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\setcounter{page}{67} \centerline{{\sc Step functions.}}

\b

\textbf{Definition.} Suppose that $c$ is a non-negative number.	 Then the \textit{unit step function} $u_c$ is defined by $u_c(t) = \cases{0 & If $t < c$. \cr 1 & If $t \geq c$.}$

\b

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The value at $t=c$ is not important; we could have defined $u_c$ to be $\cases{0 & If $t \leq c$. \cr 1 & If $t > c$.}$  We will usually not worry about which definition we are using.

\b

The function $u_0$ is sometimes called the \textit{Heaviside function}.	 Note that if $c$ is a number, $u_0(t-c) = u_c(t)$ and $u_0(c-t) = 1 - u_c(t)$.

\b

We can combine unit step functions to get other functions which are constant on intervals.	To do examples like this, we start at the left most part of the function, and look at how big the jumps are between the intervals.

\b

\textbf{Examples.}

\begin{enumerate}


\item Write the function $f(t) = \cases{0 & If $t < 5$, \cr 8 & If $t \geq 5$,}$ in terms of unit step functions.


\b

This function is just $8u_5$.

\item Write the function $f(t) = \cases{0 & If $t < 4$, \cr 1 & If $4 \leq t < 7$, \cr 0 & If $7 \leq t$,}$ in terms of unit step functions.

\b

$f(t) = u_4(t) - u_7(t)$.

\b

More generally, we get the following:

\textbf{Fact.} If $c < d$, the function	 $\chi_{[c,d)}(t) = \cases{0 & If $t < c$, \cr 1 & If $c \leq t < d$, \cr 0 & If $d \leq t$,}$ can be written as $\chi_{[c,d)}(t) = u_c(t) - u_d(t)$.	(The function $\chi_{[c,d)}(t)$ is called the \textit{indicator function} or \textit{characteristic function} of the interval $[c,d)$.)



\item Write the function $f(t) = \cases{0 & If $t < 2$, \cr 7 & If $2 \leq t < 4$, \cr -1 & $4 \leq t$,}$ in terms of unit step functions.

\b

We can view $f(t)$ as the sum of the function which is $7$ on the interval $[2,4)$ and $0$ elsewhere with the function which is $-1$ on the interval $[4, \infty)$ and $0$ elsewhere.  Therefore, $f(t) = 7\chi_{[2,4)}(t) - u_4(t) = 7(u_2(t) - u_4(t)) - u_4(t) = 7u_2(t) - 8u_4(t)$

Alternatively, we can view $f$ as starting at $0$, jumping up $7$ units at $t = 2$, and then jumping down $8$ units at $t = 4$ to give the same result.


\end{enumerate}



\b

The Laplace transform of $u_c$ is easy to compute: $\L(u_c(t))(s) = \int_0^\infty u_c(t)e^{-st} dt = \int_c^\infty e^{-st} dt = \frac{e^{-sc}}{s}$.

\b

\textbf{Example.} Let $f(t) = \cases{0 & If $t < 2$. \cr 7 & If $2 \leq t < 4$, \cr -1 & $4 \leq t$.}$	Then by the last example $\L(f(t)(s)	 = \L(7u_2(t) - 8u_4(t))(s) = 7\L(u_2(t))(s) - 8\L(u_4(t))(s) = 7 e^{-2s} - 8e^{-4s}$.





\b

We can also write other functions in terms of $u_c$.

\b

\textbf{Example.}  Write the function $h(t) = \cases{5 - t & If $t < 5$, \cr 0 & If $t \geq 5$,}$ in terms of $u_c$.

\b

The function $1 - u_5$ is $1$ if $t < 5$ and $0$ if $t \geq 5$.	 Therefore, $h(t) = (1 -  u_5(t))(5-t)$.

\b

\textbf{Note.} If $f(t)$ is a function defined for $t \geq 0$, then $u_c(t)f(t-c)$ is the function which has the same graph as that of $f(t)$ except it is $c$ units to the right, and is $0$ between $0$ and $c$.

\b

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\b

Equivalently, $u_c(t)f(t-c) = \cases{0 & If $t < c$. \cr f(t-c) & If $t \geq c$.}$


\b


\textbf{Theorem.} (i) If $f(t)$ is a reasonable function, then $$\L(u_c(t)f(t-c))(s) = e^{-cs}\L(f(t))(s).$$

\b

(ii) If $F(s)$ is a reasonable function, then $$\L^{-1}(e^{-cs}F(s)) = u_c(t)\L^{-1}(F(s))(t-c).$$



\b

\textbf{Examples.}


\begin{enumerate}

\item Let $f(t) = \cases{0 & If $t < 5$. \cr (t-5)^3 & If $t \geq 5$.}$.  Find $\L(f(t))(s)$.

 \b

 $\L(f(t))(s) = L(u_5(t)(t- 5)^3)(s) = e^{-5s}\L(t^3) = e^{-5s}(\frac{3!}{s^4}).$

\item Let $g(t) =		 \cases{5-t & If $t < 5$. \cr (t-5)^3 & If $t \geq 5$.}$ Find $\L(g(t))(s)$.

\b

Notice that $g(t) = h(t) + f(t)$ where $h(t) = (1 -	 u_5(t))(5-t) $ and $f(t)$ is the function in the last example.	 Therefore, $\L(g(t))(s) = \L(h(t))(s) + \L(f(t))(s)$.	 $h(t) = 5 - t + u_5(t)(t-5)$, so $ \L(h(t))(s) = \L(5)(s) - \L(t)(s) + \L(u_5(t)(t-5))(s) = \frac{5}{s} - \frac{1}{s^2} + e^{-5s}\L(t)(s) =
\frac{5}{s} - \frac{1}{s^2} + \frac{e^{-5s}}{s^2}$.	 Therefore, $\L(g(t))(s) = e^{-5s}(\frac{3!}{s^4}) + \frac{5}{s} - \frac{1}{s^2} + \frac{e^{-5s}}{s^2}$.




