Author Topic: BORED already? Want to learn a bit of maths? :P (Read 27010 times) Tweet Share

b^3 · « **on:** November 16, 2012, 01:05:33 pm »

Since I've finished uni exams, I've been pretty bored already to the point where I started making worked examples and notes for one of my uni/eng maths units from last semester. Anyway I thought some of you who may also be bored, and enjoy maths

, may be interested. If people are interested then I may make a few more like this, with some of the topics we covered (e.g. gauss elimination and such). So yeh, have a read through and try the questions if you want, and post the sols up if you can get them

.

I probably should point out that this is after spesh, and you won't be allowed to use these techniques in VCE maths. Although UMEP students would have probably covered this already.

Integration by Parts

We are all familar with the product rule, which we can rearrange to get the rule for integration by parts.
$\begin{alignedat}{1}\frac{d}{dx}\left(u(x)v(x)\right) & =\frac{du}{dx}v(x)+u(x)\frac{dv}{dx} \\ \int\left(\frac{du}{dx}v(x)+u(x)\frac{dv}{dx}\right)dx & =u(x)v(x) \\ \int\frac{du}{dx}v(x)dx+\int u(x)\frac{dv}{dx}dx & =uv \\ \int\frac{du}{dx}v(x)dx & =uv-\int u\frac{dv}{dx}dx \end{alignedat}$

So we have two 'functions', $u(x)$ and $v(x)$ . The goal is to end up with a simpler integral than the original. This may be done by picking $u$ as the right function, so that both $u$ and the term $u\frac{dv}{dx}$ is anti-differentiable.
e.g. $\begin{alignedat}{1}\int xe^{x}dx\end{alignedat}$
So here, if we pick $u=e^{x}$ and $v=x$ , we will end up with the term $u\frac{dv}{dx}=e^{x}(1)$ which is anti-differentiable, and a simpler integral than the first.
$\begin{alignedat}{1}\frac{d}{dx}\left(u(x)v(x)\right) & =\frac{du}{dx}v(x)+u(x)\frac{dv}{dx} \\ \int\left(\frac{du}{dx}v(x)+u(x)\frac{dv}{dx}\right)dx & =u(x)v(x) \\ \int\frac{du}{dx}v(x)dx+\int u(x)\frac{dv}{dx}dx & =uv \\ \int\frac{du}{dx}v(x)dx & =uv-\int u\frac{dv}{dx}dx \end{alignedat}$
So basically its what you would have come across in Methods, integration by recognition, but backwards. So we have two 'functions', $u(x)$ and $v(x)$ . The goal is to end up with a simpler integral than the original. This may be done by picking $u$ as the right function, so that both $u$ and the term $u\frac{dv}{dx}$ is anti-differentiable.
e.g. $\begin{alignedat}{1}\int xe^{x}dx\end{alignedat}$
So here, if we pick $u=e^{x}$ and $v=x$ , we will end up with the term $u\frac{dv}{dx}=e^{x}(1)$ which is anti-differentiable, and a simpler integral than the first.
$\begin{alignedat}{1}\int xe^{x}dx & =\int x\frac{d}{dx}\left(e^{x}\right)dx \\ & =xe^{x}-\int(1)e^{x}dx \\ & =xe^{x}-e^{x}+C \end{alignedat}$
We can also note that if we had of picked $u$ and $v$ the other way round, we would get $u\frac{dv}{dx}=\frac{1}{2}x^{2}e^{x}$ which is a worse integral than we started with.
$\begin{alignedat}{1}\int xe^{x}dx & =\int\frac{d}{dx}\left(\frac{1}{2}x^{2}\right)e^{x}dx \\ & =\frac{1}{2}x^{2}e^{x}-\int\frac{1}{2}x^{2}e^{x}dx \end{alignedat}$

So lets try another one.
e.g. $\begin{alignedat}{1}\int\ln(x)dx\end{alignedat}$
Now the little trick to this one is to look at it as $\begin{alignedat}{1}\int\left(1\right)\ln(x)dx\end{alignedat}$ , that is $u=1$ , $v=\ln(x)$
$\begin{alignedat}{1}\int\ln(x)dx & =\int\frac{d}{dx}\left(x\right)\ln(x)dx \\ & =x\ln(x)-\int x\frac{1}{x}dx \\ & =x\ln(x)-\int1dx \\ & =x\ln(x)-x+C \end{alignedat}$

We may even need to apply integration by parts twice, as sometimes the first integral won't be any simpler (note: this won't always work).
$\begin{alignedat}{1}\int x^{2}\cos(x)dx & =\int\frac{d}{dx}\left(\sin(x)\right)x^{2}dx \\ & =x^{2}\sin(x)-\int2x\sin(x)dx \\ & =x^{2}\sin(x)-2\int\frac{d}{dx}\left(-\cos(x)\right)xdx \\ & =x^{2}\sin(x)-2\left(-x\cos(x)-\int(1)(-\cos(x)dx\right) \\ & =x^{2}\sin(x)+2x\cos(x)-2\sin(x)+C \end{alignedat}$
You don't have to show all that working, but I've kept it all there for now.

One last example, sometimes when we use integration by parts twice, we can obtain the same integral, as the one we were originally looking for, in the RHS, and add it back to the other side, then solve for it to leave what the integral equals.
e.g.
$\begin{alignedat}{1}\int e^{x}\sin(x)dx & =\int e^{x}\frac{d}{dx}\left(-\cos(x)\right)dx \\ & =-e^{x}\cos(x)+\int e^{x}\cos(x)dx \\ & =-e^{x}\cos(x)+\int e^{x}\frac{d}{dx}\left(\sin(x)\right)dx \\ & =-e^{x}\cos(x)+e^{x}\sin(x)-\int e^{x}\sin(x)dx \end{alignedat}$
As we can see we know have the original integral $\begin{alignedat}{1}\int e^{x}\sin(x)dx\end{alignedat}$ on the RHS as well, so we can 'solve' for it by adding it to the left hand side and dividing by 2.
$\begin{alignedat}{1}\int e^{x}\sin(x)dx & =-e^{x}\cos(x)+e^{x}\sin(x)-\int e^{x}\sin(x)dx \\ 2\int e^{x}\sin(x)dx & =e^{x}\left(\sin(x)-\cos(x)\right)+C_{1} \\ \int e^{x}\sin(x)dx & =\frac{1}{2}e^{x}\left(\sin(x)-\cos(x)\right)+C_{2} \end{alignedat}$

Now for some questions to try (answers are in the spolier tags).
$\begin{alignedat}{1}1. & \int x^{3}\ln(x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int x^{3}\ln(x)dx & =\frac{1}{16}x^{4}\left(4\ln(x)-1\right)+C\end{alignedat}$

$\begin{alignedat}{1}2. & \int e^{2x}\cos(3x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int e^{2x}\cos(3x)dx & =\frac{1}{13}e^{2x}\left(3\sin(3x)+2\cos(3x)\right)+C\end{alignedat}$

