305 – Homework I

This homework set is due Wednesday, February 1st, at the beginning of lecture, but feel free to turn it in earlier if possible.

1. Find the general form of a $2\times 2$ matrix $X$ (with real entries) satisfying the equation $XA=B$ where $\displaystyle A=\left(\begin{array}{cc}1&2\\1&2\end{array}\right)$ and $\displaystyle B=\left(\begin{array}{cc}3&6\\5&10\end{array}\right)$, or explain why no such $X$ exists. Note that the method explained in class does not work here, since $A$ is not invertible.

2. With the same $A$ and $B$, find the general form of $X$ (or explain why it does not exist) if now we require that $AX=B$.

3. Consider the set $F$ of $2\times 2$ matrices $A$ whose entries are elements of ${\mathbb Z}/7{\mathbb Z}$, and that have the form $\displaystyle A=\left(\begin{array}{cc}r&-s\\ s&r\end{array}\right)$. How many matrices are there in the set $F$?

Show that if $B$ and $C$ are in $F$, then $BC=CB$.

Show that if $B\ne0$ and $B\in F$, then $B^{-1}$ exists and also belongs to $F$.

Show that one of these matrices, “$i$,” satisfies $i^2+I=0$.

Solve the equation $X^2-X-I=0$ with $X\in F$, and check that the solution you obtain coincides with the solution described by the quadratic formula, that in this case looks like $\displaystyle X=\frac{I\pm\sqrt{5I}}{2},$ where of course by $\sqrt{5I}$ we mean a matrix (in $F$) whose square is $5I$.

4. We examine here the solutions of the cubic $x^3+ax^2+bx+c=0$, following what is essentially Tartaglia’s method.

Show that if we define $y$ by $\displaystyle x=y-\frac a3$, then the equation now takes the form $y^3+dy+e=0$.

Show that, no matter what $d$ and $e$ are, we can always find $\alpha$ and $\beta$ such that $\alpha+\beta=-e$ and $27\alpha\beta=-d^3$.

Suppose that $p,q$ are such that $p^3=\alpha$, $q^3=\beta$, and $3pq=-d$. Show that $p+q$ is a root of $y^3+dy+e$. [Hint: Note that $(p+q)^3=p^3+q^3+3(p+q)pq$.]

Let $r$ be a cubic root of 1, $r\ne 1$. (Remember that this means that $r^3=1$.) Check that $pr+qr^2$ and $pr^2+qr$ are also roots of $y^3+dy+e$, and that $p+q,pr+qr^2,pr^2+qr$ are all the roots.

Use this method to solve $x^3-3x^2+x-3=0$. Now note that $3,\sqrt{-1},-\sqrt{-1}$ are the roots of this polynomial. Reconcile this with the expressions you have found.

Use this method to solve $x^3+x+5=0$ in ${\mathbb Z}/7{\mathbb Z}$.

5. Here we solve the cubic using trigonometry. First, prove that $\cos(3\theta)=4\cos^3\theta-3\cos\theta$.

Start with a cubic equation $x^3+ax^2+bx+c=0$. As before, we can turn it into one of the form $y^3+dy+e=0$ by means of a simple translation. Now, if $d$ happens to be 0, show how to find the roots. Suppose then that $d\ne0$. Find a value of $k$ such that $\displaystyle \frac d{k^2}=-\frac34$, and show that (for this $k$) if $y=kz$, then the equation becomes $4z^3-3z+f=0$ for some $f$.

Comparing the results from the previous two paragraphs, we see that if $\theta$ is such that $f=-\cos(3\theta)$, then $z=\cos\theta$ is a root of the cubic in $z$. Use this to solve $x^3+3x^2-3=0$.

[This method seems more limited than the previous one, because we are used to thinking of $\theta$ as a real number, in which case $\cos\theta$ must be a number between $-1$ and $1$. However, if we allow $\theta$ to be complex, then $\cos\theta$ can take any value, and the formulas we obtain by this method actually coincide with the ones found in the previous problem.]

6. This is the method discussed in class. Suppose we are given the cubic equation $x^3+3bx^2+3cx+d=0$ and it has roots $\alpha,\beta,\gamma$. [Note the main coefficient is 1 and I’m writing $3b,3c$ instead of $b,c$.]

The point here was to find a polynomial $p$ in 3 variables with the property that some power of $p$ would take only two values (rather than six) as we permute the variables. The goal was (using a quadratic) to find these values when $\alpha,\beta,\gamma$ are used as the variables,  and then use these values to find $\alpha,\beta,\gamma$ themselves.