\item Find $\L^{-1}( \frac{e^{-4s}}{s^3})$.

\b

By the theorem	$\L^{-1}( \frac{e^{-4s}}{s^3})(t) = u_4(t)\L^{-1}(\frac{1}{s^3})(t-4)$.	 From the table, $\L^{-1}(\frac{2!}{s^3}) = t^2$, so $\L^{-1}(\frac{1}{s^3}) = \frac{t^2}{2}$.	 Therefore, $\L^{-1}(\frac{1}{s^3})(t-4) = \frac{(t-4)^2}{2}$, so $\L^{-1}( \frac{e^{-4s}}{s^3})(t) = u_4(t)\frac{(t-4)^2}{2}$.



\end{enumerate}

\b

There is another theorem which is similar to the last one, but has an $e^{cs}$ term instead of a $e^{-cs}$ term.

\b

\textbf{Theorem.} (i) If $f(t)$ is a reasonable function, then $$\L(e^{ct}f(t))(s) = \L(f(t))(s-c).$$

\b

(ii) If $F(s)$ is a reasonable function, then $$\L^{-1}(F(s - c)) = e^{ct}\L^{-1}(F(s))(t).$$

\b

\textbf{Examples.}	



\begin{enumerate}


\item Find the inverse Laplace transform of the function $\frac{s}{(s - 7)^2}$.	

\b

Notice that $\frac{(s-7)}{(s - 7)^2} + \frac{7}{(s-7)^2}  =\frac{1}{s - 7} + \frac{7}{(s-7)^2}= F(s - 7)$ where $F(s) = \frac{1}{s} + 7\frac{1}{s^2}$.	 The inverse Laplace transform of $F(s)$ is $1 + 7t)$ so by the Theorem, the inverse Laplace transform of $\frac{s}{(s - 7)^2} = F(s - 7)$ is $e^{7t}(1 + 7t)$.


\item Find the inverse Laplace transform of the function $G(s) = \frac{1}{s^2 - 2s}$.

\b

We could do this with partial fractions, but notice that $G(s) = \frac{1}{s^2 - 2s + 1 - 1} = \frac{1}{(s-1)^2 - 1}.$  Therefore, $G(s) = F(s - 1)$, where $F(s) = \frac{1}{s^2 - 1}$.  By the theorem, $\L^{-1}(G(s))(t) = \L^{-1}(F(s-1))(t) = e^t(\L^{-1}(F(s)))(t) = e^t \sinh t$.




\end{enumerate}

\b

\{ Page 328 \hspace{1in} 1-4, 7-10, 13-16, 19-21 \\

\pagebreak

\centerline{{\sc Using Laplace transforms with discontinuous functions.}}

\b


The method of undetermined coefficients allows us to solve ODEs of the form \[ \dn{x}{t}{n} + a_1\dn{x}{t}{n-1} + \cdots + a_{n-1}\d{x}{t} + a_nx = h(t)\] when $h(t)$ is the sum of functions of the form $p(t)e^{\alpha t} \cos (\beta t)$ or $p(t)e^{\alpha t} \sin (\beta t)$ where $p(t)$ is a polynomial.	We can use Laplace transforms to solve some initial value problems where $h(t)$ is a piecewise continuous function.


\b


\begin{enumerate}

\item Solve the initial value problem $\d{x}{t} - 3x = \cases{0 & If $t < 2$, \cr 1 & If $t \geq 2$,}$ with initial condition $x(0) = 5$.

\b

We can write the ODE as $\d{x}{t} - 3x = u_2(t)$.  We take Laplace transforms of both sides to get $\L(x'(t))(s) - 3 \L(x(t))(s) = \L(u_2(t))(s)$.	 From the formula for the Laplace transform of a derivative, we get $\L(x'(t))(s) = s\L(x(t))(s) - x(0) = s\L(x(t))(s) - 5$.	 Since $\L(u_c(t))(s) = \frac{e^{-cs}}{s} $, we get \[s\L(x(t))(s) - 5 - 3 \L(x(t))(s) = \frac{e^{-2s}}{s}.\]	Solving for $\L(x(t))(s)$ gives \[\L(x(t))(s) = \frac{5}{s-3}+\frac{e^{-2s}}{s(s-3)}.\]	 Converting to partial fractions gives \[\L(x(t))(s)= \frac{5}{s-3} + (\frac{1}{3})\frac{e^{-2s}}{s-3} - (\frac{1}{3})\frac{e^{-2s}}{s}.\]  Taking inverse Laplace transforms (and using the fact that $\L^{-1}F(s-3) = e^{3t}\L^{-1}(F(s))$) we get \[x(t) = 5e^{3t} + (\frac{1}{3})e^{3t}u_2(t) - (\frac{1}{3})u_2(t).\]	This can be written as \[x(t) = \cases{5e^{3t} & If $t < 2$.\cr 5e^{3(t-2)}+ (\frac{1}{3})e^{3(t-2)} - \frac{1}{3} & If $t \geq 2$.}\]

\item Solve the initial value problem $\dn{x}{t}{2} - x = \cases{5 - t & If $t < 5$, \cr 0 & If $t \geq 5$,}$ $x(0) = 0$, $x'(0) = 1$.


