Elementary matrices are square matrices that are obtained by performing a single elementary row operation on the identity matrix. Elementary matrices are invertible, and their inverses are also elementary matrices of the same type. Furthermore, any invertible matrix can be written as a product of elementary matrices. This result is known as the elementary matrix decomposition theorem and is a powerful tool for solving linear systems of equations. In this lesson, you will learn about finding elementary matrices, expressing an invertible matrix as a product of elementary matrices, and how to find the inverse of a matrix using row reduction. These topics are essential in linear algebra and provide a foundation for more advanced concepts.
Add a constant times one row to another:
$R_i\to R_i +c\cdot R_j$.
Performing one of these operations on a matrix, $A$ corresponds to multiplying $A$ on the left by a matrix $E$. Such a matrix $E$ is known as an elementary matrix.
Elementary Matrices
Definition 1. Two matrices are said to be row equivalent if one can be obtained from the other by doing some row operations.
Definition 2. A square matrix is called an elementary matrix if it can be obtained by performing one elementary row operation to an identity matrix.
To find an elementary matrix corresponding to an elementary row operation, we apply the same operation on the identity matrix, and the resulting matrix is the corresponding elementary matrix.
Here are the steps to find an elementary matrix:
- Start with the identity matrix $I$.
- Apply the elementary row operation to the identity matrix. This will result in a new matrix that represents the elementary row operation.
- The resulting matrix is the elementary matrix that corresponds to the row operation you performed.
Example 1.
- The matrix $\begin{bmatrix}
1 &0 &0\\
2 &1 & 0\\
0 &0 & 1
\end{bmatrix}$ is an elementary matrix. This is the elementary matrix corresponding to the elementary row operation $R_2\to R_2+2R_1$ applied to $I_3$.
- The matrix $\begin{bmatrix}
1 &0 &0 & 0\\
0 &0 & 0 & 1\\
0 &0 & 1& 0\\
0 &1 & 0& 0
\end{bmatrix}$ is an elementary matrix. This is the elementary matrix corresponding to the elementary row operation $R_2\leftrightarrow R_4$ applied to $I_4$.
- The matrix $\begin{bmatrix}
1 &0 &0 & -9\\
0 &1 & 0 & 0\\
0 &0 & 1& 0\\
0 &0 & 0& 1
\end{bmatrix}$ is an elementary matrix. This is the elementary matrix corresponding to the elementary row operation $R_1\to R_1-9 R_4$ applied to $I_4$.
- The matrix $\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & -65
\end{bmatrix}$ is an elementary matrix. This is the elementary matrix corresponding to the elementary row operation $R_3\to -65R_3$ applied to $I_3$.
Example 2.
Let $A=\left[\begin{array}{rrrrr}
1 & -1 & -1 & 2 & 4\\
3 & -3 & -2 & 8 & 10\\
4 & -4 & -5 & 6 & 18
\end{array}\right]$ and let $B$ be the matrix obtained from $A$ by performing the row operation $R_2\to R_2-3R_1$.
- Find $B$.
- Find the corresponding elementary matrix.
- Compute $EA$.
Answer 0.1.
\begin{eqnarray*}
A=\left[\begin{array}{rrrrr}
1 & -1 & -1 & 2 & 4\\
3 & -3 & -2 & 8 & 10\\
4 & -4 & -5 & 6 & 18
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-3R_1}\\
{}
\end{array}
B=\left[\begin{array}{rrrrr}
1 & -1 & -1 & 2 & 4\\
0 & 0 & 1 & 2 & -2\\
4 & -4 & -5 & 6 & 18
\end{array}\right]
\end{eqnarray*}
- Since $A$ is a $3\times 5$, then $E$ is a $3\times 3$ matrix. Applying the same elementary operation on $I_3$, we obtain
\begin{eqnarray*}
I_3=\left[\begin{array}{rrrrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-3R_1}\\
{}
\end{array}
E=\left[\begin{array}{rrrrr}
1 & 0 & 0 \\
-3 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{eqnarray*}
- $EA=\begin{bmatrix}
1 & 0 & 0 \\
-3 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & -1 & -1 & 2 & 4\\
3 & -3 & -2 & 8 & 10\\
4 & -4 & -5 & 6 & 18
\end{bmatrix}=\begin{bmatrix}
1 & -1 & -1 & 2 & 4\\
0 & 0 & 1 & 2 & -2\\
4 & -4 & -5 & 6 & 18
\end{bmatrix}.$
Theorem 3. If the elementary matrix $E$ results from performing a certain row operation on $I_m$ and
if $A$ is an $m \times n$ matrix, then the product $EA$ is the matrix that results when this same row operation is performed on $A$.