$\begin{alignedat}{1}3. & \int\frac{\ln(x)}{x}dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int\frac{\ln(x)}{x}dx & =\frac{1}{2}\left(\ln(x)\right)^{2}+C\end{alignedat}$

$\begin{alignedat}{1}4. & \int x\ln(x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int x\ln(x)dx & =\frac{1}{4}x^{2}\left(2\ln(x)-1\right)+C\end{alignedat}$

$\begin{alignedat}{1}5. & \int x\tan^{-1}(x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int x\tan^{-1}(x)dx & =\frac{1}{2}\left(\left(x^{2}+1\right)\tan^{-1}(x)-x\right)+C\end{alignedat}$

$\begin{alignedat}{1}6. & \int x^{2}\sin^{-1}(x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int x^{2}\sin^{-1}(x)dx & =\frac{1}{3}x^{3}\sin^{-1}(x)+\frac{1}{3}\left(1-x^{2}\right)^{\frac{1}{2}}-\frac{1}{9}\left(1-x^{2}\right)^{\frac{3}{2}}+C\end{alignedat}$

$\begin{alignedat}{1}7. & \int\left(\ln(x)\right)^{2}dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int\left(\ln(x)\right)^{2}dx & =x\left(\ln(x)\right)^{2}-2x\ln(x)+2x+C\end{alignedat}$

We may also sometimes use euler's identity $e^{ix}=\cos(x)+i\sin(x)$ to use a simpler method to find the anti-derivative of some of the above. But first we are going to show one way of why $e^{ix}=\cos(x)+i\sin(x)$ . Firstly we start off with a differential equation, $\frac{d}{dx}\left(y\right)=ky$ with the initial condition of $y(0)=1$ . So we are looking for a function that basically is it's own derivative, so we can use $e^{kx}$ . Now what happens if we let $k=i$ ?
$\begin{alignedat}{1}\frac{d}{dx}\left(e^{ix}\right) & =ie^{ix}\end{alignedat}$
We can check that this satisfies $y(0)=1$ , $y(0)=e^{0i}=1$ .
Now we know that $y=e^{ix}$ will be a solution to the differential equation, but if we look at another function, $y=\cos(x)+i\sin(x)$ we can see that it is a solution as well.
$\begin{alignedat}{1}y & =\cos(x)+i\sin(x) \\ \frac{dy}{dx} & =-\sin(x)+i\cos(x) \\ & =i^{2}\sin(x)+i\cos(x) \\ & =i\left(\cos(x)+i\sin(x)\right) \\ & =iy \end{alignedat}$
Now checking with the initial condition,
$\begin{alignedat}{1}y(0) & =\cos(0)+i\sin(0) \\ & =1+0i \\ & =1 \end{alignedat}$
Since both solutions are solutions ot the same equation, and satisfy the initial condition, we can say that they are the same. $\therefore e^{ix}=\cos(x)+i\sin(x)$

So now that we have seen one way of showing euler's identity, we can use it to find some anti-derivatives.
Lets start off looking at $\begin{alignedat}{1}\int e^{2x}\cos(3x)dx\end{alignedat}$
We can find the anti-derivative of s similar function involving $\cos(3x)+i\sin(3x)$ , and then take the real part of it (we use the real part for $\cos(x)$
and the imaginary part for $\sin(x)$ ).
$\begin{alignedat}{1}\int e^{2x}\cos(3x)dx & =Re\left\{ \int e^{2x}\left(\cos(3x)+i \sin(3x)\right)\right\} \\ & =Re\left\{ \int e^{2x}e^{3ix}dx\right\} \\ & =Re\left\{ \int e^{x\left(2+3i\right)}dx\right\} \\ & =Re\left\{ \frac{1}{2+3i}\times\frac{2-3i}{2-3i}e^{2x}e^{3ix}+C_{1}\right\} \\ & =Re\left\{ \frac{2-3i}{4-9i^{2}}e^{2x}\left(\cos(3x)+i \sin(3x)\right)+C_{1}\right\} \\ & =Re\left\{ \frac{2-3i}{13}e^{2x}\left(\cos(3x)+i\sin(3x)\right)+C_{1}\right\} \end{alignedat}$
So to find the real parts we remember that a real part multiplied by a real part gives a real number, and the imaginary part multipled by an imaginary part gives a real part (as $i\times i=i^{2}=-1$ ). So that is the $\frac{2}{13}e^{2x}$ multipled by the $\cos(3x)$ and the $\frac{-3i}{13}e^{2x}$ multiplied by the $i\sin(3x)$ .
$\begin{alignedat}{1}\int e^{2x}\cos(3x)dx & =\frac{2}{13}e^{2x}\cos(3x)-\frac{3i}{13}e^{2x}\times i \sin(3x)+C_{1} \\ & =\frac{2}{13}e^{2x}\cos(3x)+\frac{3}{13}e^{2x}\sin(3x)+C_{1} \\ & =\frac{1}{13}e^{2x}\left(\cos(3x)+\sin(3x)\right)+C_{1} \end{alignedat}$
Which gives the same result as the other method, you wouldn't have to show all these steps, which means sometimes this method is quicker.

If we were looking at a function with $\sin(x)$ instead of $\cos(x)$ then we would be looking for the imaginary part instead of the real part, which means that when we get to the multiplying out line, we will be looking for the real parts multiplied by imaginary parts to give an imaginary part.
$\begin{alignedat}{1}\int e^{2x}\sin(x)dx & =Im\left\{ \int e^{2x}\left(\cos(x)+i \sin(x)\right)dx\right\} \\ & =Im\left\{ \int e^{x(2+i)}dx\right\} \\ & =Im\left\{ \frac{1}{2+i}\times\frac{2-i}{2-i}e^{2x}\left(\cos(x)+i \sin(x)\right)+C_{2}\right\} \\ & =Im\left\{ \frac{2-i}{5}e^{2x}\left(\cos(x)+i\sin(x)\right)+C_{2}\right\} \\ & =\frac{2}{5}e^{2x}\times\sin(x)-\frac{1}{5}e^{2x}\times\cos(x)+C_{3} \\ & =\frac{1}{5}e^{2x}\left(2\sin(x)-\cos(x)\right)+C_{3} \end{alignedat}$

Now for some questions to try.
$\begin{alignedat}{1}1. & \int e^{-3x}\cos(4x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int e^{-3x}\cos(4x)dx & =\frac{1}{25}e^{-3x}\left(4\sin(4x)-3\cos(4x)\right)+C\end{alignedat}$

$\begin{alignedat}{1}2. & \int e^{2x}\sin(2x)dx\end{alignedat}$

Spoiler

$\begin{alignedat}{1}\int e^{2x}\sin(2x)dx & =\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)+C\end{alignedat}$

Bhootnike · « **Reply #1 on:** November 16, 2012, 01:25:20 pm »

this is sex

polar · « **Reply #2 on:** November 16, 2012, 04:02:16 pm »

thank you. this board has been very quiet lately. anyway, here's another way of proving euler's identity by using the maclaurin series,