\b


We saw in an earlier example that the function on the right side of the ODE can be written as  $(1 -  u_5(t))(5-t)$.  Therefore, the ODE can be written as		 $\dn{x}{t}{2} - x = (1 -	u_5(t))(5-t)$ or \[\dn{x}{t}{2} - x = 5 - t + u_5(t)(5-t).\]  Taking Laplace transforms of both sides of the equation, gives $\L(x''(t))(s) - \L(x(t))(s) = 5\L(1)(s) - \L(t)(s) + \L(u_5(t)(t-5))(s)$.  From the initial conditions we get $\L(x''(t))(s) = s^2\L(x(t))(s) - sx(0)-x'(0) = s^2\L(x(t))(s) - 1$.	 Also, from the table we get $\L(1)(s) = \frac{1}{s}$, $\L(t)(s) = \frac{1}{s^2}$, and $\L(u_5(t)(t-5))(s) = e^{-5s}\L(t)(s) = \frac{e^{-5s}}{s^2}$.  Therefore, \[s^2\L(x(t))(s) - 1 - \L(x(t))(s) = \frac{5}{s} - \frac{1}{s^2}+\frac{e^{-5s}}{s^2}.\]  Solving for $\L(x(t))(s)$ gives \[\L(x(t))(s) = \frac{1}{s^2 - 1} - \frac{5}{s(s^2-1)} - \frac{1}{s^2(s^2-1)} + \frac{e^{-5s}}{s^2(s^2-1)}.\] Using Maple to find the inverse Laplace transform gives \[x(t) = t - tu_5(t) + 5u_5(t)+(\sinh (t-5))u_5(t) + 5 - 5\cosh t.\]




\end{enumerate}




\b

\{ Page 336 \hspace{1in} 1-5, 8, 9, 14 \\

\pagebreak

\centerline{{\sc Systems of first-order ODEs.}}

\b

Sometimes a situation involves more than one unknown which are described by a system of differential equations. We will look at some relatively simple cases and see how to find solutions.	In particular, we will look only at the case where only the first derivatives of the functions are involved.	 Such systems are called systems of \textit{first order}	ODEs.

\b

\textbf{Examples.}

\begin{enumerate}

\item \begin{eqnarray*}
\d{x_1}{t} & = & x_1 + 2x_2 + 3x_3\\
\d{x_2}{t} & = & x_1x_2x_3 + t^2\\
\d{x_3}{t} & = & t^3\\
\end{eqnarray*}

This is a system of three ODEs.

\item \begin{eqnarray*}
\d{x}{t} & = & 3x  + t\\
\d{y}{t} & = & 4y + e^t
\end{eqnarray*}

In this case, each of the equations involves only one dependent variable, and we can solve this system by solving each of the ODEs individually.	A system like this is said to be \textit{decoupled}.

\item \begin{eqnarray*}
\d{x}{t} & = & 3x + y^2\\
\d{y}{t} & = & 4x + y
\end{eqnarray*}

Here the independent variable does not appear on the right side.  We say that such a system is \textit{autonomous}.


\item

\begin{eqnarray*}
\d{x}{t} & = & 3x + ty + \sin t \\
\d{y}{t} & = & t^2x + y + \cos t
\end{eqnarray*}


This system has the form

\begin{eqnarray*}
\d{x}{t} & = & f_1(t)x + g_1(t)y + h_1(t). \\
\d{y}{t} & = & f_2(t)x + g_2(t)y + h_2(t).
\end{eqnarray*}

Such a system is said to be \textit{linear}.  If $h_1$ and $h_2$ are both the zero function, the linear system is said to be \textit{homogeneous}


\item The system

\begin{eqnarray*}
\d{x}{t} & = & 3x + 4y	\\
\d{y}{t} & = & 2x + 3y
\end{eqnarray*}

is a linear homogeneous system with constant coefficients.

\end{enumerate}


\b	

Systems of differential equations arise in many applications.  One way in which they arise is as a translation of higher-order differential equations.	We can do a similar thing with initial value problems.

\b

\textbf{Examples.}


\begin{enumerate}

\item Express the ODE $\dd{x}{t} - 3\d{x}{t} +4x = e^t$ as a system of first-order ODEs.

\b

Let $y = \d{x}{t}$.	 Then $\d{y}{t} = \dd{x}{t}$.  Therefore, the system can be written as	


\begin{eqnarray*}
\d{x}{t} & = & y  \\
\d{y}{t} & = & 3y - 4x + e^t
\end{eqnarray*}


\item Express the ODE $\dd{x}{t} - 3\d{x}{t} +4x = e^t$ with initial conditions $x(0) = 2$, $x'(0) = 4$ as a system of first-order ODEs.


\b

We saw that the ODE can be written as the system


\begin{eqnarray*}
\d{x}{t} & = & y  \\
\d{y}{t} & = & 3y - 4x + e^t
\end{eqnarray*}

The initial condition $x(0) = 2$ is still and initial condition for the system, but the initial condition $x'(0) = 4$ becomes $y(0) = 4$.

\item  Express the ODE $\dn{x}{t}{3} +\dd{x}{t} + 2\d{x}{t} +4x = 0$ as a system of first-order ODEs.

