4. Solving systems of linear equations via invertible matrices

There is a special matrix denoted by $I$ and referred to as the identity matrix. It's number of rows equals to it's number of columns, and it has ones down the main diagonal (from upper left cornet to the lower right corner) and zeroes elsewhere. For example,

$\begin{pmatrix}1\end{pmatrix},\ \begin{pmatrix}1&0\\0&1\end{pmatrix},\ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

are $1\times 1$ identity matrix, $2\times 2$ identity matrix, and $3\times 3$ identity matrix, respectively.

The identity matrix has the following property.

Lemma. Suppose that $A$ is an $m\times n$ matrix and $I_n$ is the $n\times n$ identity matrix. Then $AI_n=A$. If $I_m$ is the $m\times m$ identity matrix, it also follows that $I_mA=A$.

Verify the lemma in case of $A=\begin{pmatrix}1&3&2&4\\-4&-2&-3&-1\\5&8&6&7\end{pmatrix}$ by calculating

$\begin{pmatrix}1&3&2&4\\-4&-2&-3&-1\\5&8&6&7\end{pmatrix} \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\quad$ and $\quad\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} \begin{pmatrix}1&3&2&4\\-4&-2&-3&-1\\5&8&6&7\end{pmatrix}$.

Let us have a system of linear equations $A\mathbf{x}=\mathbf{b}$, where $A=(a_{ij})$, $i,j=1,\dots,n$, $\mathbf{x}=\begin{pmatrix}x_1&\dots&x_n\end{pmatrix}^T$, and $\mathbf{b}=\begin{pmatrix}b_1&\dots&b_n\end{pmatrix}^T$. And suppose that there exists an $n\times n$ matrix $D$ such that $DA=I_n$. Then the following is true:

\begin{align} A\mathbf{x}&=\mathbf{b}\Rightarrow\\ DA\mathbf{x}&=D\mathbf{b}\Rightarrow\\ I_n\mathbf{x}&=D\mathbf{b}\Rightarrow\\ \mathbf{x}&=D\mathbf{b}. \end{align}

It provides us with an alternative way of calculating $\mathbf{x}$. By knowing the matrix $D$, solving the system of linear equation means just calculating the product $Db$.

Let $A=\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}1\\2\\3\end{pmatrix}$, and $D=\begin{pmatrix}0&\frac{1}{2}&\frac{1}{2}\\1&-1&0\\1&-\frac{1}{2}&-\frac{1}{2}\end{pmatrix}$.
Verify that $DA=I$ and solve the system $A\mathbf{x}=\mathbf{b}$ by usind the matrix $D$.
Verify the obtained result by Gaussian elimination algorithm.

Solving the systems of linear equations like this seems really simple, isn't it. Moreover, by knowing $D$ we can solve the system easily for many different values of $\mathbf{b}$.

Solve the system of linear equations form previous exercise for $\mathbf{b}=\begin{pmatrix}0\\1\\3\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}1\\1\\3\end{pmatrix}$.

Answer

But the catch is, how to find the matrix $D$? For answering the question, we will need the following theorem.

Theorem. Suppose that $A$ and $D$ are $n\times n$ matrices such that $DA=I$. Then also $AD=I$.

Thus, for solving the system $A\mathbf{x}=\mathbf{b}$, we are looking for a $n\times n$ matrix $D$ such that $AD=I$. It is a matrix equation where $D$ is the unknown. From previous we know (see Gaussian elimination algorithm) how to solved similar equations where the unknown is and $n\times 1$ vector and on the right hand side is also an vector - this is were we are heading.