$\begin{aligned} \cos(\theta) &= \sum_{n= 0}^{\infty} \frac{(-1)^n\theta^{2n}}{(2n)!} \\ &= 1 - \frac{\theta^2}{2} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + ... \\ \sin(\theta) &= \sum_{n=0}^{\infty} \frac{(-1)^n\theta^{2n+1}}{(2n+1)!} \\ &= \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + ... \\ i\sin(\theta) &= i\theta - \frac{i\theta^3}{3!} + \frac{i\theta^5}{5!}- \frac{i\theta^7}{7!} + ... \\ \cos(\theta) + i\sin(\theta) &= 1 + i\theta - \frac{\theta^2}{2} - \frac{i\theta^3}{3!} + ... \end{aligned}$

$\begin{aligned} \\ e^{\theta} &= 1 + \theta + \frac{\theta^2}{2} + \frac{\theta^3}{3!} + ... \\ e^{i\theta} &= 1 + i\theta + \frac{(i\theta)^2}{2} + \frac{(i\theta)^3}{3!} \\ &= 1 + i\theta - \frac{\theta^2}{2} - \frac{i\theta^3}{3!} + ... \end{aligned}$

$\text{Thus} \; e^{i\theta} &= \cos(\theta) + i\sin(\theta)$

b^3 · « **Reply #3 on:** November 17, 2012, 12:31:55 am »

Got bored again... Anyway, hope you guys enjoy it, and I just realised I forgot to put in how to do any square sized matrix inverses by hand, may do that if I get bored again. Anyway, hope you guys enjoy it

Gaussian Elimination

Gauss Elimination can be used as a method of solving linear equations (it also leads into finding the inverse of matricies by hand). If we had the system
$\begin{alignedat}{1}x+3y & =5.....[1] \\ 2x-y & =3.....[2] \end{alignedat}$
Now to solve them by previous methods, we would multiply equation (1) by 2, then subtract the equation (2) from (3)
i.e.
$\begin{alignedat}{1}2x+6y & =10.....[3] \\ 2x-y & =3.....[2] \end{alignedat}$
$[3]-[2]$
$\begin{alignedat}{1}7y & =7 \\ \therefore y & =1 \\ 2x-(1) & =3 \\ \therefore x & =2 \end{alignedat}$
i.e. Our solutions would be $(2,1)$
So basically what we can do is, we can
• add or subtract one equation to another
• mutiply an equation by a scalar excluding zero
• swap the order of equations

Now we can apply this to gaussian elimination by doing the same things to the rows of the augmented matrix. The augmented matrix, contains the coefficients of each of the unknowns, and the furtherst right hand column has what the equation equals.
e.g. For the system
$\begin{alignedat}{1}4x+2y-z & =1 \\ 2x+2y+z & =2 \\ -2x-3y+2z & =4 \end{alignedat}$
The Augmented Matrix would be
$\left[\begin{array}{ccc|} 4 & 2 & -1 \\ 2 & 2 & 1 \\ -2 & -3 & 2 \end{array}\left.\begin{array}{c} 1 \\ 2 \\ 4 \end{array}\right]\right.$
So what we aim to do is, to make a 'pivot' in each row. A pivot will have all the entries to the left of it being 0, and each pivot must be to the right of the pivot in the row above it. So if we wanted to form a pivot in the second row, we need to try and do it in the second column, that is we need to make the 2 in the second row, column 1 be 0. To do this we can take half of row 1 away from row 2.
$\begin{alignedat}{1}r_{2}-\frac{1}{2}r_{1}\rightarrow r_{2} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{3}{2} \\ -2 & -3 & 2 \end{array}\left.\begin{array}{c} 1 \\ \vphantom{\int_{a}^{b}}\frac{3}{2} \\ 4 \end{array}\right]\right.\end{alignedat}$
Now we have a pivot at in row 1 column 1 and row 2 column 2. So we try to form the next pivot by making row 3, columns 1 and 2 become 0. Working on row 3 column 1, we can take a multiple of row 1 away, and for column 2 we can take a multiple of row 2 away, as the zero in row 2 means we won't affect what we've done to row 3 column 1.
$\begin{alignedat}{1}r_{3}+\frac{1}{2}r_{1}\rightarrow r_{3} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{3}{2} \\ 0 & -2 & \vphantom{\int_{a}^{b}}\frac{3}{2} \end{array}\left.\begin{array}{c} 1 \\ \vphantom{\int_{a}^{b}}\frac{3}{2} \\ \vphantom{\int_{a}^{b}}\frac{9}{2} \end{array}\right]\right.\end{alignedat}$
$\begin{alignedat}{1}r_{3}+2r_{2}\rightarrow r_{3} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{3}{2} \\ 0 & 0 & \vphantom{\int_{a}^{b}}\frac{9}{2} \end{array}\left.\begin{array}{c} 1 \\ \vphantom{\int_{a}^{b}}\frac{3}{2} \\ \vphantom{\int_{a}^{b}}\frac{15}{2} \end{array}\right]\right.\end{alignedat}$
What we end up with is the row echelon form of the system.
Note: That we are using row 1 to change column 1 and row 2 to change column 2, in that order, so that we don't lose the 0's that we form.
Now we come to the Back Substituion stage. We first solve for the varible in the last row, then use that result to solve for the second last variable and so on.
$\begin{alignedat}{1}\frac{9}{2}z & =\frac{15}{2} \\ \therefore z & =\frac{5}{3} \\ y+\frac{3}{2}z & =\frac{3}{2} \\ y & =\frac{3}{2}-\frac{3}{2}\left(\frac{5}{3}\right) \\ & =\frac{3}{2}-\frac{5}{2} \\ \therefore y & =-1 \\ 4x+2y-z & =1 \\ x & =\frac{1-2\left(-1\right)+\left(\frac{5}{3}\right)}{4} \\ & =\frac{\frac{9}{3}+\frac{5}{3}}{4} \\ & =\frac{14}{12} \\ \therefore x & =\frac{7}{6} \end{alignedat}$
$\therefore x=\frac{7}{6},\: y=-1,\: z=\frac{5}{3}$