\b

For aesthetic reasons, let $x_1 = x$.  Now let $x_2 = \d{x_1}{t} = \d{x}{t}$, and let $x_3 = \d{x_2}{t} = \dd{x_1}{t} = \dd{x}{t}$.  Then $\d{x_3}{t} = \dn{x_1}{t}{3} = \dn{x}{t}{3}$.  Therefore, the original ODE becomes \[\d{x_3}{t} + x_3 + 2x_2 + x_1 = 0,\] so we can write the system as


\begin{eqnarray*}
\d{x_1}{t} & = & x_2.  \\
\d{x_2}{t} & = & x_3.  \\
\d{x_3}{t} & = & -(x_3 + 2x_2 + x_1).
\end{eqnarray*}

\end{enumerate}



By a solution of a system of ODEs we mean functions which, when replaced for the unknowns, satisfy the ODEs.

\b

\textbf{Example.} A solution of the system


\begin{eqnarray*}
\d{x}{t} & = & -7x + 6y	 \\
\d{y}{t} & = & -12x + 10 y
\end{eqnarray*}

is given by $x = 3e^t - 2e^{2t}$, $y = 4e^t - 3e^{2t}$.	 (We can check this.)  This is not the only solution.	In fact, we can check that if $A$ and $B$ are any constants, then \[x = (9A - 6B)e^t -(8A-6B)e^{2t},\] \[y = (12A - 8B)e^t - (12A-9B)e^{2t}\] is also a solution. (The solution we mentioned is the case $A = B = 1$.)	We will see later how to find the solutions.




\b

\{ Page 359 \hspace{1in} 1-6, 16 \\

\pagebreak

\centerline{{\sc Some basic matrix algebra.}}

\b

An \textit{$m \times n$ matrix} is an array $A$ of the form \[ A = \pmatrix{A_{11} & A_{12} & \cdots & A_{1n} \cr A_{21} & A_{22} & \cdots & A_{2n} \cr \vdots & \vdots & \vdots & \vdots \cr
A_{m1} & A_{m2} & \cdots & A_{mn}}.\]  The $A_{ij}$s are called \textit{entries} of the matrix.	 Two matrices are equal if corresponding entries are equal.

\b

In the context of matrices, numbers are sometimes called \textit{scalars}.

\b

\textbf{Examples.}

\begin{enumerate}

\item $\pmatrix{1 & -1 & 0 & 3 \cr 4 & 2 & 7 & 0 \cr}$ is a $2 \times 4$ matrix.

\item $\pmatrix{2 & 4 & -3 \cr 1 & 0 & 0 \cr 2 & 3 & 7\cr}$ is a  $3 \times 3$ matrix.	





\end{enumerate}

A matrix in which the number of rows is the same as the number of columns is called a \textit{square matrix}.  An $1 \times n$ matrix is called a \textit{row matrix} or \textit{row vector}.	 An $m \times 1$ matrix is called a \textit{column matrix} or \textit{column vector}.



\b


If $A = \pmatrix{A_{11} & A_{12} & \cdots & A_{1n} \cr A_{21} & A_{22} & \cdots & A_{2n} \cr \vdots & \vdots & \vdots & \vdots \cr
A_{m1} & A_{m2} & \cdots & A_{mn}}$ is an $m \times n$ matrix, then the \textit{transpose} $A^T$ of $A$ is the $n \times m$ matrix \[ A^T = \pmatrix{A_{11} & A_{21} & \cdots & A_{m1} \cr A_{12} & A_{22} & \cdots & A_{m2} \cr \vdots & \vdots & \vdots & \vdots \cr
A_{1n} & A_{2m} & \cdots & A_{mn}}\] obtained by making the rows of $A$ into columns.

\b

\textbf{Example.} $\pmatrix{1 & -1 & 0 & 3 \cr 4 & 2 & 7 & 0 \cr}^T = \pmatrix{1 & 4 \cr -1 & 2 \cr 0 & 7 \cr 3 & 0 \cr}.$


\b

We can add two $m \times n$ matrix by adding the corresponding entries. Note that we can add two matrices only if they have the same numbers of rows and the same number of columns.	 For example, we cannot add a $3 \times 5$ matrix to a $3 \times 4$ matrix. We can multiply a matrix $A$ by a scalar $c$ by multiplying each entry by $c$.

\b

\textbf{Examples.}

\begin{enumerate}

\item $\pmatrix{1 & -1 & 0 & 3 \cr 4 & 2 & 7 & 0 \cr} + \pmatrix{3 & 4 & 1 & -3 \cr 2 & -2 & 1 & 1 \cr} = \pmatrix{4 & 3 & 1 & 0 \cr 6 & 0 & 8 & 1 \cr}$.

\item $6\pmatrix{1 & -1 & 0 & 3 \cr 4 & 2 & 7 & 0 \cr} = \pmatrix{6 & -6 & 0 & 18 \cr 24 & 12 & 14 & 0 \cr}$.


\end{enumerate}


We can also multiply certain matrices. Suppose that $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix.  In other words, the number of columns of $A$ is the same as the number of rows of $B$.  Then $AB$ is the $m \times p$ matrix whose $ij^{th}$ entry	is $\sum_{k=1}^nA_{ik}B_{kj}$.	

\b

Another way of describing this is that the $ij^{th}$ entry of $AB$ is the dot product of the $i^{th}$ row of $A$ with the transpose of the $j^{th}$ column of $B$.

\b

If $A$ is a square matrix, we can define $AA$, which can be written as $A^2$.  Similarly, we define $A^n$ when $n$ is a positive integer.


