Today we start Section 2.2. Continue reading Section 2.2 for next class. Work through suggested exercises.
Homework 3 is due on Friday at 11:55pm on WeBWorK.
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Annual General Meeting of Women in Science Western: September 24, 6-7pm, Chemistry Building Rm 9.
Definition: A system of linear equations is a finite set of linear equations, each with the same variables. A solution to the system is a vector that satisfies all of the equations.
Example: $ \begin{aligned} x + y\ &= 2\\ -x + y\ &= 4 \end{aligned} $
$[1, 1]$ is not a solution, but $[-1, 3]$ is. Geometrically, this corresponds to finding the intersection of two lines in $\R^2$.
A system is consistent if it has one or more solutions, and inconsistent if it has no solutions. We'll see later that a consistent system always has either one solution or infinitely many.
We solved it by doing row operations, such as replacing $R_2$ with $R_2 - 3 R_1$ or exchanging rows 2 and 3 until we got it to the form: $$ \kern-6ex \begin{aligned} \ph x - \ph y - \ph z\ &= 2 \\ y + 3 z\ &= 5 \\ 5 z\ &= 10 \end{aligned} \qquad\qquad \bmat{rrr|r} 1 & -1 & -1 & 2 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & 5 & 10 \\ \emat $$ This system is easy to solve, because of its triangular structure. The method is called back substitution: $$ \begin{aligned} z\ &= 2\\ y\ &= 5 - 3z = 5 - 6 = -1\\ x\ &= 2 + y + z = 2 - 1 + 2 = 3. \end{aligned} $$ So the unique solution is $[3, -1, 2]$. We can check this in the original system to see that it works!
Definition: An $m \times n$ matrix is in an array of real numbers with $m$ rows and $n$ columns. For example, this is a $2 \times 3$ matrix: $$ \bmat{rrr} 3 & 2 & 6\\ 4 & -1 & 2 \emat $$
Example: These matrices are in row echelon form: $$ \kern-6ex \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & 0 \emat \qquad \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & \red{4} \emat \qquad \bmat{rrrrr} 0 & \red{3} & 2 & 0 & 4 \\ 0 & 0 & 0 & \red{-1} & 2\\ 0 & 0 & 0 & 0 & \red{4} \emat $$
Example: These matrices are not in row echelon form: $$ \kern-6ex \bmat{rrr} {\bf 0} & {\bf 0} & {\bf 0} \\ \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ \emat \qquad \bmat{rrr} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & {\bf 2} & 4 \emat \qquad \bmat{rrrrr} 0 & \red{3} & 2 & 0 & 4 \\ 0 & 0 & 0 & \red{-1} & 2\\ 0 & 0 & {\bf 2} & 0 & 4 \emat $$ This terminology makes sense for any matrix, but we will usually apply it to the augmented matrix of a linear system. The conditions apply to the entries to the right of the line as well.
Question 8-1: For a $2 \times 3$ matrix, in what ways can the leading entries be arranged? Board.
Just as for triangular systems, we can solve systems in row echelon form using back substitution.
Example 8-2: Solve the system whose augmented matrix is: $$ \bmat{rrr|r} \red{3} & 2 & 2 & 0\\ 0 & 0 & \red{-1} & 2\\ 0 & 0 & 0 & 0 \emat $$ How many variables? How many equations? Solution on board.
Example: Solve the system whose augmented matrix is: $$ \bmat{rr|r} \red{3} & 2 & 0\\ 0 & \red{-1} & 2\\ 0 & 0 & \red{4} \emat $$ How many variables? How many equations?
Note: This is the general pattern for an augmented matrix in row echelon form:
Example 8-3: Board. Reduce the given matrix to row echelon form: $$ \bmat{rrr} -2 & 6 & -7 \\ 3 & -9 & 10 \\ 1 & -3 & 3 \emat $$ Note that there are many ways to proceed, and the row echelon form is not unique.
Note that for a random matrix, row reduction will often lead to many awkward fractions. Sometimes, by choosing the appropriate operations, one can avoid some fractions, but sometimes they are inevitable.
Example: Here's another example: $$ \begin{aligned} \bmat{rrrr} 0 & 4 & 2 & 3 \\ 2 & 4 & -2 & 1 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \xrightarrow{R_1 \leftrightarrow R_2} &\bmat{rrrr} \red 2 & \red 4 & \red{-2} & \red 1 \\ \red 0 & \red 4 & \red 2 & \red 3 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \\ \lra{\frac{1}{2}R_1} \bmat{rrrr} \red 1 & \red 2 & \red{-1} & \red{1/2} \\ 0 & 4 & 2 & 3 \\ -3 & 2 & 2 & 1/2 \\ 0 & 0 & 10 & 8 \emat \lra{R_3 + 3R_1} &\bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ \red 0 & \red 8 & \red{-1} & \red 2 \\ 0 & 0 & 10 & 8 \emat \\ \lra{R_3 - 2R_2} \bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ 0 & \red 0 & \red{-5} & \red{-4} \\ 0 & 0 & 10 & 8 \emat \lra{R_4 + 2R_3} &\bmat{rrrr} \phm 1 & 2 & -1 & 1/2 \\ 0 & 4 & 2 & 3 \\ 0 & 0 & -5 & -4 \\ 0 & 0 & \red 0 & \red 0 \emat \\ \end{aligned} $$
Example 8-4: (On board, if time.) $\bmat{rrr} 2 & 4 & 6 \\ 1 & 2 & 4 \\ -3 & -6 & 4 \emat$
Gaussian elimination: This means to do row reduction on the augmented matrix of a linear system until you get to row echelon form, and then use back substitution to find the solutions.
Example 2.11: Board. Solve the system $$ \begin{aligned} \ph w - \ph x - y + 2 z \ &= \phm 1 \\ 2 w - 2 x - y + 3 z \ &= \phm 3 \\ - w + \ph x - y \phantom{{} + 3 z}\ &= -3 \end{aligned} $$ We call the variables corresponding to a column with a leading entry the leading variables, and the remaining variables the free variables. We solve for the leading variables in terms of the free variables, and assign parameters to the free variables.