Example 3.
Let $A=\begin{bmatrix}
1 & -1 & -1 \\
3 & -3 & -2 \\
4 & -4 & -5\\
0 & -2 & -9 \\
1 & 8 & 5
\end{bmatrix}$ and let $B$ be the matrix obtained from $A$ by performing the row operation $R_2\leftrightarrow R_3$.
- Find $B$.
- Find the corresponding elementary matrix.
- Compute $EA$ and compare with $B$.
Answer 0.2.
\begin{eqnarray*}
A=\begin{bmatrix}
1 & -1 & -1 \\
3 & -3 & -2 \\
4 & -4 & -5\\
0 & -2 & -9 \\
1 & 8 & 5
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_2\leftrightarrow R_5}\\
{}
\end{array}
B=\begin{bmatrix}
1 & -1 & -1 \\
1 & 8 & 5 \\
4 & -4 & -5\\
0 & -2 & -9 \\
3 & -3 & -2
\end{bmatrix}
\end{eqnarray*}
- Since $A$ is a $5\times 3$, then $E$ is a $5\times 5$ matrix. Applying the same elementary operation on $I_3$, we obtain
\begin{eqnarray*}
I_5=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_2\leftrightarrow R_5}\\
{}
\end{array}
E=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 0
\end{bmatrix}
\end{eqnarray*}
- $EA=\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 1 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
1 & -1 & -1 \\
3 & -3 & -2 \\
4 & -4 & -5\\
0 & -2 & -9 \\
1 & 8 & 5
\end{bmatrix}=\begin{bmatrix}
1 & -1 & -1 \\
1 & 8 & 5 \\
4 & -4 & -5\\
0 & -2 & -9 \\
3 & -3 & -2
\end{bmatrix}=B.$
Theorem 4. Every elementary matrix is invertible, and the inverse is also an elementary matrix.
To find the inverse of an elementary matrix $E$, we have the following:
- If $E$ is the elementary matrix resulting from applying the elementary row operation $R_j\leftrightarrow R_i$ to $I_m$, then $E^{-1}$ is the elementary matrix resulting from applying the elementary row operation $R_j\leftrightarrow R_i$ to $I_m$. In this case $E^{-1}=E$.
- If $E$ is the elementary matrix resulting from applying the elementary row operation $R_j\to cR_j$ to $I_m$, then $E^{-1}$ is the elementary matrix resulting from applying the elementary row operation $R_j\to \frac{1}{c}R_j$ to $I_m$.
- If $E$ is the elementary matrix resulting from applying the elementary row operation $R_j\to R_j+cR_i$ to $I_m$, then $E^{-1}$ is the elementary matrix resulting from applying the elementary row operation $R_j\to R_j-cR_i$ to $I_m$.
Example 4.
- The elementary matrix $E=\begin{bmatrix}
1 &0 &0\\
2 &1 & 0\\
0 &0 & 1
\end{bmatrix}$ results from applying the elementary row operation $R_2\to R_2+2R_1$ to $I_3$. Therefore its inverse is the matrix resulting from applying the elementary row operation $R_2\to R_2-2R_1$ to $I_3$. We have
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 &0 &0\\
1 &1 & 0\\
0 &0 & 1
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-2R_1}\\
{}
\end{array}
E^{-1}=\begin{bmatrix}
1 &0 &0\\
-2 &1 & 0\\
0 &0 & 1
\end{bmatrix}
\end{eqnarray*}
Moreover,
$$\begin{bmatrix}
1 &0 &0\\
2 &1 & 0\\
0 &0 & 1
\end{bmatrix}\begin{bmatrix}
1 &0 &0\\
-2 &1 & 0\\
0 &0 & 1
\end{bmatrix}=\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & 1
\end{bmatrix}$$
and
$$\begin{bmatrix}
1 &0 &0\\
-2 &1 & 0\\
0 &0 & 1
\end{bmatrix}\begin{bmatrix}
1 &0 &0\\
2 &1 & 0\\
0 &0 & 1
\end{bmatrix}=\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & 1
\end{bmatrix}$$
which shows that they are inverse to each other.