As mentioned in class, if we let $\epsilon$ be a cubic root of 1, $\epsilon\ne 1$, then the polynomial $p(r,s,t)=r+\epsilon s+\epsilon^2t$ works because $p(r,s,t)^3$ only takes 2 values, namely $(r+\epsilon s+\epsilon^2t)^3$ and $(r+\epsilon^2 s+\epsilon t)^3$.

When $\alpha,\beta,\gamma$ are used in place of $r,s,t$, we get the expressions $(\alpha+\epsilon\beta+\epsilon^2\gamma)^3$ and $(\alpha+\epsilon^2\beta+\epsilon\gamma)^3.$ Explain how to find $\alpha,\beta,\gamma$ if we know these two expressions.

Show that these two numbers are the roots of the quadratic

$y^2+27(d-3bc+2b^3)y+729(b^2-c)^3=0$.

Use this to solve the cubic $2x^3+6x^2+18x+10=0$.

Extra credit problems:

(Extra credit problems can be turned in by February 8 at the latest.)

7. To solve the quartic equation, one would use the same procedure: Start with a quartic, find a polynomial in 4 variables a power of which takes at most 3 values (rather than $4!=24$), Evaluate these powers when the roots of the quartic are used as the variables, and use a cubic to find these three values. Then use these three values to find the actual roots.

Suppose the quartic is $f(x)=x^4+ax^3+bx^2+cx+d$. Under the substitution $\displaystyle x=y-\frac a4$ this becomes $g(y)=y^4+py^2+qy+r$. Find $p,q,r$ in terms of $a,b,c,d$.

Let the roots of $g$ be $\alpha_1,\alpha_2,\alpha_3,\alpha_4$. Check that $\alpha_1+\alpha_2+\alpha_3+\alpha_4=0$.

Consider $p(r,s,t,u)=(r+s)(t+u)$ and prove that $p$ takes only 3 values as we permute $r,s,t,u$. When evaluated at $\alpha_1,\dots,\alpha_4$, these 3 values are

$\theta_1=(\alpha_1+\alpha_2)(\alpha_3+\alpha_4)$, $\theta_2=(\alpha_1+\alpha_3)(\alpha_2+\alpha_4)$, and $\theta_3=(\alpha_1+\alpha_4)(\alpha_2+\alpha_3)$.

Prove that the cubic $h(z)=z^3-2pz^2+(p^2-4r)z+q^2$ has roots $\theta_1,\theta_2,\theta_3$.

Check that we get

$2\alpha_1=\sqrt{-\theta_1}+\sqrt{-\theta_2}+\sqrt{-\theta_3}$,

$2\alpha_2=\sqrt{-\theta_1}-\sqrt{-\theta_2}-\sqrt{-\theta_3}$,

$2\alpha_3=-\sqrt{-\theta_1}+\sqrt{-\theta_2}-\sqrt{-\theta_3}$,

and

$2\alpha_4=-\sqrt{-\theta_1}-\sqrt{-\theta_2}+\sqrt{-\theta_3}$.

Use this to solve $x^4-4x^3+x^2-4x+1=0$.

What Galois proved can be phrased in these terms as saying that if we want to solve in general the equation of degree $n$, then we need to be able to find a polynomial in $n$ variables, a power of which takes fewer than $n$ values as the order of the variables is permuted, and using these values we should be able to recover the roots of the original polynomial of degree $n$. He proved that this is not possible for $n\ge5$.

8. The Arabian mathematicians of the middle ages where able to solve quadratic equations but not cubics, and could not understand equations of degree 4 or higher. This is because they understood the equations geometrically, so squares represented areas and cubes volumes. In addition, they only understood positive numbers. So, an equation such as $x^2-3x-5=0$ necessarily had to be presented as $x^2=3x+5$, while something like $x^2-2x+5=0$ would be written as $x^2+5=2x$ and thought of as being meaningless. Their main other drawback was that their arguments were rhetorical, meaning that they never used variables, which greatly complicated their exposition. For example, instead of asking to solve $x^2=3x+5$, they would say

“If three times an unknown added to 5 is equal to the square of that unknown, what is the value of the unknown?”

Investigate how they used geometric diagrams to solve quadratic equations, and write a (short) exposition of their techniques. Since numbers are non-negative, there are at least three cases, that need slightly different techniques: For equations of the form $x^2=bx+c$, of the form $x^2+bx=c$, and of the form $x^2+c=bx$. If we insist that quantities cannot be zero, then there are additional cases $x^2=bx$, $x^2=c$, etc.