Note, that we can express the product $AD$ in the following way

$AD=\begin{pmatrix}A\mathbf{x_1}&A\mathbf{x_2}&\dots&A\mathbf{x_n}\end{pmatrix}$

where $\mathbf{x_i}$ is the $i^{th}$ column of matrix $D$ (i.e. $D=\begin{pmatrix}\mathbf{x_1}&\mathbf{x_2}&\dots&\mathbf{x_n}\end{pmatrix}$). For example,

$$\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix} \begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}= \begin{pmatrix}\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}a\\d\\g\end{pmatrix}& \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}b\\e\\h\end{pmatrix}& \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}c\\f\\i\end{pmatrix}\end{pmatrix}. \qquad (\ast)$$

Calculate the second column of the product

$$\begin{pmatrix}1&2&2\\1&0&2\\2&2&4\end{pmatrix}\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}$$

without calculating the whole product.

Answer

Hence,

$AD=\begin{pmatrix}A\mathbf{x_1}&A\mathbf{x_2}&\dots&A\mathbf{x_n}\end{pmatrix}= \begin{pmatrix}\mathbf{e_1}&\mathbf{e_2}&\dots&\mathbf{e_n}\end{pmatrix}$,

where $\mathbf{e_i}$ is the $i^{th}$ column of matrix $I$ (i.e. $I=\begin{pmatrix}\mathbf{e_1}&\mathbf{e_2}&\dots&\mathbf{e_n}\end{pmatrix}$).

Since two matrices are equal if and only if all the corresponding elements (or groups of elements, e.g., column vectors) are equal, solving the matrix equation $AD=I$ is equivalent to solving the following systems of linear equations

$A\mathbf{x_1}=\mathbf{e_1},\ A\mathbf{x_2}=\mathbf{e_2},\ \dots,\ A\mathbf{x_n}=\mathbf{e_n}$.

In the example $(\ast)$, the matrix equation

$$\begin{pmatrix}\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}a\\d\\g\end{pmatrix}& \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}b\\e\\h\end{pmatrix}& \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}c\\f\\i\end{pmatrix}\end{pmatrix}= \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$

is equivalent to solving the systems

$$\begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}a\\d\\g\end{pmatrix}= \begin{pmatrix}1\\0\\0\end{pmatrix},\ \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}b\\e\\h\end{pmatrix}= \begin{pmatrix}0\\1\\0\end{pmatrix},\ \begin{pmatrix}1&0&1\\1&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}c\\f\\i\end{pmatrix}= \begin{pmatrix}0\\0\\1\end{pmatrix}. $$

Now, for finding the $\mathbf{x_i},\ i=1,\dots,n$, we can apply $n$ times Gaussian elimination algorithm:

\begin{align} (A|\mathbf{e_1})&\longrightarrow (I|\mathbf{c_1})\\ (A|\mathbf{e_2})&\longrightarrow (I|\mathbf{c_2})\\ \dots\\ (A|\mathbf{e_n})&\longrightarrow (I|\mathbf{c_n}). \end{align}

Thus, the matrix $D=\begin{pmatrix}\mathbf{c_1}&\mathbf{c_2}&\dots&\mathbf{c_n}\end{pmatrix}$.

Let us view how it looks like on our example $(\ast)$: $$ \left(\begin{array}{rrr|r} 1&0&1&1\\ 1&-1&1&0\\ 1&1&-1&0 \end{array}\right) \overset{\color{red}{R_2-R_1}}{\overset{\color{red}{R_3-R_1}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&1&1\\ 0&-1&0&-1\\ 0&1&-2&-1 \end{array}\right) \overset{\color{red}{R_3+R_2}}{\longrightarrow} \left(\begin{array}{rrr|r} 1&0&1&1\\ 0&-1&0&-1\\ 0&0&-2&-2 \end{array}\right) \underset{\color{red}{-\frac{1}{2}R_3},\color{red}{\ -1R_2}}{\overset{\color{red}{R_1+\frac{1}{2}R_3}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&0&\color{blue}0\\ 0&1&0&\color{blue}1\\ 0&0&1&\color{blue}1 \end{array}\right), $$