So lets do another example.
e.g. Find the solution to the following system of equations.
$\begin{alignedat}{1}4x+2y-z & =2 \\ 2x+y+2z & =-1 \\ x+y+z & =-1 \end{alignedat}$
$\begin{alignedat}{1} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{array}\left.\begin{array}{c} 2 \\ -1 \\ -1 \end{array}\right]\right. \\ \begin{aligned}r_{2}-\frac{1}{2}r_{1} & \rightarrow r_{2} \\ r_{3}-\frac{1}{4}r_{1} & \rightarrow r_{3} \end{aligned} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & 0 & \vphantom{\int_{a}^{b}}\frac{5}{2} \\0 & \vphantom{\int_{a}^{b}}\frac{1}{2} & \vphantom{\int_{a}^{b}}\frac{5}{4} \end{array}\left.\begin{array}{c} 2 \\ -2 \\ \vphantom{\int_{a}^{b}}-\frac{3}{2} \end{array}\right]\right. \end{alignedat}$
Now since the pivot in the second row is to the left of a non-zero element in the third row, we want to swap the second and third rows, which will leave us with a pivot in each row.
$\begin{alignedat}{1}r_{2}\leftrightarrow r_{3} & \left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & \vphantom{\int_{a}^{b}}\frac{1}{2} & \vphantom{\int_{a}^{b}}\frac{5}{4} \\ 0 & 0 & \vphantom{\int_{a}^{b}}\frac{5}{2} \end{array}\left.\begin{array}{c} 2 \\ \vphantom{\int_{a}^{b}}-\frac{3}{2} \\ -2 \end{array}\right]\right.\end{alignedat}$
$\begin{alignedat}{1}\frac{5}{2}z & =-2 \\ \therefore z & =-\frac{4}{5} \\ \frac{1}{2}y+\frac{5}{4}z & =-\frac{3}{2} \\ y & =\frac{-5z-6}{2} \\ & =\frac{-5\left(-\frac{4}{5}\right)-6}{2} \\ & =\frac{-2}{2} \\ \therefore y & =-1 \\ 4x+2y-z & =2 \\ x & =\frac{2+z-2y}{4} \\ & =\frac{\frac{10}{5}-\frac{4}{5}+\frac{10}{5}}{4} \\ & =\frac{16}{20} \\ \therefore x & =\frac{4}{5} \end{alignedat}$
$\therefore x=\frac{4}{5},\: y=-1,\: z=-\frac{4}{5}$

Now for a few problems to try
$\begin{alignedat}{1}1. & \begin{alignedat}{1}6x+3y-z & =3 \\ -3x-2y+2z & =-1 \\ x+y+z & =1 \end{alignedat} \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 6 & 3 & -1 \\ 0 & \vphantom{\int_{a}^{b}}-\frac{1}{2} & \vphantom{\int_{a}^{b}}\frac{3}{2} \\ 0 & 0 & \vphantom{\int_{a}^{b}}\frac{8}{3} \end{array}\left.\begin{array}{c} 3 \\ \vphantom{\int_{a}^{b}}\frac{1}{2} \\ 1 \end{array}\right]\right.$ , Solution: $\therefore x=\frac{1}{2},\: y=\frac{1}{8},\: z=\frac{3}{8}$

$\begin{alignedat}{1}2. & \begin{alignedat}{1}2x+4y-z & =2 \\ x+y+z & =1 \\ 8x+4y & =1 \end{alignedat} \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 2 & 4 & -1 \\ 0 & -1 & \vphantom{\int_{a}^{b}}\frac{3}{2} \\ 0 & 0 & -2 \end{array}\left.\begin{array}{c} 2 \\ 0 \\ -1 \end{array}\right]\right.$ , Solution: $\therefore x=-\frac{1}{4},\: y=\frac{3}{4},\: z=\frac{1}{2}$

$\begin{alignedat}{1}3. & \begin{alignedat}{1}4x-2z & =3 \\ y+z & =1 \\ 2x+4y & =5 \end{alignedat} \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 4 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & -3 \end{array}\left.\begin{array}{c} 3 \\ 1 \\ -\frac{1}{2} \end{array}\right]\right.$ , Solution: $\therefore x=\frac{5}{6},\: y=\frac{5}{6},\: z=\frac{1}{6}$

$\begin{alignedat}{1}4. & \begin{alignedat}{1}x+y+z+u+v & =1 \\ x+2y-z+2v & =4 \\ u+v & =2 \\ z+2u-v & =1 \\ x+y+2z+4u+5v & =5 \end{alignedat} \end{alignedat}$

Spoiler

$\left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & -1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 1 & 1 & 2 & 4 & 5 \end{array}\left.\begin{array}{c} 1 \\ 4 \\ 2 \\ 1 \\ 5 \end{array}\right]\right.$ , Solution: $\therefore x=\frac{5}{4},\: y=0,\: z=-\frac{9}{4},\: u=\frac{7}{4},\: v=\frac{1}{4}$
$\begin{alignedat}{1} & \left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & -1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 1 & 1 & 2 & 4 & 5 \end{array}\left.\begin{array}{c} 1 \\ 4 \\ 2 \\ 1 \\ 5 \end{array}\right]\right. \\ \begin{aligned}r_{2}-r_{1} & \rightarrow r_{2} \\r_{5}-r_{1} & \rightarrow r_{5} \end{aligned} & \left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\0 & 1 & -2 & -1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 1 & 3 & 4 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 2 \\ 1 \\ 4 \end{array}\right]\right. \\ r_{3}\leftrightarrow r_{4} & \left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & -2 & -1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 3 & 4 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 1 \\ 2 \\ 4 \end{array}\right]\right. \\ r_{5}-r_{3}\rightarrow r_{5} & \left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & -2 & -1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 5 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 1 \\ 2 \\ 3 \end{array}\right]\right. \\ r_{5}-r_{4}\rightarrow r_{5} & \left[\begin{array}{ccccc|} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & -2 & -1 & 1 \\ 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 4 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 1 \\ 2 \\ 1 \end{array}\right]\right. \end{alignedat}$

Now what we've been dealing with is mostly the intersection of planes and lines at a point, but we can have planes intersection along a line, as in the picture below. (We can also have the intersection being a plane aswell)

So that means that we would have our solutions in terms of a parameter, which we will use as $t$ . The variable that we set as $t$ will be the one that is 'free', that is the variable whose column doesn't have a pivot in it.
e.g. If we are using the variables $x$ , $y$ and $z$ , and we have
$\left[\begin{array}{ccc|} 2 & 1 & -1 \\ 0 & 1 & 3 \end{array}\left.\begin{array}{c} 1 \\ 2 \end{array}\right]\right.$
Then we have pivots int he first and second columns, so $z$ is free, so we set $z=t$ .
$\left[\begin{array}{ccc|} 4 & 2 & -1 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{array}\left.\begin{array}{c} 2 \\ -1 \\ -1 \end{array}\right]\right.$
Then we have pivots in the first and third columns, so that means the variable $y$ is free, so we set $y=t$ .

So lets do an example.
e.g. We have the system
$\begin{alignedat}{1}x-y+4z & =1 \\ 2x-y-2z & =-1 \end{alignedat}$
$\begin{alignedat}{1} & \left[\begin{array}{ccc|} 1 & -1 & 4 \\ 2 & -1 & -2 \end{array}\left.\begin{array}{c} 1 \\ -1 \end{array}\right]\right. \\ \begin{aligned}r_{2}-2r_{1} & \rightarrow r_{2}\end{aligned} & \left[\begin{array}{ccc|} 1 & -1 & 4 \\ 0 & 1 & -10 \end{array}\left.\begin{array}{c} 1 \\ -3 \end{array}\right]\right. \end{alignedat}$
$z$ is free, let $z=t$
$\begin{alignedat}{1}y-10z & =-3 \\ \therefore y & =-3+10t \\ x-y+4z & =1 \\ x & =1-4z+y \\ & =1-4t-3+10t \\ \therefore x & =-2+6t \end{alignedat}$
$\therefore x=-2+6t,\: y=-3+10t,\; z=t$
So taking a point on the line (lets say for $t=0$ ), we get $\left(-2,-3,0\right)$ and the direction vector $\left(6,10,1\right)$ , so the solution to the system is a line with equation
$\underset{\sim}{r}\left(t\right)=\left(-2,-3,0\right)+t\left(6,10,1\right)$

