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### AuthorTopic: "Challenge" Math Qs - Can You Figure It Out?  (Read 4354 times) Tweet Share

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#### TheBigC

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##### "Challenge" Math Qs - Can You Figure It Out?
« on: December 20, 2018, 11:17:34 am »
+5
Hello everyone,

In my spare time, I have constructed some mathematics questions for those who want to push themselves... the following is Question Set #1. Good luck.
PM me your solutions and I will identify whether or not you are correct! Solutions will be posted in a couple of days.
« Last Edit: December 20, 2018, 11:19:38 am by TheBigC »

#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #1 on: December 20, 2018, 11:34:14 am »
+4
Should Q2 also say "leading up to and including"? The way it's written would make the answer just be $a+b$.

(Similarly for Q3.)

Also, is it meant to be assumed that Q1 begins counting from 1?

#### TheBigC

• Guest
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #2 on: December 20, 2018, 12:24:19 pm »
+1
Should Q2 also say "leading up to and including"? The way it's written would make the answer just be $a+b$.

(Similarly for Q3.)

Also, is it meant to be assumed that Q1 begins counting from 1?

lol. oops! haha. poor wording - you are correct! (up to and including!)

#### TheBigC

• Guest
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #3 on: December 21, 2018, 11:58:13 am »
+3
Congratulation to all winners (solvers):

RuiAce

Our entrances were short and sweet with neater solutions than my own! Feel free to post your own solutions if you desire
My (convoluted) solution is as follows in the attachment.

Please Note: Some non-standard contrived notation is used at times.

#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #4 on: December 21, 2018, 12:46:08 pm »
+7
I'll post up my solution as well for anyone who's interested. I basically just did a combinatorics bash on it. I considered how many times each digit appeared to add them up.

Q2: Suppose the number was $[ab]$. (For visual purposes, I will set $[ab] = 57$.)
Units digits:
- For each $k \in \{0, \dots, a\}$, the number $1, \dots, 9$ all appeared exactly once, contributing $\sum_{k=0}^9 k$, $a$ times. This adds $a \times \frac{9\times 10}{2}$ to the sum. These $k$'s represent all the tens values prior to $[a0]$.
- Then, for that current tens value, we only add the units from $0$ up to $b$, which basically contributes $\sum_{k=0}^b k = \frac{b(b+1)}{2}$ to the sum.

So in the number 57,
For each $k \in \{0, \dots, 5\}$ we add $\frac{9\times 10}{2}$ to the sum. One lot comes from 0 to 9, the next lot from 10 to 19, and so on up to 40 to 49.
Then we just add 0+1+...+7

Tens digits:
- For each of the previous 10s values, i.e. each $\ell \in \{0, \dots, a-1\}$, the number $\ell$ gets contributed exactly $10$ times to the sum. So this gives $10\times \sum_{\ell = 0}^{a-1} k = 10\times \frac{(a-1)a}{2}$.
- Then, for the current tens value (i.e. $a$), we just add $a$ to the count the number of times as specified by $b$. This contributes $\sum_{k=0}^b a = (b+1)a$ to the sum.

So in the number 57,
For the units, we just add 0 ten times, For $[1b]$, we add 1 ten times. For $[2b]$, we add 2 ten times and so on until we reach 4.
Then, because we have 50, 51, ..., 57, we need to add 5 eight times.
$\text{Answer to Q2: }\boxed{45a + \frac{b(b+1)}{2} + 5(a-1)a + (b+1)a}$
Slight subtlety: If $a=0$, i.e. we tried applying this formula on a one digit number, the formula will work but the part where it says $\ell \in \{ 0, \dots, a-1\}$ will appear nonsensical. Essentially for this case, just interpret that set on the right to be empty, i.e. $\ell \in \emptyset$. (This is a bit of an abuse of notation but still commonly used in combinatorics to represent an empty sum.)
___________________________________________________________________________

Q3: The idea now follows similarly to the previous one so I won't do a numeric example anymore, but it's important to keep track with all the progress.

Units:
- Starts the exact same way, but now we need to go up to the two-digit number $[ab]$ now. So this adds $\boxed{[ab] \sum_{k=1}^9 k}$ to the sum.
- Finishes the exact same way in that now we just sum the integers up to $z$ instead, hence contributing $\boxed{\sum_{k=0}^z k }$ to the sum.