- The elementary matrix $\begin{bmatrix}
1 &0 &0 & 0\\
0 &0 & 0 & 1\\
0 &0 & 1& 0\\
0 &1 & 0& 0
\end{bmatrix}$ results from applying the elementary row operation $R_2\leftrightarrow R_4$ applied to $I_4$. Therefore its inverse is the matrix resulting from applying the row elementary row operation $R_2\leftrightarrow R_4$ to $I_4$.
\begin{eqnarray*}
I_4=\begin{bmatrix}
1 &0 &0 & 0\\
0 &1 & 0 & 0\\
0 &0 & 1& 0\\
0 &0 & 0& 1
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_2\leftrightarrow R_4}\\
{}
\end{array}
E^{-1}=\begin{bmatrix}
1 &0 &0 & 0\\
0 &0 & 0 & 1\\
0 &0 & 1& 0\\
0 &1 & 0& 0
\end{bmatrix}
\end{eqnarray*}
One can verify that they are inverse to each other.
The elementary matrix $\begin{bmatrix}
1 &0 &0 & -9\\
0 &1 & 0 & 0\\
0 &0 & 1& 0\\
0 &0 & 0& 1
\end{bmatrix}$ results from applying the elementary row operation $R_1\to R_1-9R_4$ applied to $I_4$. Therefore its inverse is the matrix resulting from applying the row elementary row operation $R_1\to R_1+9R_4$ to $I_4$.
\begin{eqnarray*}
I_4=\begin{bmatrix}
1 &0 &0 & 0\\
0 &1 & 0 & 0\\
0 &0 & 1& 0\\
0 &0 & 0& 1
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_1\to R_1+9R_4}\\
{}
\end{array}
E^{-1}=\begin{bmatrix}
1 &0 &0 & 9\\
0 &1 & 0 & 0\\
0 &0 & 1& 0\\
0 &0 & 0& 1
\end{bmatrix}
\end{eqnarray*}
The elementary matrix $\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & -65
\end{bmatrix}$ results from applying the elementary row operation $R_3\to -65R_3$ applied to $I_3$. Therefore its inverse is the matrix resulting from applying the row elementary row operation $R_3\to \dfrac{1}{-65}R_3$ to $I_3$.
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & 1
\end{bmatrix}
\begin{array}{r}
\\
\underrightarrow{R_3\to \dfrac{1}{-65}R_3}\\
{}
\end{array}
E^{-1}=\begin{bmatrix}
1 &0 &0\\
0 &1 & 0\\
0 &0 & \dfrac{1}{-65}
\end{bmatrix}
\end{eqnarray*}
The Equivalence Theorem
The aim is this section is to state a theorem that relates to results we have obtained about the invertibility of matrices, homogeneous linear systems, and reduced rows. There is a fundamental result in linear algebra that relates these concepts. It is known as “The Equivalence Theorem”.
Theorem 5 (
Equivalent statements)
. If $A$ is an $n\times n$ matrix, then the following statements are equivalent, that is, all true or all false.
- $A$ is invertible.
- $AX = 0$ has only the trivial solution.
- The reduced row echelon form of $A$ is $I_n$.
- $A$ is expressible as a product of elementary matrices.
This means that if any one of these statements is true for a matrix $A$, then the other three statements are also true. Furthermore, if any one of these statements is false for matrix $A$, then the other three statements are also false.
A method for inverting matrices
In this section, we show a method to find the inverse of an invertible matrix by decomposing it into a product of elementary matrices. Also, we give an algorithm to find the inverse of any invertible matrix.
Algorithm to express $A^{-1}$ and $A$ as a product of elementary matrices:
To find the inverse of an invertible matrix $A$, we have to follow the following steps:
- find a sequence of elementary row operations that reduces $A$ to the identity,
- for each elementary for operation, find the corresponding elementary matrix $E_i$,
- write $E_kE_{k-1}\cdots E_2E_1A=I_n$,
- then $A^{-1}=E_kE_{k-1}\cdots E_2E_1$ so that $$A=\left(E_kE_{k-1}\cdots E_2E_1\right)^{-1}=E_1^{-1}E_2^{-1}\cdots E_{k-1}^{-1}E_{k}^{-1}$$
The following theorem gives a direct method to find the inverse of any invertible matrix $A$.
Theorem 6. If an $n\times n$ matrix $A$ has an inverse, it can be found by performing elementary row operations on the augmented matrix $[A|I_n]$ to reduce $A$ to $I_n$. In the process $I_n$ becomes $A^{-1}$.
Algorithm to find $A^{-1}$:
An algorithm to find the inverse of a matrix involves the following steps:
- Augment the original matrix $A$ with an identity matrix $I$ to form an augmented matrix $[A|I]$.
- Use elementary row operations to transform $[A|I]$ into $[I|B]$, where $B$ is the inverse of $A$.
- If $A$ is not invertible, the row reduction process will lead to a row of zeros on the diagonal of $[A|I]$. In this case, the matrix is singular, and it does not have an inverse.
Example 5. Let $A=\begin{bmatrix}
1 & 0 & 2\\
2 & 1 & 1\\
-1 & -2 & 1
\end{bmatrix}$.