\b


\textbf{Examples.}


\begin{enumerate}

\item If $A = \pmatrix{1 & 2 \cr 3 & 4 \cr 5 & 6 \cr}$ and $B = \pmatrix{3 & -4 \cr  2 & 4 \cr}$, then \[AB = \pmatrix{3 + 4 & -4 + 8 \cr 9 + 8 & -12 + 16 \cr 15 + 12 & -20 + 24 \cr} = \pmatrix{7 & 4 \cr 17 & 4 \cr 27 & 4 \cr}.\]	 However, $BA$ is not defined.



\item If $A = \pmatrix{1 & 2 \cr 3 & -4 \cr}$ then $A^2 = \pmatrix{7 & -6 \cr -9 & 22 \cr}$.



\end{enumerate}

\b

The $n \times n$ matrix $I_n$ (or just $I$ if $n$ is clear) is defined by $I_{ij} = \cases{1 & If $i=j$, \cr 0 & Otherwise,}$ is called the \textit{$n \times n$ identity matrix}.	

\b

\textbf{Fact.} If $A$ is an $m \times n$ matrix, then $I_mA = A$ and $AI_n = A$.	

\b

If $A$ is a square matrix, $A^0$ is defined to be $I$.

\b

A square matrix $A$ is \textit{invertible} or \textit{nonsingular} if there is a matrix $B$ such that $AB = BA = I$.  If $A$ is invertible, then there is only one such $B$ and it is called the \textit{inverse} of $A$ and is denoted $A^{-1}$.

\b

\textbf{Examples.}

\begin{enumerate}

\item The matrix $A = \pmatrix{ 1 & -3 \cr -2 & 6 \cr}$ is not invertible. To see this, suppose that $B = \pmatrix{ a & b \cr c & d \cr}$ were an inverse.  Then \[\pmatrix{ a -3c & b - 3d \cr -2a + 6c & -2b + 6d \cr} = \pmatrix{ 1 & 0 \cr 0 & 1 \cr}.\] Then $a - 3c = 1$ and $-2a + 6c = 0$.	 But dividing the second equation by $-2$ gives that $a - 3c = 0$, which cannot be, because it is $1$.

\item Let $A = \pmatrix{ 1 & -3 \cr -2 & 5 \cr}$.  Then $A$ is invertible. To see this, let $B = \pmatrix{ -5 & -3 \cr -2 & -1 \cr}$. Then $AB = I = BA$.	 Therefore, $A^{-1} = \pmatrix{ -5 & -3 \cr -2 & -1 \cr}$.


\end{enumerate}


The question is how to determine if a matrix is invertible, and if it is, how do we find its inverse.

\b

\textbf{Theorem.} A (square) matrix $A$ is invertible if and only if $det(A) \ne 0$.

\b

This leaves the question of how to find the inverse.  For $2 \times 2$ matrices, there is a simple formula.

\b

Suppose $A = \pmatrix{a & b \cr c & d \cr}$ is a matrix such that $det(A) \ne 0$.	 Then \[A^{-1} = \left( \frac{1}{det(A)}\right)\pmatrix{d & -b \cr -c & a}= \left( \frac{1}{ad - bc}\right)\pmatrix{d & -b \cr -c & a} .\]

\b

\textbf{Example.} $\pmatrix{2 & 4 \cr 1 & 3 \cr}^{-1} = (\frac{1}{2})\pmatrix{3 & -4 \cr -1 & 2 \cr} = \pmatrix{\frac{3}{2} & -2 \cr -\frac{1}{2} & 1 \cr}.$


\b

For larger matrices, the easiest way is to use \textit{row reduction} or \textit{Gauss-Jordan elimination}.	This process starts with a matrix and repeatedly applies the following operations, called \textit{elementary row operations}:

\begin{enumerate}
\item Interchange two rows.
\item Multiply a row by a non-zero constant.
\item Add a multiple of one row to a different row.
\end{enumerate}


Now suppose that $A$ is an $n \times n$ matrix whose determinant is not $0$.  (You do not have to compute the determinant--if the determinant is $0$, this method automatically will not work.)  Form the $n \times 2n$ matrix $A\vert I$ whose first $n$ columns are the columns of $A$ and whose last $n$ columns are the columns of the $n \times n$ identity matrix.  Now perform the elementary row operations until you get a matrix of the form $I\vert B$.  Then $B$ is the inverse of $A$.

\b

\textbf{Example.} Let $A = \pmatrix{ 0 & 2 & 2 \cr 4 & 0 & 4 \cr 1 & 1 & 0}$.  Find $A^{-1}$ if it exists.

\b

First form the matrix \[\left( \begin{array}{ccc|ccc}
0&2&2&1&0&0\\
4&0&4&0&1&0\\
1&1&0&0&0&1\\
\end{array}\right).\]

\b

Interchange the first and third rows.

\[\left( \begin{array}{ccc|ccc}
1&1&0&0&0&1\\
4&0&4&0&1&0\\
0&2&2&1&0&0\\
\end{array}\right).\]



\b

Subtract $4$ times the first row from the second row.




\[\left( \begin{array}{ccc|ccc}
1&1&0&0&0&1\\
0&-4&4&0&1&-4\\
0&2&2&1&0&0\\
\end{array}\right).\]

Notice that now the first column is the first column of the identity matrix.  We want to make the second column into the second column of the identity matrix without disturbing the first column.  Multiply the second column by $\frac{-1}{4}$.


\[\left( \begin{array}{ccc|ccc}
1&1&0&0&0&1\\
0&1&-1&0&\frac{-1}{4}&1\\
0&2&2&1&0&0\\
\end{array}\right).\]

\b

Now we will do two operations at once.	We subtract the second row from the first, and we subtract twice the second row from the third.

