Linear Equations with Special Coefficients

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In this chapter we have shown how to use elementary row operations to solve systems of linear equations. We have assumed that each linear equation in the system has the form $a_{j1}x_1 + \cdots + a_{jn}x_n = b_j,$ where the $a_{ji}$ s and the $b_j$ s are real numbers. For simplicity, in our examples we have only chosen equations with integer coefficients — such as: $2x_1 - 3x_2 +15x_3 = -1.$

Systems with Nonrational Coefficients

In fact, a more general choice of coefficients for a system of two equations might have been

$\begin{eqnarray} \sqrt{2}x_1 + 2\pi x_2 & = & 22.4 \nonumber \\ 3x_1+36.2 x_2 & = & e. \label{e:irrat} \end{eqnarray}$

Suppose that we solve (??) by elementary row operations. In matrix form we have the augmented matrix $\left (\begin {array}{cc|c} \sqrt {2} & 2\pi & 22.4\\ 3 & 36.2 & e\end {array}\right ).$ Proceed with the following elementary row operations. Divide the $1^{st}$ row by $\sqrt {2}$ to obtain $\left (\begin {array}{cc|c} 1 & \pi \sqrt {2} & 11.2\sqrt {2}\\ 3 & 36.2 & e\end {array}\right ).$ Next, subtract $3$ times the $1^{st}$ row from the $2^{nd}$ row to obtain: $\left (\begin {array}{cc|c} 1 & \pi \sqrt {2} & 11.2\sqrt {2}\\ 0 & 36.2- 3\pi \sqrt {2} & e- 33.6\sqrt {2}\end {array}\right ).$ Then divide the $2^{nd}$ row by $36.2-3\pi \sqrt {2}$ , obtaining: $\left (\begin {array}{cc|c} 1 & \pi \sqrt {2} & 11.2\sqrt {2}\\ 0 & 1 & \frac {e-33.6\sqrt {2}}{36.2-3\pi \sqrt {2}}\end {array}\right ).$ Finally, multiply the $2^{nd}$ row by $\pi \sqrt {2}$ and subtract it from the $1^{st}$ row to obtain: $\left (\begin {array}{cc|c} 1 & 0 & 11.2\sqrt {2}-\pi \sqrt {2}\frac {e-33.6\sqrt {2}}{36.2-3\pi \sqrt {2}} \\ 0 & 1 & \frac {e-33.6\sqrt {2}}{36.2-3\pi \sqrt {2}}\end {array}\right ).$ So

$\begin{eqnarray} x_1 & = & 11.2\sqrt{2}-\pi\sqrt{2}\frac{e-33.6\sqrt{2}}{36.2-3\pi\sqrt{2}} \nonumber \\ & & \label{e:irratans} \\ x_2 & = & \frac{e-33.6\sqrt{2}}{36.2-3\pi\sqrt{2}} \nonumber \end{eqnarray}$

which is both hideous to look at and quite uninformative. It is, however, correct.

Both $x_1$ and $x_2$ are real numbers — they had to be because all of the manipulations involved addition, subtraction, multiplication, and division of real numbers — which yield real numbers.

If we wanted to use MATLAB to perform these calculations, we have to convert $\sqrt {2}$ , $\pi$ , and $e$ to their decimal equivalents — at least up to a certain decimal place accuracy. This introduces errors — which for the moment we assume are small.

To enter $A$ and $b$ in MATLAB , type

                                                                  

                                                                  
A = [sqrt(2) 2*pi; 3 36.2];
 
b = [22.4; exp(1)];

Now type A to obtain:

A =
1.4142 6.2832
3.0000 36.2000

As its default display, MATLAB displays real numbers to four decimal place accuracy. Similarly, type b to obtain

Next use MATLAB to solve this system by typing:

A\b

to obtain

ans =
24.5417
-1.9588

The reader may check that this answer agrees with the answer in (??) to MATLAB output accuracy by typing

x1 = 11.2*sqrt(2)-pi*sqrt(2)*(exp(1)-33.6*sqrt(2))/(36.2-3*pi*sqrt(2))
 
x2 = (exp(1)-33.6*sqrt(2))/(36.2-3*pi*sqrt(2))

to obtain

x1 =
 
   24.5417

and

x2 =
 
   -1.9588

More Accuracy

MATLAB can display numbers in machine precision (15 digits) rather than the standard four decimal place accuracy. To change to this display, type

format long

Now solve the system of equations (??) again by typing

A\b

and obtaining

ans =
24.54169560069650
-1.95875151860858

Integers and Rational Numbers

Now suppose that all of the coefficients in a system of linear equations are integers. When we add, subtract or multiply integers — we get integers. In general, however, when we divide an integer by an integer we get a rational number rather than an integer. Indeed, since elementary row operations involve only the operations of addition, subtraction, multiplication and division, we see that if we perform elementary row operations on a matrix with integer entries, we will end up with a matrix with rational numbers as entries.