Hundreds:
- Starts the exact same way for the tens in the previous one. We basically do it for each of the previous hundreds values $\ell \in \{ 0, \dots, a-1\}$, and the idea is that each of these numbers will appear 100 times instead, giving $\boxed{\sum_{k=0}^{a-1}k}$
- Finishes pretty much the exact same way as well now, but because we're working modulo 100 we want to terminate at $[bz]$ instead. So this contributes $\boxed{\sum_{k=0}^{[bz]} a}$

Tens: This is the new one - it sorta combines and mixes the methods used in the other cases.
- For each of the previous hundreds values, i.e. $\ell \in \{0, \dots, a-1\}$, each of the values 0 to 9 will have appeared ten times. (Ten 0's from 0-9, ten 1's from 10-19 and so on, up till ten 9's from 90-99.) This therefore contributes $10 \sum_{\ell=0}^{a-1} \sum_{k=0}^9 k$, but since the outer sum really represents summing over constants, I'll at least write $\boxed{10 (a-1) \sum_{k=0}^9 k}$.
- Then we can narrow our focus to the current hundreds value $a$. Once we're here, we repeat what we did for the two digit case. For each tens digit $k \in \{0, \dots, b-1\}$ the same $k$ appears 10 times, so we have $\boxed{10 \sum_{k=0}^{b-1} k}$
- Lastly, keep adding the current tens value $b$, as required by the current value of $z$: $\boxed{\sum_{k=0}^z b}$.

Again, add and simplify using $\sum_{k=0}^n k = \frac{n(n+1)}{2}$ to get the answer. Note that this formula can easily be derived as it is the partial sum of an arithmetic progression.

#### TheBigC

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #5 on: December 21, 2018, 09:22:39 pm »
+4
Great stuff Rui! A much nicer solution than my own

A note to anyone: I recently received a message that requested to post a fun question up. If anyone feels like posting a question - please feel free to do so! This is all about promoting the fun and creativity that mathematics has to offer! I might impose limits on daily novel question posts, however only if things start getting hectic (in an effort to substantiate a notion of time to complete and submit answers to problem sets!).

#### AlphaZero

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #6 on: December 21, 2018, 10:20:33 pm »
+4
Hey everyone. (It was me that asked to post a problem lol).

Perhaps if we can get some regulars here, we could rotate posting problems? (Just a thought).

Anyway, here is my problem

« Last Edit: December 21, 2018, 10:28:07 pm by dantraicos »
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#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #7 on: December 21, 2018, 10:30:49 pm »
0
Is this thread intended to go beyond VCE maths and into university level stuff?

#### TheBigC

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #8 on: December 21, 2018, 10:35:00 pm »
+1
Is this thread intended to go beyond VCE maths and into university level stuff?

Preferably, it'd be great if we could make it solvable at a high school level (even though it may require some level of advanced understanding), however university level is fine as well - though it just limits the demographic of those who can solve the problem.

It would be awesome if everyone who posts could delineate their target audiences from now on

#### AlphaZero

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #9 on: December 21, 2018, 10:46:23 pm »
+1
Is this thread intended to go beyond VCE maths and into university level stuff?

Preferably, it'd be great if we could make it solvable at a high school level (even though it may require some level of advanced understanding), however university level is fine as well - though it just limits the demographic of those who can solve the problem.

It would be awesome if everyone who posts could delineate their target audiences from now on

I think it would be best to keep it within VCE, but as @TheBigC, it is limiting.

I mean, we technically already exceeded VCE level with the first question set when we introduced summation notation (I actually can't believe it isn't taught/required in VCE).

So that people know, the question I posted requires no university level knowledge other than what's given in the "Some useful information" paragraph about $\mathbb{F}_2$. In fact, other than that, we only require knowledge of Year 11 Specialist Maths. The question is structured in a way so that you only have to know: how to multiply matrices, what an inverse is, and some basic knowledge of logic and proof.

In saying that though, the question does require a very high understanding of those topics. (It also still gives me nightmares lol )
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#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #10 on: December 22, 2018, 12:31:51 pm »
+3
Hey everyone. (It was me that asked to post a problem lol).

