 5y + 9z -   w   = -9