MATLAB can display calculations using rational numbers rather than decimal numbers. To display calculations using only rational numbers, type

format rational

For example, let

$\begin{equation} \label{e:A4} A=\left(\begin{array}{ccrc} 2 & 2 & 1 & 0\\ 1 & 3 & -5 & 1\\ 4 & 2 & 1 & 3 \\ 2 & 1 & -1 & 4\end{array}\right) \end{matlabEquation} and let \begin{matlabEquation} \label{e:b4} b=\left(\begin{array}{r} 1 \\ 1 \\ -5 \\2 \end{array} \right). \end{matlabEquation} Enter $A$ and $b$ into \Matlab by typing \begin{verbatim} e2_5_3 e2_5_4 \end{verbatim} Solve the system by typing \begin{verbatim} A\b \end{verbatim} to obtain \begin{verbatim} ans = -357/41 309/41 137/41 156/41 \end{verbatim} To display the answer in standard decimal form, type \begin{verbatim} format A\b \end{verbatim} obtaining \begin{verbatim} ans = -8.7073 7.5366 3.3415 3.8049 \end{verbatim} The same logic shows that if we begin with a system of equations whose coefficients are rational numbers, we will obtain an answer consisting of rational numbers --- since adding, subtracting, multiplying and dividing rational numbers yields rational numbers. More precisely: \begin{theorem} Let $A$ be an $n\times n$ matrix that is row equivalent to $I_n$, and let $b$ be an $n$ vector. Suppose that all entries of $A$ and $b$ are rational numbers. Then there is a unique solution to the system corresponding to the augmented matrix $(A|b)$ and this solution has rational numbers as entries. \end{theorem} \begin{proof} Since $A$ is row equivalent to $I_n$, Corollary~\ref{consistent} states that this linear system has a unique solution $x$. As we have just discussed, solutions are found using elementary row operations --- hence the entries of $x$ are rational numbers. \end{proof} \subsection*{Complex Numbers} \index{complex numbers} In the previous parts of this section, we have discussed why solutions to linear systems whose coefficients are rational numbers must themselves have entries that are rational numbers. We now discuss solving linear equations whose coefficients are more general than real numbers; that is, whose coefficients are complex numbers. First recall that addition, subtraction, multiplication and division of complex numbers yields complex numbers. Suppose that \begin{eqnarray*} a & = & \alpha + i\beta \\ b & = & \gamma + i\delta \end{eqnarray*} where $\alpha,\beta,\gamma,\delta$ are real numbers and $i=\sqrt{-1}$. Then \begin{eqnarray*} a+b & = & (\alpha+\gamma) + i(\beta+\delta) \\ a-b & = & (\alpha-\gamma) + i(\beta-\delta) \\ ab & = & (\alpha\gamma-\beta\delta)+i(\alpha\delta+\beta\gamma)\\ \frac{a}{b} & = & \frac{\alpha\gamma+\beta\delta}{\gamma^2+\delta^2}+ i\frac{\beta\gamma-\alpha\delta}{\gamma^2+\delta^2} \end{eqnarray*} \Matlab has been programmed to do arithmetic with complex numbers using exactly the same instructions as it uses to do arithmetic with real and rational numbers. For example, we can solve the system of linear equations \begin{eqnarray*} (4-i)x_1+2x_2 & = & 3-i \\ 2x_1 +(4-3i)x_2 & = & 2+i \end{eqnarray*} in \Matlab by typing \begin{verbatim} A = [4-i 2; 2 4-3i]; b = [3-i; 2+i]; A\b \end{verbatim} \index{\computer!$\backslash$} The solution to this system of equations is: \begin{verbatim} ans = 0.8457 - 0.1632i -0.1098 + 0.2493i \end{verbatim} \index{\computer!i} \footnotesize \noindent {\bf Note:} Care must be given when entering complex numbers into arrays in \Matlabp. For example, if you type \begin{verbatim} b = [3 -i; 2 +i] \end{verbatim} then \Matlab will respond with the $2\times 2$ matrix \begin{verbatim} b = 3.0000 0 - 1.0000i 2.0000 0 + 1.0000i \end{verbatim} Typing either {\tt b = [3-i; 2+i]} or {\tt b = [3 - i; 2 + i]} will yield the desired $2\times 1$ column vector. \normalsize All of the theorems concerning the existence and uniqueness of row echelon form --- and for solving systems of linear equations --- work when the coefficients of the linear system are complex numbers as opposed to real numbers. In particular: \begin{theorem} \label{T:complexcoeff} If the coefficients of a system of $n$ linear equations in $n$ unknowns are complex numbers and if the coefficient matrix is row equivalent to $I_n$, then there is a unique solution to this system whose entries are complex numbers. \end{theorem} \index{row!equivalent} \index{matrix!identity} \subsubsection*{Complex Conjugation} Let $a=\alpha+i\beta$ be a complex number. Then the {\em complex conjugate\/} \index{complex conjugation} of $a$ is defined to be \[ \overline{a} = \alpha - i\beta. \] Let $a=\alpha+i\beta$ and $c=\gamma+i\delta$ be complex numbers. Then we claim that \begin{equation} \begin{array}{rcl} \overline{a+c} & = & \overline{a} + \overline{c}\\ \overline{ac} & = & \overline{a}\;\overline{c} \end{array} \end{equation}$ To verify these statements, calculate $\begin{align*} \overline{a+c} &= \overline{(\alpha+\gamma)+i(\beta+\delta)} = (\alpha+\gamma)-i(\beta+\delta) \\ &= (\alpha-i\beta) + (\gamma-i\delta) = \overline{a} + \overline{c} \end{align*}$