Lets try another example, this time we three planes to start off with.
$\begin{alignedat}{1}3x+2y+z & =1 \\ 3x+2y & =3 \\ 6x+4y+z & =4 \end{alignedat}$
$\begin{alignedat}{1} & \left[\begin{array}{ccc|} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 6 & 4 & 1 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 4 \end{array}\right]\right. \\ \begin{aligned}r_{2}-r_{1} & \rightarrow r_{2} \\ r_{3}-2r_{1} & \rightarrow r_{3} \end{aligned} & \left[\begin{array}{ccc|} 3 & 2 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & -1 \end{array}\left.\begin{array}{c} 1 \\ 2 \\ 2 \end{array}\right]\right. \\ r_{3}-r_{2}\rightarrow r_{3} & \left[\begin{array}{ccc|} 3 & 2 & 1 \\0 & 0 & -1 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} 1 \\ 2 \\ 0 \end{array}\right]\right. \end{alignedat}$
Now since we have a row of zeros, we aren't getting any new information from them, so we can get rid of it just to end up with
$\left[\begin{array}{ccc|} 3 & 2 & 1 \\ 0 & 0 & -1 \end{array}\left.\begin{array}{c} 1 \\ 2 \end{array}\right]\right.$
$y$ is free, let $y=t$
$\begin{alignedat}{1}-z & =2 \\ \therefore z & =-2 \\ 3x+2y+z & =1 \\ x & =\frac{1-2y-z}{3} \\ & =\frac{1-2t+2t}{3} \\ \therefore x & =\frac{1}{3} \end{alignedat}$
$\therefore x=\frac{1}{3},\: y=t,\: z=-2$

Sometimes we will have a system that has no solutions, such as parallel planes or lines. When we convert to row echelon form, systems with no solutions will have a row of zeros on the left hand side, and a non zero element on the right hand side (as this cannot happen, we can't have $0x+0y+0z$ giving a number). That is they contain
$\left[\begin{array}{ccc|} 0 & 0 & 0\end{array}\left.\begin{array}{c} \alpha\end{array}\right]\right.$ , where $\alpha$ is a non-zero element.
So if we had the system
$\begin{alignedat}{1}x+2y+z & =2 \\ x+3y+z & =4 \\ 2x+y+2z & =1 \end{alignedat}$
$\begin{alignedat}{1} & \left[\begin{array}{ccc|} 1 & 2 & 1 \\ 1 & 3 & 1 \\ 2 & 1 & 2 \end{array}\left.\begin{array}{c} 2 \\ 4 \\ 1 \end{array}\right]\right. \\ \begin{aligned}r_{2}-r_{1} & \rightarrow r_{2} \\ r_{3}-2r_{1} & \rightarrow r_{3} \end{aligned} & \left[\begin{array}{ccc|} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & -4 & 0 \end{array}\left.\begin{array}{c} 2 \\ 2 \\ -3 \end{array}\right]\right. \\ r_{3}+4r_{2}\rightarrow r_{3} & \left[\begin{array}{ccc|} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} 2 \\ 2 \\ 5 \end{array}\right]\right. \end{alignedat}$
Now we have that row of zeros resulting in a number on the right hand side, so this system has no soultions. We call this system inconsistent.

Lets try some problems.
1. $\begin{alignedat}{1}x+3y+2z & =-2 \\ x+3y+4z & =-1 \\ -x-3y & =3 \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 1 & 3 & 2 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} -2 \\ 1 \\ 0 \end{array}\right]\right.$ , Solution: $x=-3-3t,\: y=t,\: z=\frac{1}{2}$

2. $\begin{alignedat}{1}2x+2y-z & =1 \\ 2x+2y & =4 \\ -x-y+z & =1 \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 2 & 2 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} 1 \\ 3 \\ 0 \end{array}\right]\right.$ , Solution: $x=2-t,\: y=t,\: z=3$

3. $\begin{alignedat}{1}x+y+z & =1 \\ x+3y & =2 \\ -x+y-2z & =0 \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 1 & 1 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} 1 \\ 1 \\ 0 \end{array}\right]\right.$ , Solution: $x=\frac{1}{2}-\frac{3}{2}t,\: y=\frac{1}{2}+\frac{1}{2}t,\: z=t$

4. $\begin{alignedat}{1}2x+y-3z & =1 \\ -2x+4z & =1 \\ 2x+2y-2z & =5 \end{alignedat}$

Spoiler

$\left[\begin{array}{ccc|} 2 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\left.\begin{array}{c} 1 \\ 2 \\ 2 \end{array}\right]\right.$ , No Solutions, the system is inconsistent.

Special At Specialist · « **Reply #4 on:** November 17, 2012, 12:50:00 am »

Thank you for teaching me Gaussian Elimination! I learnt integration by parts from Khan Academy a while ago, but I've never learnt Gaussian Elimination until just now.

Quote from: b^3 on November 17, 2012, 12:31:55 am

What we end up with is the row echelon form of the system.
Note: That we are using row 1 to change column 1 and row 2 to change column 2, in that order, so that we don't lose the 0's that we form.
Now we come to the Back Substituion stage. We first solve for the varible in the last row, then use that result to solve for the second last variable and so on.
[tex]\begin{alignedat}{1}\frac{9}{2}z & =\frac{15}{2}
\\ \therefore z & =\frac{5}{3}
\\ y+\frac{3}{2}z & =\frac{3}{2}
\\ y & =\frac{3}{2}-\frac{3}{2}\left(\frac{5}{3}\right)
\\ & =\frac{3}{2}-\frac{5}{2}
\\ \therefore y & =-1
\\ 4z+2y-z & =1

Isn't it meant to be:
4x + 2y - z = 1

b^3 · « **Reply #5 on:** November 17, 2012, 12:59:52 am »

Fixed Now.

Jenny_2108 · « **Reply #6 on:** November 17, 2012, 02:03:37 am »

^ Didn't see this thread until now

Hey b^3, if you get bored during hols, can you post 1 new topic everyday please?

and if we have any ques about it, we will post here to ask
Finally, there is something to do

pi · « **Reply #7 on:** November 17, 2012, 01:54:29 pm »

Quote from: b^3 on November 16, 2012, 01:05:33 pm

We may also sometimes use euler's identity $e^{ix}=\cos(x)+i\sin(x)$ ...

I found this doc useful last year when I toyed with this technique, it slightly extends on b^3's post too

Hancock · « **Reply #8 on:** November 17, 2012, 01:58:17 pm »

Oh my god, so sick of row reduction of matrices after Linear Algebra this semester.

Biceps · « **Reply #9 on:** November 17, 2012, 03:05:33 pm »

http://www.youtube.com/watch?v=CpM1jJ0lob8
Good video to understand complexifying the integral

BubbleWrapMan · « **Reply #10 on:** November 17, 2012, 03:46:59 pm »

Thought I'd share a bit more on matrix methods and expand on the whole row-echelon thing.

