Gauss’s Method

Use Gauss’s Method to find the unique solution for each system.

• (a)  $\begin{array}{rcrcr}2x&+&3y&=&13\\ x&-&y&=&-1\end{array}$     (b)  $\begin{array}{rcrcrcr}x&&&-&z&=&0\\ 3x&+&y&&&=&1\\ -x&+&y&+&z&=&4\end{array}$

Each system is in echelon form. For each, say whether the system has a unique solution, no solution, or infinitely many solutions.

• (a)  $\begin{array}{rcrcr}-3x&+&2y&=&0\\ &&-2y&=&0\end{array}$     (b)  $\begin{array}{rcrcrcr}x&+&y&&&=&4\\ &&y&-&z&=&0\\ \end{array}$     (c)  $\begin{array}{rcrcrcr}x&+&y&&&=&4\\ &&y&&-z&=&0\\ &&&&0&=&0\end{array}$     (d)  $\begin{array}{rcrcr}x&+&y&=&4\\ &&0&=&4\end{array}$     (e)  $\begin{array}{rcrcrcr}3x&+&6y&+&z&=&-0.5\\ &&&&-z&=&2.5\\ \end{array}$     (f)  $\begin{array}{rcrcr}x&-&3y&=&2\\ &&0&=&0\end{array}$     (g)  $\begin{array}{rcrcr}2x&+&2y&=&4\\ &&y&=&1\\ &&0&=&4\end{array}$     (h)  $\begin{array}{rcrcr}2x&+&y&=&0\end{array}$     (i)  $\begin{array}{rcrcr}x&-&y&=&-1\\ &&0&=&0\\ &&0&=&4\end{array}$     (j)  $\begin{array}{rcrcrcr}x&+&y&-&3z&=&-1\\ &&y&-&z&=&2\\ &&&&z&=&0\\ &&&&0&=&0\end{array}$

Use Gauss’s Method to solve each system or conclude ‘many solutions’ or ‘no solutions’.

• (a)  $\begin{array}{rcrcr}2x&+&2y&=&5\\ x&-&4y&=&0\end{array}$     (b)  $\begin{array}{rcrcr}-x&+&y&=&1\\ x&+&y&=&2\end{array}$     (c)  $\begin{array}{rcrcrcr}x&-&3y&+&z&=&1\\ x&+&y&+&2z&=&14\end{array}$     (d)  $\begin{array}{rcrcr}-x&-&y&=&1\\ -3x&-&3y&=&2\end{array}$     (e)  $\begin{array}{rcrcrcr}&&4y&+&z&=&20\\ 2x&-&2y&+&z&=&0\\ x&&&+&z&=&5\\ x&+&y&-&z&=&10\end{array}$     (f)  $\begin{array}{rcrcrcrcr}2x&&&+&z&+&w&=&5\\ &&y&&&-&w&=&-1\\ 3x&&&-&z&-&w&=&0\\ 4x&+&y&+&2z&+&w&=&9\end{array}$

Solve each system or conclude ‘many solutions’ or ‘no solutions’. Use Gauss’s Method.

• (a)  $\begin{array}{rcrcrcr}x&+&y&+&z&=&5\\ x&-&y&&&=&0\\ &&y&+&2z&=&7\end{array}$     (b)  $\begin{array}{rcrcrcr}3x&&&+&z&=&7\\ x&-&y&+&3z&=&4\\ x&+&2y&-&5z&=&-1\end{array}$     (c)  $\begin{array}{rcrcrcr}x&+&3y&+&z&=&0\\ -x&-&y&&&=&2\\ -x&+&y&+&2z&=&8\end{array}$

We can solve linear systems by methods other than Gauss’s. One often taught in high school is to solve one of the equations for a variable, then substitute the resulting expression into other equations. Then we repeat that step until there is an equation with only one variable. From that we get the first number in the solution and then we get the rest with back-substitution. This method takes long…

For which values of $k$ are there no solutions, many solutions, or a unique solution to this system?

This system is not linear in that it says $\sin\alpha$ instead of $\alpha$

and yet we can apply Gauss’s Method. Do so. Does the system have a solution?