\[\left( \begin{array}{ccc|ccc}
1&0&1&0&\frac{1}{4}&0\\
0&1&-1&0&\frac{-1}{4}&1\\
0&0&4&1&\frac{1}{2}&-2\\
\end{array}\right).\]

\b

Now the first two columns have the correct form and we work on the third column.	 Multiply the third column by $\frac{1}{4}$.

\[\left( \begin{array}{ccc|ccc}
1&0&1&0&\frac{1}{4}&0\\
0&1&-1&0&\frac{-1}{4}&1\\
0&0&1&\frac{1}{4}&\frac{1}{8}&\frac{-1}{2}\\
\end{array}\right).\]

\b

Finally, we subtract the third row from the first and add the third row to the second.


\[\left( \begin{array}{ccc|ccc}
1&0&0&\frac{-1}{4}&\frac{1}{8}&\frac{1}{2}\\
0&1&0&\frac{1}{4}&\frac{-1}{8}&\frac{1}{2}\\
0&0&1&\frac{1}{4}&\frac{1}{8}&\frac{-1}{2}\\
\end{array}\right).\]

\b

Therefore, \[A^{-1} = \left( \begin{array}{ccc}
\frac{-1}{4}&\frac{1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{-1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{1}{8}&\frac{-1}{2}\\
\end{array}\right).\]

\b

It is straightforward to check that this is the inverse.

\b

\textbf{Proposition.} Suppose that $A$ and $B$ are invertible $n \times n$ matrices.  Then $AB$ is invertible and $(AB)^{-1} = B^{-1}A^{-1}$.

\b

Sometimes we look at matrix-valued functions $A(t)$.  Then we can talk about derivatives, integrals, continuity, and so on by looking at each entry.

\b

\textbf{Example.} If $A(t) = \pmatrix{t^2 & \sin t \cr \cos t & e^t \cr}$, then $A'(t) = \pmatrix{2t & \cos t \cr -\sin t & e^t \cr}$.


\b

With this notation, systems of differential equations can be written in matrix form.

\b

\textbf{Example.} The system \begin{eqnarray*}
\d{x}{t} & = & 4x + y  \\
\d{y}{t} & = & 3x - 5y + e^t
\end{eqnarray*} can be written as $\mathbf{x'} = \pmatrix{4 & 1 \cr 3 & -5 \cr}\mathbf{x} + \pmatrix{0 \cr e^t \cr}$, where $\mathbf{x} = \pmatrix{x \cr y}$.

\b

\{ Page 371 \hspace{1in} 1-6, 10-15, 21, 22 \\

\pagebreak

\centerline{{\sc Systems of linear equations and eigenvalues.}}

\b

Let $\mathbf{0}$ denote the $m \times n$ vector all of whose entries are $0$.  We will let the context indicate what $m$ and $n$ are.

\b

Recall that $n \times 1$ matrices are sometimes called column vectors. If $\mathbf{v}_1, \cdots ,	 \mathbf{v}_k$ are vectors, we say that they are \textit{linearly independent} if $c_1 \mathbf{v}_1 + \cdots c_k \mathbf{v}_k = \mathbf{0}$ implies $c_1 = \cdots = c_k = 0$.

\b

\textbf{Example.}  The vectors $\pmatrix{2\cr 1\cr 0\cr 4}$, $\pmatrix{1 \cr 0 \cr 0 \cr 0}$ and $\pmatrix{1 \cr 0 \cr 1 \cr 0}$ are linearly independent because if $c_1\pmatrix{2\cr 1\cr 0\cr 4}+ c_2\pmatrix{1 \cr 0 \cr 0 \cr 0} + c_3\pmatrix{1 \cr 0 \cr 1 \cr 0} = \pmatrix{0 \cr 0 \cr 0 \cr 0}$, then it is easy to check that $c_1 = c_2 = c_3 = 0$.

\b

\textbf{Theorem.} If $\mathbf{v}_1, \cdots, \mathbf{v}_n$ are linearly independent $n \times 1$ column vectors, then the ($n \times n$) matrix whose columns are $\mathbf{v}_1, \cdots, \mathbf{v}_n$ is invertible.

\b

The system of linear equations

\[\left\{ \begin{array}{ccc}
A_{11}x_1 + \cdots + A_{1n}x_n &  = & y_1 \\
\vdots & \vdots & \vdots \\
A_{n1}x_1 + \cdots + A_{nn}x_n &  = & y_n \\
\end{array}
\right.\]

can be written as \[A \mathbf{x} = \mathbf{y}\] where $A = \pmatrix{A_{11}& \cdots & A_{1n} \cr
\vdots & \vdots & \vdots \cr
A_{n1} & \cdots & A_{nn}x_n \cr }$, $\mathbf{x} = \pmatrix{x_1 \cr \vdots \cr x_n}$, and $\mathbf{y} = \pmatrix{y_1 \cr \vdots \cr y_n}$.	 Note that if $A$ is invertible, then $\mathbf{x} = A^{-1}\mathbf{y}$, so the system is easy to solve.

\b

\textbf{Example.} Solve the system $\left\{ \begin{array}{ccc}
2y + 2z & = & 0. \\
4x + 4z & = & 1. \\
x + y & = & 4. \\
\end{array} \right. $

We saw that if $A = \pmatrix{ 0 & 2 & 2 \cr 4 & 0 & 4 \cr 1 & 1 & 0}$, then $A^{-1} = \left( \begin{array}{ccc}
\frac{-1}{4}&\frac{1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{-1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{1}{8}&\frac{-1}{2}\\
\end{array}\right).$


\b

The given system is $A \mathbf{x} = \pmatrix{0 \cr 1 \cr 4}$, where $\mathbf{x} = \pmatrix{x \cr y \cr z \cr}$.	 Therefore, $\mathbf{x} = \left( \begin{array}{ccc}
\frac{-1}{4}&\frac{1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{-1}{8}&\frac{1}{2}\\
\frac{1}{4}&\frac{1}{8}&\frac{-1}{2}\\
\end{array}\right)\pmatrix{0 \cr 1 \cr 4} = \pmatrix{ \frac{17}{8} \cr \frac{15}{8} \cr \frac{-15}{8}}$.

