Section 2.4 Elementary row operations
Gaussian elimination, which we shall describe in detail presently, is an algorithm (a well-defined procedure for computation that eventually completes) that finds all solutions to any \(m\times n\) system of linear equations. It is defined via certain operations carried out on the augmented matrix. These operations change the matrix (and hence the system of linear equations associated with it), but they leave the set of solutions unchanged. There are three of them, which we now describe.
Subsection 2.4.1 First operation: interchanging two rows
Exchanging two rows in the augmented matrix is the same as writing the the equations in a different order. The equations themselves are unchanged, as is the set of all solutions. When we interchange row \(i\) and row \(j\text{,}\) we denote it by \(R_i\leftrightarrow R_j\text{.}\)
Example 2.4.1. Augmented matrix change (interchange rows).
The system of equations
corresponds to the augmented matrix
Now we interchange rows 1 and 3 (\(R_1\leftrightarrow R_3\)) to get the matrix
which corresponds to the equations
Subsection 2.4.2 Second operation: multiplying a row by a nonzero constant
If we have a linear equation and multiply both sides of the equation by a nonzero constant \(\lambda\text{,}\) (The symbol \(\lambda\) is the Greek letter lambda; it is the traditional name for this constant.) then the linear equation
becomes
and so
This would apply to any equation in a system, of course, and so when we apply this operation to row \(i\) we denote it by \(R_i \gets \lambda R_i\) (The symbol \(\gets\) means is replaced by). As long as \(\lambda\) is nonzero, this operation leaves the set of solutions unchanged.
Example 2.4.2. Augmented matrix change (multiply row by \(\lambda\)).
The system of equations
corresponds to the augmented matrix
Now we multiply row \(2\) by \(\lambda=-3\) (\(R_2\gets -3R_2\)) to get the matrix
which corresponds to the system of equations
Subsection 2.4.3 Third operation: adding a multiple of one row to another
Recall that one of our basic techniques for solving a system of equations is to pick a variable, make the coefficients of that variable equal in two equations and then take the difference of the equations to create a new one with that variable eliminated. In terms of the associated augmented matrix, this means we get a new row formed by subtracting the corresponding entries in the rows of the two equations. In other words, we replace one row by the difference of two rows. We use the notation \(R_i\gets R_i-R_j\) to indicate that row \(i\) is replaced by the difference of row \(i\) and row \(j\text{.}\) The set of solutions is unchanged. We have already seen that multiplying a row by a nonzero constants leaves the set of solutions unchanged; it is often convenient to combine these two steps as one: \(R_i\gets R_i+\lambda R_j.\) Note that \(\lambda=-1\) is the case of subtracting one row from another.
Example 2.4.3. Augmented matrix change (add multiple of one row to another).
Eliminating \(x\) from the second equation:
corresponds to the augmented matrix
Now we replace row 2 by the sum of row 2 and twice row 1 \((R_2\gets R_2+2R_1)\) to get the matrix
which corresponds to the system of linear equations
Subsection 2.4.4 Summary of the three elementary row operations
In summary, here is a table of the three elementary row operations:
Elementary Operation | Notation |
Interchange rows | \(R_i\leftrightarrow R_j\) |
Multiply row by a nonzero constant | \(R_i\gets\lambda R_i\text{,}\) \(\lambda\not=0\) |
Add multiple of one row to another | \(R_i\gets R_i+\lambda R_j\) |
Each operation changes the matrix and the associated system of linear equations, but it leaves the set of solutions unchanged.
Subsection 2.4.5 Changing a specific entry of a matrix using elementary row operations
We want to show that it is possible to change specific entries in a matrix in an advantageous way using elementary row operations. Specifically, we will show two things:
Any \(a_{i,k}\not=0\) may be changed to \(1\) with one elementary row operation.
If \(a_{i,k}\not=0\text{,}\) then any other entry in the same column may be changed to \(0\) with one elementary row operation.
Theorem 2.4.5. Change a matrix entry to \(0\) or \(1\) with elementary row operations.
