Objectives
-
Define a elementary matrix.
-
Show that the row operations from the previous chapter can be written as matrix products with elementary matrices.
-
Reexamination a reduced row-echelon form of a matrix with elementary matrices.
Section 2.4 Elementary Matrices
In Chapter 1, linear systems and solutions were introduced. The standard way to solve these is to use elementary row operations as were shown in Section 1.3. In this section, we learn that we can write all of the row operations as matrix products.
We will denote by the matrix \(E\) any matrix that performs an elementary row operations. For example,
\begin{equation*}
E = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix}
\end{equation*}
will swap the rows in a 2 by 2 matrix. The matrix
\begin{equation*}
E = \begin{bmatrix} 2 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}
\end{equation*}
will multiply the first row of a 3 by 3 matrix by 2. The matrix
\begin{equation*}
E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ -2 \amp 0 \amp 1 \end{bmatrix}
\end{equation*}
is the operation that multiplies the first row of a 3 by 3 matrix by \(-2\) and add to the third row.
Definition 2.4.1. Elementary Matrices and Row Operations.
Let \(E\) be a \(n \times n \) matrix. The matrix \(E\) is an elementary matrix is the result of applying one row operation to \(I_n\text{,}\) the \(n \times n\) identity matrix.
An elementary matrix that involves switching rows is also called a permutation matrix .
Although we have examples above, here are the specifics of the related row operations and the elementary matrix.
- \(R_i \leftrightarrow R_j\)
- \(c R_i \)
- \(E_{cR_i}\) is the identity matrix where the element is row \(i\text{,}\) column \(i\) is replaced with \(c\text{.}\)
- \(c R_i + R_j\)
- \(E_{cR_i + R_j}\) is the identity matrix with the element in row \(j\text{,}\) column \(i\) has the number \(c\text{.}\)
Example 2.4.2.
Consider \(4 \times 4\) elementary matrices. Find the following:
-
\(\displaystyle E_{1 \leftrightarrow 3}\)
-
\(\displaystyle E_{-2 R_2 + R_4}\)
-
\(\displaystyle E_{7 R_3}\)
Solution.
- \(E_{1 \leftrightarrow 3}\)
- \begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
- \(E_{-2 R_2 + R_4}\)
- \begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp -2 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
- \(E_{7 R_3}\)
- \begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 7 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
There are a lot of nice properties of elementary matrices. We now show that all elementary matrices are invertible.
Theorem 2.4.3.
Every elementary matrix is invertible and its inverse is also an elementary matrix. In paticular,
-
The inverse of \(E_{i \leftrightarrow i}\) is itself.
Example 2.4.4.
Find the inverses of the following elementary matrices and the determine the row operation the inverse represents
-
\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \\ -2 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
-
\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
-
\begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \end{bmatrix} \end{equation*}
Solution.
-
Note that this represents the operation \(-2R_1+R_2\text{,}\) so the inverse is \(E_{2R_1+R_2}\) or\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 2 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
-
\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1/3 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
-
This is the operation \(R_1 \leftrightarrow R_3\text{,}\) so the inverse is also the matrix \(E_{1 \leftrightarrow 3}\) or\begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \end{bmatrix} \end{equation*}
We next look at putting matrix is reduced row-echelon form using elementary matrices. The following lemma indicates that we can do this.
Lemma 2.4.5.
\begin{equation*}
B = UA
\end{equation*}
where \(U\) is the product of elementary matrices representing the row operations performed on \(A\) to obtain \(B\text{.}\)
Let’s take a look at an example.
Example 2.4.6.
Consider the matrix
\begin{equation*}
A = \begin{bmatrix}
1 \amp 2 \amp 0 \\
0 \amp -2 \amp 2 \\
3 \amp 0 \amp 1
\end{bmatrix}
\end{equation*}
Find all elementary matrices to put \(A\) in row-echelon form and show that the product of the matrices form \(U\text{.}\)
Solution.
The first step is \(-3R_1+R_3 \to R_3\text{,}\) which has the corresponding elementary matrix
\begin{equation*}
E_1 = \begin{bmatrix}
1 \amp 0 \amp 0 \\
0 \amp 1 \amp 0 \\
-3 \amp 0 \amp 1
\end{bmatrix}
\end{equation*}
And the product \(E_1A\) is
\begin{equation*}
E_1 A = \begin{bmatrix}
1 \amp 2 \amp 0 \\
0 \amp -2 \amp 2 \\
0 \amp -6 \amp 1
\end{bmatrix}
\end{equation*}
Next, the step \(-3R_2 + R_3 \to R_3\) which correspond to the elementary matrix
\begin{equation*}
E_2 =\begin{bmatrix}
1 \amp 0 \amp 0 \\
0 \amp 1 \amp 0 \\
0 \amp -3 \amp 1
\end{bmatrix}
\end{equation*}
And now the product of the elementary matrices and \(A\) is
\begin{equation*}
E_2 E_1 A = \begin{bmatrix}
1 \amp 0 \amp 2 \\
0 \amp -2 \amp 2 \\
0 \amp 0 \amp -5
\end{bmatrix}
\end{equation*}
And since this matrix is an upper triangular matrix, we stop here. The matrix on the right hand side is \(B\) and the matrix \(U\) is
\begin{equation*}
U = E_2 E_1 = \begin{bmatrix}
1 \amp 0 \amp 0 \\
0 \amp 1 \amp 0 \\
-3 \amp -3 \amp 1
\end{bmatrix}
\end{equation*}