Perhaps if we can get some regulars here, we could rotate posting problems? (Just a thought).

Anyway, here is my problem

(Image removed from quote.)
Not saying this is the best approach and I needed some external help for one tiny bit in the middle but here goes. Parts a)-c):\text{Let: }\\ \begin{align*}\mathbb{Z}_2^{n\times n} \ni M &= \begin{pmatrix} m_{11} & m_{12} & \dots & m_{1n}\\ m_{21} & m_{22} & \dots & m_{2n}\\ \vdots & \vdots&\ddots& \vdots\\ m_{n1} &m_{n2}& \dots & m_{nn} \end{pmatrix} \\ \mathbb{Z}_2^{n\times n}\ni K &= \begin{pmatrix} 1 &1& \dots & 1\\1&1&\dots & 1\\ \vdots & \vdots&\ddots& \vdots\\ 1 & 1&\dots & 1 \end{pmatrix} \end{align*}
$\text{Then }\boxed{MK = \begin{pmatrix} \sum_{i=1}^n m_{1i} & \sum_{i=1}^n m_{1i}& \dots & \sum_{i=1}^n m_{1i}\\ \sum_{i=1}^n m_{2i} & \sum_{i=1}^n m_{2i}& \dots & \sum_{i=1}^n m_{2i}\\ \vdots &\vdots& \ddots& \vdots\\ \sum_{i=1}^n m_{ni} &\sum_{i=1}^n m_{ni} & \dots & \sum_{i=1}^n m_{ni} \end{pmatrix}}$
_________________________________________________________________________________________________
$\text{Recall that over }\mathbb{Z}_2 \text{ subtraction and addition are equivalent.}$
$\text{Firstly given }M + M^{-1} = K\text{ we have }\boxed{M^2 = MK + I}\\ \text{so only displaying row 1 and column 1:}\\ \boxed{M^2 = \begin{pmatrix} 1 + \sum_{i=1}^n m_{1i} & \sum_{i=1}^n m_{1i} & \dots & \sum_{i=1}^n m_{1i} \\ \sum_{i=1}^n m_{2i} \\ \vdots & & \ddots \\ \sum_{i=1}^n m_{ni} \end{pmatrix} }$
$\text{Given }M + M^{-1} = K\text{ we consequently have }\boxed{M^{-1} = K + M}\\ \text{so }M^{-1} = \begin{pmatrix}1+m_{11} & 1+m_{12} & \dots & 1+m_{1n}\\ 1+m_{21} & 1+m_{22} & \dots & 1+m_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ 1+m_{n1} & 1+m_{n2} & \dots & 1+m_{nn} \end{pmatrix}$
$\text{Then }M^{-1} K = \begin{pmatrix} n + \sum_{i=1}^n m_{1i} & n + \sum_{i=1}^n m_{1i} & \dots & n + \sum_{i=1}^n m_{1i}\\ n + \sum_{i=1}^n m_{2i} & n + \sum_{i=1}^n m_{2i} & \dots & n + \sum_{i=1}^n m_{2i}\\ \vdots & \vdots & \ddots & \vdots\\ n + \sum_{i=1}^n m_{ni} & n + \sum_{i=1}^n m_{ni} &\dots& n + \sum_{i=1}^n m_{ni}\end{pmatrix}\\ \text{and we note that for a fixed }n \in \mathbb{Z},\, n\text{ should be replaced by }n\mod 2.$
$\text{We find this, because given }M+M^{-1}K\text{ we also have }\boxed{M^{-2} = I + M^{-1}K}.\\ \text{So only displaying row 1 and column 1:}\\ M^{-2}=\begin{pmatrix} 1+n + \sum_{i=1}^n m_{1i} & n+\sum_{i=1}^n m_{1i} & \dots &n+ \sum_{i=1}^n m_{1i} \\ n+\sum_{i=1}^n m_{2i} \\ \vdots & & \ddots \\ n+\sum_{i=1}^n m_{ni} \end{pmatrix}$