What conditions must the constants, the $b$’s, satisfy so that each of these systems has a solution? Hint. Apply Gauss’s Method and see what happens to the right side.

• (a)  $\begin{array}{rcrcr}x&-&3y&=&b_{1}\\ 3x&+&y&=&b_{2}\\ x&+&7y&=&b_{3}\\ 2x&+&4y&=&b_{4}\end{array}$     (b)  $\begin{array}{rcrcrcr}x_{1}&+&2x_{2}&+&3x_{3}&=&b_{1}\\ 2x_{1}&+&5x_{2}&+&3x_{3}&=&b_{2}\\ x_{1}&&&+&8x_{3}&=&b_{3}\end{array}$

True or false: a system with more unknowns than equations has at least one solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must produce a counterexample.)

Must any Chemistry problem like the one that starts this subsection — a balance the reaction problem — have infinitely many solutions?

Find the coefficients $a$, $b$, and $c$ so that the graph of $f(x)=ax^{2}+bx+c$ passes through the points $(1,2)$, $(-1,6)$, and $(2,3)$.

After Theorem 1.5 we note that multiplying a row by $0$ is not allowed because that could change a solution set. Give an example of a system with solution set $S_{0}$ where after multiplying a row by $0$ the new system has a solution set $S_{1}$ and $S_{0}$ is a proper subset of $S_{1}$, that is, $S_{0}\neq S_{1}$. Give an example where $S_{0}=S_{1}$.

Gauss’s Method works by combining the equations in a system to make new equations.

• Can we derive the equation $3x-2y=5$ by a sequence of Gaussian reduction steps from the equations in this system?

• Can we derive the equation $5x-3y=2$ with a sequence of Gaussian reduction steps from the equations in this system?

• Can we derive $6x-9y+5z=-2$ by a sequence of Gaussian reduction steps from the equations in the system?

Prove that, where $a,b,c,d,e$ are real numbers with $a\neq 0$, if this linear equation

has the same solution set as this one

then they are the same equation. What if $a=0$?

Show that if $ad-bc\neq 0$ then

has a unique solution.

In the system

each of the equations describes a line in the $xy$-plane. By geometrical reasoning, show that there are three possibilities: there is a unique solution, there is no solution, and there are infinitely many solutions.

Finish the proof of Theorem 1.5.

Is there a two-unknowns linear system whose solution set is all of $\mathbb{R}^{2}$?

Are any of the operations used in Gauss’s Method redundant? That is, can we make any of the operations from a combination of the others?

Prove that each operation of Gauss’s Method is reversible. That is, show that if two systems are related by a row operation $S_{1}\rightarrow S_{2}$ then there is a row operation to go back $S_{2}\rightarrow S_{1}$.

A box holding pennies, nickels and dimes contains thirteen coins with a total value of $83$ cents. How many coins of each type are in the box? (These are US coins; a penny is $1$ cent, a nickel is $5$ cents, and a dime is $10$ cents.)

Four positive integers are given. Select any three of the integers, find their arithmetic average, and add this result to the fourth integer. Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers is:

• (a)  19     (b)  21     (c)  23     (d)  29     (e)  17

Laugh at this: $\text{AHAHA}+\text{TEHE}=\text{TEHAW}$. It resulted from substituting a code letter for each digit of a simple example in addition, and it is required to identify the letters and prove the solution unique.

The Wohascum County Board of Commissioners, which has 20 members, recently had to elect a President. There were three candidates ($A$, $B$, and $C$); on each ballot the three candidates were to be listed in order of preference, with no abstentions. It was found that 11 members, a majority, preferred $A$ over $B$ (thus the other 9 preferred $B$ over $A$). Similarly, it was found that 12 members preferred $C$ over…

“This system of $n$ linear equations with $n$ unknowns,” said the Great Mathematician, “has a curious property.”

“Good heavens!” said the Poor Nut, “What is it?”

“Note,” said the Great Mathematician, “that the constants are in arithmetic progression.”

“It’s all so clear when you explain it!” said the Poor Nut. “Do you mean like $6x+9y=12$ and $15x+18y=21$?”

“Quite so,” said the Great Mathematician, pulling out his basso…