\b

\textbf{Theorem.} If $A\mathbf{x} = \mathbf{y}$ is a system of $n$ linear equations in $n$ unknowns, then the system has a unique solution if and only if the matrix $A$ is invertible.  Therefore, the system has a unique solution if and only if $det(A) \ne 0$.

\b

A special case is important.  To see this notice that the system $A\mathbf{x} = \pmatrix{0 \cr \vdots \cr 0}$ always has one solution, namely, $\mathbf{x} =  \pmatrix{0 \cr \vdots \cr 0}$.	This is called the \textit{trivial solution}.  Other solutions are \textit{non-trivial solutions}.	 By the theorem, the trivial solution is the only solution if and only if $det(A) = 0$. Therefore, we get the following:

\b

\textbf{Corollary.} The system	$A\mathbf{x} = \pmatrix{0 \cr \vdots \cr 0}$ has a non-trivial solution if and only if $det(A) = 0$.




\b


\textbf{Example.} If $A = \pmatrix{1 & -2 \cr -3 & 6 \cr}$, then one solution of $A\mathbf{x} = \pmatrix{0 \cr 0 \cr}$ is $\mathbf{x} = \pmatrix{0 \cr 0 \cr}$ but any solution of the form $\mathbf{x} = \pmatrix{ 2a \cr a \cr}$ is a solution.



\b


Suppose that $A$ is an $n \times n$ matrix. If $\lambda$ is a number such $A \mathbf{x} = \lambda \mathbf{x}$ for some \textit{non-zero} column vector $\mathbf{x}$, then $\lambda$ is called an \textit{eigenvalue} of $A$.  Any non-zero vector $\mathbf{x}$ such that $A \mathbf{x} = \lambda \mathbf{x}$ is called an \textit{eigenvector} associated with $\lambda$.

\b

\textbf{Example.} Find all eigenvalues of $\pmatrix{2 & 1 \cr -1 & 4}$.

\b

We want to find numbers $\lambda$ such that the equation $\pmatrix{2 & 1 \cr -1 & 4} \mathbf{x} = \lambda \mathbf{x}$ has a non-trivial solution.  This equation is the same as the equation  $\pmatrix{2 - \lambda & 1 \cr -1 & 4 - \lambda} \mathbf{x} = \mathbf{0}$.	This has a non-trivial solution exactly if $\det\pmatrix{2 - \lambda & 1 \cr -1 & 4 - \lambda} = 0$, that is, if and only if $9 - 6\lambda + \lambda^2 = 0$.	The only solution of this equation is $\lambda = 3$.	 Therefore, the only eigenvalue is $\lambda = 3$.

We can find a corresponding eigenvector by solving the equation $\pmatrix{2 & 1 \cr -1 & 4} \mathbf{x} = 3 \mathbf{x}$ for $\mathbf{x} = \pmatrix{x \cr y}$.  This can be written as $\pmatrix{-1 & 1 \cr -1 & 1} \mathbf{x} = \mathbf{0}$, or \begin{eqnarray*} - x + y &= & 0. \\ -x + y & = & = 0. \end{eqnarray*}  This is satisfied by any vector of the form $\pmatrix{a \cr a \cr}$, so, for example,  $\pmatrix{7 \cr 7 \cr}$ is an eigenvector associated with $\lambda = 3$.

\b

Note that we had to find the roots of the polynomial $\det(A - \lambda I)$. This polynomial is called the \textit{characteristic polynomial} of the matrix $A$.

\b

\textbf{Theorem.} The eigenvalues of the matrix $A$ are the roots of the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.

\b

\textbf{Example.} The eigenvalues of the matrix $A = \pmatrix{1 & 2 \cr 3 & 2}$ are the roots of the polynomial $p(\lambda) = det\pmatrix{1 - \lambda & 2 \cr 3 & 2-\lambda} = \lambda^2 - 3\lambda - 4$, so the eigenvalues are $-1$ and $4$.

\b

If $A$ and $B$ are $n \times n$ matrices, then $A$ is \textit{similar} to $B$ if there is an invertible matrix $P$ such that $A = P^{-1}BP$.

\b

\textbf{Example.} $A = \pmatrix{7 & -12 \cr 3 & -4}$ is similar to $B= \pmatrix{1 & -2 \cr 4 & 4}$ because $A = P^{-1}BP$ where $P = \pmatrix{1 & -2 \cr -1 & 3}$ (so $P^{-1} = \pmatrix{3 & 2 \cr 1 & 1}$).

\b

Note that if $A$ is similar to $B$, then $B$ is similar to $A$, because if $A = P^{-1}BP$ , then $B =Q^{-1}AQ$ where $Q = P^{-1}$.

\b

\textbf{Proposition.} If $A = P^{-1}BP$, then for every positive integer $n$, $A^n = P^{-1}B^nP$.

\b

An $n \times n$ matrix $D$ is a \textit{diagonal matrix} if all entries $D_{ij}$ where $i \ne j$ (that is, all entries not on the main diagonal) are $0$. An $n \times n$ matrix $A$ is \textit{diagonalizable} if $A$ is similar to a diagonal matrix.