_________________________________________________________________________________________________
\text{Suppose now that }n\text{ is odd, so that}\\ \begin{align*}1+n &\equiv 0 \mod 2\\ n &\equiv 1 \mod 2 \end{align*}\\ \text{Then, }M^{-2}\text{ simplifies to}\\ \boxed{M^{-2} = \begin{pmatrix} \sum_{i=1}^n m_{1i} & 1+\sum_{i=1}^n m_{1i} & \dots &1+ \sum_{i=1}^n m_{1i} \\ 1+\sum_{i=1}^n m_{2i} \\ \vdots & & \ddots \\ 1+\sum_{i=1}^n m_{ni} \end{pmatrix}}
$\text{We now work with the facts }M^2M^{-2} = I\text{ and }M^{-2}M^2 = I.$
$\text{The entry in row 1, column 1 of }M^2M^{-2}\text{, obtained through matrix multiplication is}\\ \left(1 + \sum_{i=1}^n m_{1i}\right) \sum_{i=1}^n m_{1i} + \sum_{i=1}^n m_{1i} \left( n + \sum_{j=2}^n \sum_{i=1}^n m_{ji} \right).$
$\text{To simplify this, note that the product of two consecutive numbers in }\mathbb{Z}_2\text{ is zero, and thus the first term vanishes.}\\ \text{Also, as }n\text{ is odd, }n \equiv 1\mod 2\text{, so it simplifies to}\\ \sum_{i=1}^n m_{1i} \left(1+ \sum_{j=2}^n \sum_{i=1}^n m_{ji} \right).$
$\text{But }M^2 M^{-2} = I\text{, and the entry in row 1, column 1 of }I\text{ is just }1.\\ \text{So equating components, we have}\\ \boxed{1 = \sum_{i=1}^n m_{1i} \left(1+ \sum_{j=2}^n \sum_{i=1}^n m_{ji} \right)}$
$\text{But in }\mathbb{Z}_2, \,\, ab = 1 \text{ if and only if }a = b = 1.\\ \text{Consequently, we deduce }\boxed{\sum_{i=1}^n m_{1i} = 1}\text{ and }\boxed{\sum_{j=2}^n \sum_{i=1}^n m_{ji} = 0}$
$\text{Then, following a similar procedure, the entry in row 1, column 1 of }M^{-2}M^2\text{ is}\\ 0 + \left( 1 + \sum_{i=1}^n m_{1i} \right)\sum_{j=2}^n \sum_{i=1}^n m_{ji}\\ \text{which, subbing in the results from above, is thus equal to }0.$
$\text{But then the entry in row 1, column 1 of }I\text{ is }1\\ \text{so upon equating, we end up with }\boxed{0 = 1}\text{ - a contradiction!}$
$\text{Backtracking, the first major assumption we made was that}\\ \text{there exists an invertible such that }M + M^{-1} = K.\\ \text{Hence this assumption was false, i.e. such a matrix cannot exist.}$
Haven't had a proper go at part d) yet - trying to stick to 2x2 matrices didn't work out easily enough for me. Can probably take advantage of these constructions to produce an example of a 4x4 matrix though. (Otherwise there's a 2x2 that I just missed the first time round.)