If a matrix has an entry \(a_{i,k}\not=0\text{,}\) then, with one elementary row operation, it may be changed to \(1\text{.}\)
If some column \(k\) of a matrix has two non-zero entries \(a_{i,k}\) and \(a_{j,k}\) then, with one elementary row operation, (either) one of them may be changed to \(0\text{.}\)
Proof.
If a matrix has an entry \(a_{i,k}\not=0\text{,}\) then multiply the row \(R_i\) by \(\frac 1{a_{i,k}}\text{,}\) that is, carry out the elementary row operation \(R_i\gets \frac 1{a_{i,k}}R_i\text{.}\) In the resulting matrix, the \(i\)-\(k\) entry is \(\frac {a_{i,k}}{a_{i,k}}=1\text{.}\)
Suppose a matrix has two non-zero entries \(a_{i,k}\) and \(a_{j,k}\text{.}\) First, do the elementary row operation \(R_i\gets \frac 1{a_{i,k}}R_i\) to change \(a_{i,k}\) to \(1\text{.}\) Then, do the elementary row operation \(R_j\gets R_j-a_{j,k}R_i\text{.}\) The matrix now has a new value in the \(j\)-\(k\) position which is \(a_{j,k}-a_{j,k}=0\text{,}\) and we have accomplished our goal using two elementary operations. Now we note that these two operations can be carried out with the single operation \(R_j\gets R_j-\frac {a_{j,k}}{a_{i,k}}R_i \text{.}\)
Exercises Exercises
For these exercises, let \(A= \begin{bmatrix} 1\amp2\amp3\amp4\\ 5\amp6\amp7\amp-6\\ -5\amp-4\amp-3\amp-2 \end{bmatrix} \text{.}\)
1.
Find the elementary row operation that
changes the \(2\) in the first row to a \(1\)
changes the \(7\) in the second row to a \(1\)
changes the \(-4\) in the third row to a \(1\)
In each case give the matrix that results from the application of your elementary row operation.
- \(\displaystyle R_1\gets \frac12 R_1\colon \begin{bmatrix} \frac12\amp1\amp\frac32\amp\frac12\\ 5\amp6\amp7\amp-6\\ -5\amp-4\amp-3\amp-2 \end{bmatrix}\)
- \(\displaystyle R_2\gets \frac17 R_2\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ \frac57\amp\frac67\amp1\amp-\frac67\\ -5\amp-4\amp-3\amp-2 \end{bmatrix}\)
- \(\displaystyle R_3\gets -\frac14 R_3\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ 5\amp6\amp7\amp-6\\ \frac54\amp1\amp\frac34\amp\frac12 \end{bmatrix}\)
2.
Find an elementary row operation that
- changes the \(5\) in the second row to a \(0\)
- changes the \(6\) in the second row to a \(0\)
- changes the \(-4\) in the third row to a \(0\)
- changes the \(7\) in the second row to a \(0\)
In each case give the matrix that results from the application of your elementary row operation.
- \(\displaystyle R_2\gets R_2+R_3\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ 0\amp2\amp4\amp-8\\ -5\amp-4\amp-3\amp-2 \end{bmatrix}\)
- \(\displaystyle R_2\gets R_2-3R_1\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ 2\amp0\amp-2\amp-18\\ -5\amp-4\amp-3\amp-2 \end{bmatrix}\)
- \(\displaystyle R_3\gets R_3+2R_1\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ 5\amp6\amp7\amp-6\\ -3\amp0\amp3\amp6 \end{bmatrix}\)
- \(\displaystyle R_2\gets R_2-\frac73 R_1\colon \begin{bmatrix} 1\amp2\amp3\amp4\\ \frac83\amp\frac43\amp0\amp-\frac{46}3\\ -5\amp-4\amp-3\amp-2 \end{bmatrix}\)
3.
Use two elementary row operations to make the first column of the resulting matrix \(\begin{bmatrix} 1\\0\\0 \end{bmatrix}\)
\(R_2\gets R_2-5R_1\) and \(R_3\gets R_3+5R_1\) give