What b^3 has been doing is getting the matrices in row-echelon form, i.e. manipulating the augmented matrices so that each row-leader only has zeros below them. However, the definition of row-echelon form varies from place to place, and some definitions have more requirements. The most common extra requirement is that each row-leader show be 1, and another is that the row-leaders should also have zeros above them, as well as below (this is often called reduced row-echelon form). The main advantage of going this much further is to make it extremely easy to read off the solution set. I'll give a brief example of this:

Say we have the system:
$\left[\begin{array}{ccc}1&-1&7\\1&3&-1\\3&2&11\end{array}\right|\left.\begin{array}{ccc}-6\\6\\-3\end{array}\right]$
The first row begins with 1, which is what we want, so we will leave it like that. To eliminate the other non-zero entries in the first column we will add/subtract multiples of row 1 to the other two. It's important that when you're adding or subtracting a row that you leave that row unchanged in the process, since otherwise you may destroy information.

So now we have:
$\left[\begin{array}{ccc}1&-1&7\\0&4&-8\\0&5&-10\end{array}\right|\left.\begin{array}{ccc}-6\\12\\15\end{array}\right] \begin{array}{ccc}\\R_{2}'=R_{2}-R_{1}\\R_{3}'=R_{3}-3R_{1}\end{array}$

The second row looks nicely divisible by 4 so:

$\left[\begin{array}{ccc}1&-1&7\\0&1&-2\\0&5&-10\end{array}\right|\left.\begin{array}{ccc}-6\\3\\15\end{array}\right] \begin{array}{ccc}\\R_{2}'=\frac{1}{4}R_{2}\\\end{array}$

Now we want to get rid of the non-zero entries in the second column, so we will add/subtract row 2 from the others:

$\left[\begin{array}{ccc}1&0&5\\0&1&-2\\0&0&0\end{array}\right|\left.\begin{array}{ccc}-3\\3\\0\end{array}\right] \begin{array}{ccc}R_{1}'=R_{1}+R_{2}\\\\R_{3}'=R_{3}-5R_{2}\end{array}$

This is now in [reduced] row-echelon form. We ended up with two equations which means there is an infinite solution set (it would have been empty if the bottom-right entry was non-zero, as we would have had the equation 0x + 0y + 0z = c).

We have the two equations:

$x+5z=-3$
and $y-2z=3$

Which can be rearranged to give:

$x=-3-5z$
and $y=3+2z$

So now, this solution set can be described by a parameter, so I'll use t as the parameter. Letting z = t, the solution set is:

$\left \{ (-3-5t,3+2t,t):t\in\mathbb{R} \right \}$ , which describes a line in $\mathbb{R}^{3}$ .

A single linear equation in 3 variables always represents a plane in $\mathbb{R}^{3}$ , so the intersection of two of these linear equations will represent a line, unless the planes are parallel. Planes will either intersect completely (if they are the same plane), or never intersect (they are parallel and do not share a point), or they intersect along a line. This is analogous to lines in $\mathbb{R}^{2}$ having a unique intersection except when they are parallel.

Is there another way to describe/visualise this line we found? Yeah, most definitely. Consider the position vector of any point in the solution set:

$(-3-5t)\underset{\sim}i+(3+2t)\underset{\sim}j+t\underset{\sim}k$

$=-3\underset{\sim}i+3\underset{\sim}j-5t\underset{\sim}i+2t\underset{\sim}j+t\underset{\sim}k$

$=-3\underset{\sim}i+3\underset{\sim}j+t(-5\underset{\sim}i+2\underset{\sim}j+\underset{\sim}k)$

So from this we can see it is the line through the point $(-3,3,0)$ and in the direction of $-5\underset{\sim}i+2\underset{\sim}j+\underset{\sim}k$ .

b^3 · « **Reply #11 on:** November 17, 2012, 04:01:57 pm »

EDIT: Post number 2222

I don't know if I will do anymore after this, depends, but anyway. Also thanks for the contributions to the thread guys (Also it kind of leads into what I had coming next

)

Matrix Inverses (by hand)

We can use the methods forward eliminiation and backwards elimination methods used in gaussian elimination to find the inverse of matricies by hand, although it is quite tedious.

We start off with an augmented matrix $\bigl[\begin{array}{c|c} A & I\end{array}\bigr]$ and then use row echelon operations to try and turn the left hand side into the inverse matrix $I$ , which will leave the right hand side as our inverse, i.e. $\bigl[\begin{array}{c|c} I & A^{-1}\end{array}\bigr]$ .
If we get a row entirely of zeros or a column entirely of zeros then the matrix has no inverse and we don't need to go any further, as we can't turn the left hand side into the identity matrix.
e.g. Find the inverse of A where
$A=\begin{bmatrix}1 & 2 & 0 \\ -1 & 3 & 2 \\ -1 & 2 & 0 \end{bmatrix}$
So we start with $\bigl[\begin{array}{c|c} A & I\end{array}\bigr]$ , we work our ways down using the same forward elimination as before, so we get a triangle of zeros in the bottom left hand corner, then we work back up using the bottom rows instead of the top rows, to get a triangle of zeros in the top right hand corner.
$\begin{alignedat}{1} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ -1 & 3 & 2 & 0 & 1 & 0 \\ -1 & 2 & 0 & 0 & 0 & 1 \end{array}\right] \\ \begin{alignedat}{1}r_{2}+r_{1} & \rightarrow r_{2} \\ r_{3}+r_{1} & \rightarrow r_{2} \end{alignedat} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 5 & 2 & 1 & 1 & 0 \\ 0 & 4 & 0 & 1 & 0 & 1 \end{array}\right] \\ \frac{1}{5}r_{2}\rightarrow r_{2} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{2}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & 0 \\ 0 & 4 & 0 & 1 & 0 & 1 \end{array}\right] \\ r_{3}-4r_{2}\rightarrow r_{3} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{2}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & 0 \\ 0 & 0 & \vphantom{\int_{a}^{b}}-\frac{8}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & \vphantom{\int_{a}^{b}}-\frac{4}{5} & 1 \end{array}\right] \\ \left(-\frac{5}{8}\right)r_{3}\rightarrow r_{3} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & \vphantom{\int_{a}^{b}}\frac{2}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & \vphantom{\int_{a}^{b}}\frac{1}{5} & 0 \\ 0 & 0 & 1 & \vphantom{\int_{a}^{b}}-\frac{1}{8} & \vphantom{\int_{a}^{b}}\frac{1}{2} & -\frac{5}{8} \end{array}\right] \end{alignedat}$
Now that is the forward elimination process finished, now for the backwards elimination working. So we start with the bottom rows and work our way back up.
$\begin{alignedat}{1}r_{2}-\frac{2}{5}r_{3}\rightarrow r_{2} & \left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \vphantom{\int_{a}^{b}}\frac{1}{4} & 0 & \frac{1}{4} \\ 0 & 0 & 1 & \vphantom{\int_{a}^{b}}-\frac{1}{8} & \vphantom{\int_{a}^{b}}\frac{1}{2} & -\frac{5}{8} \end{array}\right] \\ r_{1}-2r_{2}\rightarrow r_{1} & \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \vphantom{\int_{a}^{b}}\frac{1}{2} & 0 & \vphantom{\int_{a}^{b}}-\frac{1}{2} \\ 0 & 1 & 0 & \vphantom{\int_{a}^{b}}\frac{1}{4} & 0 & \vphantom{\int_{a}^{b}}\frac{1}{4} \\ 0 & 0 & 1 & \vphantom{\int_{a}^{b}}-\frac{1}{8} & \vphantom{\int_{a}^{b}}\frac{1}{2} & \vphantom{\int_{a}^{b}}-\frac{5}{8} \end{array}\right] \end{alignedat}$
Now we have the identity matrix formed on the LHS, so the right hand side is our inverse of the original, that is we now have $\bigl[\begin{array}{c|c} I & A^{-1}\end{array}\bigr]$
$\begin{alignedat}{1}A^{-1} & =\begin{bmatrix}\vphantom{\int_{a}^{b}}\frac{1}{2} & 0 & \vphantom{\int_{a}^{b}}-\frac{1}{2} \\ \vphantom{\int_{a}^{b}}\frac{1}{4} & 0 & \vphantom{\int_{a}^{b}}\frac{1}{4} \\ \vphantom{\int_{a}^{b}}-\frac{1}{8} & \vphantom{\int_{a}^{b}}\frac{1}{2} & \vphantom{\int_{a}^{b}}-\frac{5}{8} \end{bmatrix} \\ \therefore A^{-1} & =\frac{1}{8}\begin{bmatrix}4 & 0 & -4 \\ 2 & 0 & 2 \\ -1 & 4 & -5 \end{bmatrix} \end{alignedat}$