and

$\begin{align*} \overline{ac} &= \overline{(\alpha\gamma-\beta\delta)+i(\alpha\delta+\beta\gamma)} \\ &= (\alpha\gamma-\beta\delta)-i(\alpha\delta+\beta\gamma) \\ &= (\alpha-i\beta)(\gamma-i\delta)=\overline{a}\;\overline{c}. \end{align*}$

Exercises

Solve the system of equations $\begin {array}{rcrcr} x_1 & - & ix_2 & = & 1\\ ix_1 & + & 3x_2 & = & -1 \end {array}$ Check your answer using MATLAB. $\left (\begin {array}{c} x_1 \\ x_2\end {array}\right ) = \left (\begin {array}{r} \answer {\frac {3}{2} - \frac {1}{2}i} \\ \answer {-\frac {1}{2} - \frac {1}{2}i}\end {array}\right ).$

Solve the systems of linear equations given in Exercises ?? – ?? and verify that the answers are rational numbers.

$\begin {array}{rcrcrcr} x_1 & + & x_2 & - & 2x_3 & = & 1 \\ x_1 & + & x_2 & + & x_3 & = & 2 \\ x_1 & - & 7x_2 & + & x_3 & = & 3 \end {array}$ $\left (\begin {array}{c} x_1 \\ x_2 \\ x_3\end {array}\right ) = \left (\begin {array}{r} \answer {\frac {43}{24}} \\ \answer {-\frac {1}{8}} \\ \answer {\frac {1}{3}} \end {array}\right )$ .

$\begin {array}{rcrcr} x_1 & - & x_2 & = & 1\\ x_1 & + & 3x_2 & = & -1 \end {array}$ $\left (\begin {array}{c} x_1 \\ x_2\end {array}\right ) = \left (\begin {array}{r} \answer {\frac {1}{2}} \\ \answer {-\frac {1}{2}} \end {array}\right )$ .

In Exercises ?? – ?? use MATLAB to solve the given system of linear equations to four significant decimal places.

$\begin {array}{rcrcrcr} 0.1 x_1 & + & \sqrt {5}x_2 & - & 2 x_3 & = & 1 \\ -\sqrt {3}x_1 & + & \pi x_2 & - & 2.6 x_3 & = & 14.3 \\ x_1 & - & 7 x_2 & + & \frac {\pi }{2}x_3 & = & \sqrt {2} \end {array}.$

$\begin {array}{rcrcr} (4-i)x_1 & + & (2+3i)x_2 & = & -i \\ i x_1 & - & 4 x_2 & = & 2.2 \end {array}.$

$\begin {array}{rcrcrcr} (2+i) x_1 & + & (\sqrt {2}-3i)x_2 & - & 10.66 x_3 & = & 4.23 \\ 14 x_1 & - & \sqrt {5}i x_2 & + & (10.2-i) x_3 & = & 3-1.6i \\ -4.276 x_1 & + & 2 x_2 & - & (4-2i) x_3 & = & \sqrt {2}i \end {array}.$

Hint: When entering $\sqrt {2}i$ in MATLAB you must type sqrt(2)*i, even though when you enter $2i$ , you can just type 2i.