$$ \left(\begin{array}{rrr|r} 1&0&1&0\\ 1&-1&1&1\\ 1&1&-1&0 \end{array}\right) \overset{\color{red}{R_2-R_1}}{\overset{\color{red}{R_3-R_1}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&1&0\\ 0&-1&0&1\\ 0&1&-2&0 \end{array}\right) \overset{\color{red}{R_3+R_2}}{\longrightarrow} \left(\begin{array}{rrr|r} 1&0&1&0\\ 0&-1&0&1\\ 0&0&-2&1 \end{array}\right) \underset{\color{red}{-\frac{1}{2}R_3},\color{red}{\ -1R_2}}{\overset{\color{red}{R_1+\frac{1}{2}R_3}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&0&\color{blue}{\frac{1}{2}}\\ 0&1&0&\color{blue}{-1}\\ 0&0&1&\color{blue}{-\frac{1}{2}} \end{array}\right), $$

$$ \left(\begin{array}{rrr|r} 1&0&1&0\\ 1&-1&1&0\\ 1&1&-1&1 \end{array}\right) \overset{\color{red}{R_2-R_1}}{\overset{\color{red}{R_3-R_1}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&1&0\\ 0&-1&0&0\\ 0&1&-2&1 \end{array}\right) \overset{\color{red}{R_3+R_2}}{\longrightarrow} \left(\begin{array}{rrr|r} 1&0&1&0\\ 0&-1&0&0\\ 0&0&-2&1 \end{array}\right) \underset{\color{red}{-\frac{1}{2}R_3},\color{red}{\ -1R_2}}{\overset{\color{red}{R_1+\frac{1}{2}R_3}}{\longrightarrow}} \left(\begin{array}{rrr|r} 1&0&0&\color{blue}{\frac{1}{2}}\\ 0&1&0&\color{blue}0\\ 0&0&1&\color{blue}{-\frac{1}{2}} \end{array}\right), $$

$D=\begin{pmatrix}0&\frac{1}{2}&\frac{1}{2}\\1&-1&0\\1&-\frac{1}{2}&-\frac{1}{2}\end{pmatrix}$.

Notice, that we repeat three times the same elementary row operations, and only the last column changes. This observation allows us to solve the three systems of linear equation simultaneously by making elementary row operations on the matrix

$$ \left(\begin{array}{rrr|rrr} 1&0&1&1&0&0\\ 1&-1&1&0&1&0\\ 1&1&-1&0&0&1 \end{array}\right). $$

Perform the above mentioned elementary row operations on the matrix

$$ \left(\begin{array}{rrr|rrr} 1&0&1&1&0&0\\ 1&-1&1&0&1&0\\ 1&1&-1&0&0&1 \end{array}\right). $$

Compare the result with the result above obtained by solving three different systems of linear equations.

Let $A$ be an $n\times n$ matrix. If

$$(A|I_n) \qquad\overset{\text{elementary row operations}}{\longrightarrow \cdots\longrightarrow}\qquad (I_n|D), $$

then $AD=I_n$ and $DA=I_n$.

Such a $D$ is called the inverse of $A$ and is denoted by $A^{-1}$.

However, if it is not possible to obtain the matrix $(I_n|D)$ by using elementary row operations, then $A$ is not inverible (i.e. $A^{-1}$ does not exist).

In the case of system of linear equations $A\mathbf{x}=\mathbf{b}$, if $A$ is invertible, $\mathbf{x}=A^{-1}\mathbf{b}$. It is also often used for denoting solution of the system even tough $A^{-1}$ is not calculated.

Find the inverse of the matrix \begin{align*} &\begin{pmatrix} 1 &1 &1\\ 1 &2 &4\\ 1 &5 &4 \end{pmatrix} \end{align*}

over $\mathbb{Z}_7$.

Modify your IPython script for simple Gaussian elimination algorithm for finding the inverse of a given matrix.

linear_algebra/04_invertible_matrices.txt ยท Last modified: 2014/01/20 11:39 by marje