- Find $A^{-1}$ if it exists.
- Express $A^{-1}$ as a product of elementary matrices.
- Express $A$ as a product of elementary matrices.
Answer 0.3.
-
\begin{eqnarray*}
& & \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 & 0 &0\\
2 & 1 & 1 & 0 & 1 &0\\
-1 & -2 & 1 & 0 & 0 &1
\end{array}\right]
\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-2R_1}\\
R_3\to R_3+R_1
{}
\end{array}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 & 0 &0\\
0 & 1 &-3 &-2 & 1 &0\\
0 &-2 & 3 & 1 & 0 &1
\end{array}\right]\\
& &\begin{array}{r}
\\
\\
\underrightarrow{R_3\to R_3+2R_2}\\
{}
\end{array}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 & 0 &0\\
0 & 1 &-3 &-2 & 1 &0\\
0 & 0 &-3 &-3 & 2 &1
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_3\to -\frac{1}{3} R_3}\\
{}
\end{array}\\
&&
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 & 0 &0\\
0 & 1 &-3 &-2 & 1 &0\\
0 & 0 & 1 & 1 & -\frac{2}{3} &-\frac{1}{3}
\end{array}\right]
\begin{array}{r}
\\
\\
\underrightarrow{R_2\to R_2+3R_3}\\
{}\\
{}
\end{array}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 & 0 &0\\
0 & 1 & 0 & 1 &-1 &-1\\
0 & 0 & 1 & 1 & -\frac{2}{3} &-\frac{1}{3}
\end{array}\right]\\
&&\begin{array}{r}
\\
\\
\underrightarrow{R_1\to R_1-2 R_3}\\
{}
\end{array}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & \frac{4}{3} &\frac{2}{3}\\
0 & 1 & 0 & 1 &-1 &-1\\
0 & 0 & 1 & 1 & -\frac{2}{3} &-\frac{1}{3}
\end{array}\right]
\end{eqnarray*}
This proves that $A$ is invertible and $A^{-1}=\begin{bmatrix}
-1 & \frac{4}{3} &\frac{2}{3}\\
1 &-1 &-1\\
1 & -\frac{2}{3} &-\frac{1}{3}
\end{bmatrix}$.
- For each of the elementary operations used above to reduce $A$ to $I_3$, we find its corresponding elementary matrix.
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_2\to R_2-2R_1}\\
{}
\end{array}E_1=\begin{bmatrix}
1 & 0 &0\\
-2 & 1 &0\\
0 & 0 &1
\end{bmatrix}
\end{eqnarray*}
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_3\to R_3+R_1}\\
{}
\end{array}E_2=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
1 & 0 &1
\end{bmatrix}
\end{eqnarray*}
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_3\to R_3+2R_2}\\
{}
\end{array}E_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 2 &1
\end{bmatrix}
\end{eqnarray*}
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_3\to -\frac{1}{3} R_3}\\
{}
\end{array}E_4=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &-\frac{1}{3}
\end{bmatrix}
\end{eqnarray*}
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_2\to R_2+3R_3}\\
{}
\end{array}E_5=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &3\\
0 & 0 &1
\end{bmatrix}
\end{eqnarray*}
\begin{eqnarray*}
I_3=\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{array}{r}
\\
\underrightarrow{R_1\to R_1-2R_3}\\
{}
\end{array}E_6=\begin{bmatrix}
1 & 0 &-2\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}
\end{eqnarray*}
Therefore $A^{-1}=E_6E_5E_4E_3E_2E_1$. That is,
\begin{eqnarray*}
A^{-1}&=&\begin{bmatrix}
1 & 0 &-2\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
0 & 1 &3\\
0 & 0 &1
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &-\frac{1}{3}
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 2 &1
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
1 & 0 &1
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
-2 & 1 &0\\
0 & 0 &1
\end{bmatrix}
\end{eqnarray*}
- $A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}E_5^{-1}E_6^{-1}$.
That is,
\begin{eqnarray*}
A&=&
\begin{bmatrix}
1 & 0 &0\\
2 & 1 &0\\
0 & 0 &1
\end{bmatrix}
\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
-1 & 0 &1
\end{bmatrix}
\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & -2 &1
\end{bmatrix}
\begin{bmatrix}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &-3
\end{bmatrix}
\begin{bmatrix}
1 & 0 &0\\
0 & 1 &-3\\
0 & 0 &1
\end{bmatrix}
\begin{bmatrix}
1 & 0 &2\\
0 & 1 &0\\
0 & 0 &1
\end{bmatrix}
\end{eqnarray*}
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