Now for some questions, they may be a fair bit tedious.....
1. $A=\begin{bmatrix}1 & -1 & 2 \\ 3 & 3 & 1 \\ 0 & 1 & 0 \end{bmatrix}$

Spoiler

Solution: $A^{-1}=\frac{1}{5}\begin{bmatrix}-1 & 2 & -7 \\ 0 & 0 & 5 \\ 3 & -1 & 6 \end{bmatrix}$

2. $B=\begin{bmatrix}3 & -3 & 1 \\ 1 & 2 & -1 \\ 2 & 1 & -2 \end{bmatrix}$

Spoiler

Solution: $B^{-1}=\frac{1}{12}\begin{bmatrix}3 & 5 & -1 \\ 0 & 8 & -4 \\ 3 & 9 & -9 \end{bmatrix}$

3. $C=\begin{bmatrix}2 & 0 & -5 \\ -1 & 0 & 2 \\ 1 & -4 & -2 \end{bmatrix}$

Spoiler

Solution: $C^{-1}=\frac{1}{4}\begin{bmatrix}-8 & -20 & 0 \\ 0 & -1 & -1 \\ -4 & -8 & 0 \end{bmatrix}$

4. $D=\begin{bmatrix}1 & -1 & 2 & 0 & 1 \\ 2 & 0 & 0 & 1 & -1 \\ 2 & 2 & 1 & -2 & -1 \\ 1 & 0 & -2 & -1 & 0 \\ 0 & 0 & 1 & 2 & -1 \end{bmatrix}$

Spoiler

Solution: $D^{-1}=\frac{1}{2}\begin{bmatrix}-2 & 6 & -1 & -6 & -7 \\ -10 & 22 & -4 & -26 & -28 \\ 2 & -4 & 1 & 4 & 5 \\ -6 & 14 & -3 & -16 & -17 \\ -10 & 24 & -5 & -28 & -31 \end{bmatrix}$

pi · « **Reply #12 on:** November 17, 2012, 04:45:19 pm »

As an exercise and after readng b^3's and polar's posts:
Simplify $i^i$

Spoiler

$e^{ix} = \cos(x) + i \sin(x)$
$When \ x= \frac{\pi}{2}$
$e^{\frac{i\pi}{2}} = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = i$
$Hence \ i^i = e^{({\frac{i\pi}{2}})^i} = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$

However, there are actually infinite solutions, had we let $x= \frac{\pi}{2} + 2k\pi$ where $k \in Z$ , we would arrive at the more generalised solution of $i^i = e^{-(\frac{\pi}{2} + 2k\pi)}$ where $k \in Z$

When I was in yr11, my methods teacher made a passing comment on Gamma Functions (what fascinated me was how we can get the factorials of non-integers)

Here is a nice doc explaining how the basics of them work

-> "gamma functions.pdf"

edit: added a more advance doc, still very understandable (well, mostly

) -> "GammaAndStirling.pdf"

charmanderp · « **Reply #13 on:** November 17, 2012, 04:58:40 pm »

Only on ATARNotes

You guys never cease to amaze me.

BubbleWrapMan · « **Reply #14 on:** November 17, 2012, 06:30:19 pm »

Finding (n x n) Matrix Determinants

You're all familiar with the concept of a matrix determinant, particularly those of (2 x 2) matrices, and their role in invertibility - namely, the fact that a square matrix has an inverse if, and only if, its determinant is non-zero.

What is a determinant, though? There are a few ways to think about it, though it's not really a simple thing to describe - I couldn't really tell you definitively what a determinant is. But these days I associate it with a sort of 'scale factor' when applying a matrix transformation to n-dimensional space. What do I mean by that? Well, consider some now-familiar transformations:

Say we applied the following transformation to $\mathbb{R}^{2}$ :

$T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}a&0\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$ , where $a>0$

This is a dilation by a factor of $a$ parallel to the x-axis. The determinant of this coefficient matrix is $a$ , which make sense since the entire plane is scaled by a factor of a.

Now consider this transformation:

$T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

This is a reflection of the plane in the line $y=x$ , or an inversion of a function, if you will. The determinant of this coefficient matrix is -1. This tells you that the plane hasn't been scaled (the magnitude of the determinant is 1), but the negative indicates the plane's orientation is reversed.

The third and final example is:

$T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T\left (\begin{bmatrix}x\\y\end{bmatrix}\right )=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

This is unfamiliar, due to the fact that this time, the determinant of the coefficient matrix is zero. What does that mean, though? What happens to the plane?

Well, we end up with $\begin{bmatrix}x\\0\end{bmatrix}$ after multiplication. That tells you that every x-coordinate stays where it is, but every y-coordinate becomes 0. So the ENTIRE plane is mapped onto the line y = 0. Needless to say, when you squash the whole plane onto a line, you essentially scale by a factor of 0, and this transformation cannot be undone. It's like multiplying something by zero -- you can't find the original number afterwards. Thusly, we cannot reverse the application of this coefficient matrix.

So, we have some idea of what determinants are. Now let's see how we can find them.