\b

\textbf{Example.} The matrix $A = \pmatrix{-1 & -2 \cr 10 & 8 }$ is diagonalizable because $P^{-1}AP$ is the diagonal matrix $\pmatrix{3 & 0 \cr 0 & 4}$, where $P =\pmatrix{1 & -2 \cr -2 & 5}$.

\b

One nice thing about diagonal matrices is that if $D$ and $E$ are $n \times n$ diagonal matrices, then	 $DE$ is also a diagonal matrix and the entries are just the product of the corresponding entries in $D$ and $E$.  In particular, if $D$ is a diagonal matric the $ij^{th}$ entry of $D^k$ is just the $k^{th}$ power of the $ij^{th}$ entry of $D$.

\b

\textbf{Example.} If $D = \pmatrix{2 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 5}$, then $D^2 = \pmatrix{4 & 0 & 0 \cr 0 &	1 & 0 \cr 0 & 0 & 25}$ and $D^3 = \pmatrix{8 & 0 & 0 \cr 0 &	-1 & 0 \cr 0 & 0 & 125}$.


\b

\textbf{Theorem.} If an $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors, then $A$ is diagonalizable.


\b

\textbf{Theorem.} Suppose that $A$ is an $n \times n$ matrix.  If $A$ has $n$ distinct eigenvalues, then $A$ is diagonalizable.  Furthermore, if the distinct eigenvalues are $\lambda_1, \cdots , \lambda_n$ and $\mathbf{v}_1, \cdots , \mathbf{v}_n$ are corresponding eigenvectors, then $A$ is similar to the diagonal matrix $D$ that has $\lambda_1, \cdots , \lambda_n$ on the main diagonal, and $D = P^{-1}AP$, where $P$ is the matrix whose columns are $\mathbf{v}_1, \cdots , \mathbf{v}_n$.

\b

\textbf{Example.} Let $A = \pmatrix{-13 & -4 \cr 42 & 13}$.	 Then the characteristic polynomial of $A$ is $\lambda^2 - 1$ which has roots $\pm 1$.  Therefore, by the theorem, $A$ is similar to the matrix $\pmatrix{-1 & 0 \cr 0 & 1}$.	An eigenvector corresponding to $\lambda = -1$ is $\pmatrix{-1 \cr 3}$. An eigenvector	 corresponding to $\lambda = 1$ is $\pmatrix{-2 \cr 7}$.	 By the theorem, if $P = \pmatrix{-1 & -2 \cr 3 & 7}$, then $P^{-1}AP = \pmatrix{-1 & 0 \cr 0 & 1}$.

\b

If $A$ is an $n \times n$ matrix, $exp(A) = e^A$ is defined to be $\sum_{k=0}^\infty \left( \frac{1}{k!} \right) A^k$.

\b

\textbf{Examples.}

\begin{enumerate}


\item $exp(\pmatrix{3 & 0 \cr 0 & -1}) = \sum_{k=0}^\infty \left( \frac{1}{k!} \right) \pmatrix{3 & 0 \cr 0 & -1}^k = \sum_{k=0}^\infty \left( \frac{1}{k!} \right) \pmatrix{3^k & 0 \cr 0 & (-1)^k} = \pmatrix{ \sum_{k=0}^\infty\frac{3^k}{k!} & 0 \cr 0 & \sum_{k=0}^\infty\frac{(-1)^k}{k!}} = \pmatrix{e^3 & 0 \cr 0 & e^{-1}}$.


    \b

In general, if $D$ is a diagonal matrix, $exp(D)$ is the diagonal matrix whose $ii^{th}$ entry is $e$ raised to the $ii^{th}$ entry of $D$.		

\item Let $A = \pmatrix{-13 & -4 \cr 42 & 13}$.	 Find $exp(A)$.	 $A$ is not diagonal, so we can't proceed as we did in the last example.  However, we saw in an earlier example that $P^{-1}AP$ is the diagonal matrix $D = \pmatrix{-1 & 0 \cr 0 & 1}$ where $P = \pmatrix{-1 & -2 \cr 3 & 7}$.	 Therefore, $A = PDP^{-1}$.	Therefore, $exp(A) = \sum_{k=0}^\infty \left( \frac{1}{k!} \right) (PDP^{-1})^k = \sum_{k=0}^\infty \left( \frac{1}{k!} \right) (PD^kP^{-1}) = P^{-1}(\sum_{k=0}^\infty \left( \frac{1}{k!} \right) D^k)P = P^{-1}\pmatrix{e^{-1} & 0 \cr 0 & e}P = \pmatrix{-7 & -2 \cr 3 & 1}\pmatrix{e^{-1} & 0 \cr 0 & e}\pmatrix{-1 & -2 \cr 3 & 7} =$\\ $\pmatrix{7e^{-1}-6e & 14e^{-1}-14e \cr -3e^{-1}+3e & 6e^{-1}+7e}$.

\end{enumerate}



\b

\{ Page 383 \hspace{1in} 7-9, 15, 16-21 \\

\b

Also, for each of the following matrices $A$, find $exp(A)$.

\begin{enumerate}

\item $A = \pmatrix{2 & 0 \cr 0 & 3}$.

\item $A = \pmatrix{0 & 1 \cr 1 & 0}$.

\item $A = \pmatrix{0 & -1 \cr 1 & 0}$.


\item $A = \pmatrix{2 & 1 \cr 2 & 3}$.




\end{enumerate}


\end{document}