I'll start with a general (3 x 3) matrix:

$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$

What we do is pick one row or one column (I'll use the first row), and expand it like so (this won't make sense immediately so bear with me):

$\det\left(\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\right )=\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=a\begin{vmatrix}e&f\\h&i\end{vmatrix}-b\begin{vmatrix}d&f\\g&i\end{vmatrix}+c\begin{vmatrix}d&e\\g&h\end{vmatrix}$

(note: the use of straight lines on the matrices is a shorthand for the determinant -- you may not have seen it before, I hadn't seen it before uni...)

So we're left with a few (2 x 2) determinants, which we know how to find. But what exactly did I do there? Well, the $a, b$ , and $c$ came from the first row. The $\begin{vmatrix}e&f\\h&i\end{vmatrix}$ matrix determinant came from temporarily 'deleting' the row and column that $a$ came from:

$\begin{vmatrix}(a)&(b)&(c)\\(d)&e&f\\(g)&h&i\end{vmatrix}\rightarrow \begin{vmatrix}e&f\\h&i\end{vmatrix} $

And similarly, the matrix determinant that $b$ is multiplied by came from 'deleting' the row and column that $b$ is in:

$\begin{vmatrix}(a)&(b)&(c)\\d&(e)&f\\g&(h)&i\end{vmatrix}\rightarrow \begin{vmatrix}d&f\\g&i\end{vmatrix}$

But there's one more thing: the negative sign in front of the $b$ . Whether the coefficient of each element is 1 or -1 can be determined from the following representation:

$\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}$

So basically, starting at the first element (first row, first column), it's positive, then it alternates between positive and negative. This can be extended to higher dimensions, e.g. for a (6 x 6):

$\begin{bmatrix}+&-&+&-&+&-\\-&+&-&+&-&+\\+&-&+&-&+&-\\-&+&-&+&-&+\\+&-&+&-&+&-\\-&+&-&+&-&+\end{bmatrix}$

So, for example, if we expanded the second row instead, we would have:

$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=-d\begin{vmatrix}b&c\\h&i\end{vmatrix}+e\begin{vmatrix}a&c\\g&i\end{vmatrix}-f\begin{vmatrix}a&b\\g&h\end{vmatrix}$

If you expand this out you'll find it's exactly the same as the other expression.

The best way to solidify understanding of this method is to do an example, so I'll use the matrix from my last post:

$\begin{bmatrix}1&-1&7\\1&3&-1\\3&2&11\end{bmatrix}$

We found that there was no unique solution to the system, so that should give a clue as to what the determinant will be...

If we expand along the last column (I'll emphasise again that you can use any row or column) we get:

$\begin{vmatrix}1&-1&7\\1&3&-1\\3&2&11\end{vmatrix}$

$=7\begin{vmatrix}1&3\\3&2\end{vmatrix}-(-1)\begin{vmatrix}1&-1\\3&2\end{vmatrix}+11\begin{vmatrix}1&-1\\1&3\end{vmatrix}$

$=7(2-9)+(2+3)+11(3+1)$

$=-49+5+44$

$=0$ , as expected.

This is usually a relatively quick procedure for (3 x 3)s, so it's a good idea to find the determinant of a matrix before you set out to find its inverse, in case it doesn't have one.

Now, I'll steal the (5 x 5) from b^3's last post to make an important point:

$\begin{bmatrix}1&-1&2&0&1\\2&0&0&1&-1\\2&2&1&-2&-1\\1&0&-2&-1&0\\0&0&1&2&-1\end{bmatrix}$

Note that, since we can use any row or column we like, we generally try to use the one with the most zeros in it (if there are any zeros at all), since the calculations are made a lot faster. Here, we can see the second column has three zero entries, so we will use this column:

$\begin{vmatrix}1&-1&2&0&1\\2&0&0&1&-1\\2&2&1&-2&-1\\1&0&-2&-1&0\\0&0&1&2&-1\end{vmatrix}=-(-1)\begin{vmatrix}2&0&1&-1\\2&1&-2&-1\\1&-2&-1&0\\0&1&2&-1\end{vmatrix}+0-2\begin{vmatrix}1&2&0&1\\2&0&1&-1\\1&-2&-1&0\\0&1&2&-1\end{vmatrix}+0-0$

So we're left with only two (4 x 4) determinants to find, rather than the maximum of five (4 x 4) determinants, if we had used, say, the third row. But of course, you won't always be that lucky, and you'll have to find five of them. Anyway, expanding out the (4 x 4) determinants:

$\begin{vmatrix}1&-1&2&0&1\\2&0&0&1&-1\\2&2&1&-2&-1\\1&0&-2&-1&0\\0&0&1&2&-1\end{vmatrix}=\begin{vmatrix}2&0&1&-1\\2&1&-2&-1\\1&-2&-1&0\\0&1&2&-1\end{vmatrix}-2\begin{vmatrix}1&2&0&1\\2&0&1&-1\\1&-2&-1&0\\0&1&2&-1\end{vmatrix}$

$=\left(2\begin{vmatrix}1&-2&-1\\-2&-1&0\\1&2&-1\end{vmatrix}-0+\begin{vmatrix}2&1&-1\\1&-2&0\\0&1&-1\end{vmatrix}-(-1)\begin{vmatrix}2&1&-2\\1&-2&-1\\0&1&2\end{vmatrix}\right)-2\left(\begin{vmatrix}0&1&-1\\-2&-1&0\\1&2&-1\end{vmatrix}-2\begin{vmatrix}2&1&-1\\1&-1&0\\0&2&-1\end{vmatrix}+0-\begin{vmatrix}2&0&1\\1&-2&-1\\0&1&2\end{vmatrix}\right)$

$=2\left((-1)\begin{vmatrix}-2&-1\\1&2\end{vmatrix}-0+(-1)\begin{vmatrix}1&-2\\-2&-1\end{vmatrix}\right)+\left(0-\begin{vmatrix}2&-1\\1&0\end{vmatrix}+(-1)\begin{vmatrix}2&1\\1&-2\end{vmatrix}\right)+\left(0-\begin{vmatrix}2&-2\\1&-1\end{vmatrix}+2\begin{vmatrix}2&1\\1&-2\end{vmatrix}\right)-2\left(0-\begin{vmatrix}-2&0\\1&-1\end{vmatrix}+(-1)\begin{vmatrix}-2&-1\\1&2\end{vmatrix}\right)+4\left(0-2\begin{vmatrix}2&-1\\1&0\end{vmatrix}+(-1)\begin{vmatrix}2&1\\1&-1\end{vmatrix}\right)+2\left(2\begin{vmatrix}-2&-1\\1&2\end{vmatrix}-0+\begin{vmatrix}1&-2\\0&1\end{vmatrix}\right)$

$=2(-(-4+1)-(-1-4))+(-(0+1)-(-4-1))+(-(-2+2)+2(-4-1))-2(-(2-0)-(-4+1))+4(-2(0+1)-(-2-1))+2(2(-4+1)+(1-0))$

$=16+4-10-2+4-10$

$=2$

So yeah, it's pretty tedious as you go higher than (3 x 3)... but the good thing is the method itself is fairly simple. I didn't indicate which rows I used in each case above -- try and work it out, or just pick them on your own and see that you get the same number.

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