Edit: Ran a hard-coded program on R to find a matrix for me. Turns out this one works but I wouldn't know how to construct it at this stage.
$M = \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\end{pmatrix}$
I did do some math before attempting hard-coding though. Basically I saw that if $n$ were even, we would have $M^2 = M^{-2}$. This would give $M^{-1} = M^3$ and also $M^4 = I$. So I wrote a program to check each 4x4 matrix over $\mathbb{Z}_2$, and told it to stop if it found a matrix satisfying $\boxed{M^4= I}$ and $\boxed{M + M^3 = K}$ simultaneously.

The code is ugly. It stabs my soul as a programmer repeatedly. It's literally just a 16-nested loop.

Code: (test.R) [Select]
require(expm)identity <- diag(4)kk <- matrix(rep(1,16), nrow=4, ncol=4)for(A in 0:1) {  for (B in 0:1) {    for (C in 0:1) {      for (D in 0:1) {        for (E in 0:1) {          for (FF in 0:1) {            for (G in 0:1) {              for (H in 0:1) {                for (I in 0:1) {                  for (J in 0:1) {                    for (K in 0:1) {                      for (L in 0:1) {                        for (M in 0:1) {                          for (N in 0:1) {                            for (O in 0:1) {                              for (P in 0:1) {                                found <- TRUE                                mm <- matrix(c(A,B,C,D,E,FF,G,H,I,J,K,L,M,N,O,P), nrow=4, ncol=4)                                mm4 <- (mm %^% 4) %% 2                                mm3 <- (mm %^% 3) %% 2                                for (i in 1:4) {                                  for (j in 1:4) {                                    if (mm4[i,j] != identity[i,j]) found <- FALSE                                    if (mm[i,j] + mm3[i,j] != 1) found <- FALSE                                  }                                }                                if (found) stop("stopped!")                              }                            }                          }                        }                      }                    }                  }                }              }            }          }        }      }    }  }}
« Last Edit: December 22, 2018, 01:24:43 pm by RuiAce »

#### AlphaZero

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #11 on: December 22, 2018, 03:10:16 pm »
+1
...
Not saying this is the best approach and I needed some external help for one tiny bit in the middle but here goes.
...
Haven't had a proper go at part d) yet - trying to stick to 2x2 matrices didn't work out easily enough for me. Can probably take advantage of these constructions to produce an example of a 4x4 matrix though. (Otherwise there's a 2x2 that I just missed the first time round.)
...
I did do some math before attempting hard-coding though. Basically I saw that if $n$ were even, we would have $M^2 = M^{-2}$. This would give $M^{-1} = M^3$ and also $M^4 = I$. So I wrote a program to check each 4x4 matrix over $\mathbb{Z}_2$, and told it to stop if it found a matrix satisfying $\boxed{M^4= I}$ and $\boxed{M + M^3 = K}$ simultaneously.
...

Super job! You did the exact approach that was intended by the question. I'm not actually sure there's a more economical/efficient solution.

I think part d is really interesting, and you're right to not find any $2\times 2$ matrices that work - I'm quite sure there are none, and proving it is pretty straight forward. (It's pretty much a brute force method since there are only 16 $2\times2$ matrices over $\mathbb{Z}_2$) that actually exist.

Start by noticing that if $a+b=1$, then exactly one of $a$ and $b$ must be $1$. That is, [$a=1$ and $b=0$] or [$a=0$ and $b=1$]. So, to produce $K$, the matrices $M$ and $M^{-1}$ must have opposite entries. This fact actually allows us to eliminate all possible $2\times 2$ matrices.

We'll start with the 0 matrix and consider replacing entries with ones. First, let's get rid of any matrices that are not invertible. \begin{align*}\text{Replacing no entries: }&M=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\text{ is not invertible.}\\
\text{Replacing one entry: }&M=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},\ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix},\ \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix},\ \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\ \text{are not invertible (always exists a zero row)}\\
\text{Replacing two entries: }&M=\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix},\ \begin{bmatrix}1 & 0 \\ 1 & 0\end{bmatrix},\ \begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix},\ \begin{bmatrix}0 & 0 \\ 1 & 1\end{bmatrix}\ \text{are not invertible (always exists a zero row or zero column)}\\
\text{Replacing four entries: }&M=\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\ \text{is not invertible (zero determinant / two rows are the same)}\end{align*}
Now, let's use the property we discussed above. Replacing three entries so that $M=\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix},\ \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix},\ \begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix},\ \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ would require $M^{-1}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix},\ \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix},\ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},\ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\text{ respectively,}$ if $M+M^{-1}=K$. But, none of the stated $M^{-1}$ are invertible since there always exists a zero row.

There are only two matrices left to investigate: $M=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\text{ and }\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ If $M$ is the former, then $M+M^{-1}=I+I=\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\neq K.$If $M$ is the latter, then we would require $M^{-1}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I$ so that $M+M^{-1}=K$, but this is nonsense since $(M^{-1})^{-1}=M=I$, forming a contradiction.

Hence, there are no $2\times 2$ matrices $M$ over $\mathbb{Z}_2$ that have the property $M+M^{-1}=K$.

Part d is interesting. The original question didn't actually have it. I just added it because I thought it was cool to investigate. So, I don't actually have a neat solution to constructing $4\times4$ matrices without technology. I literally just used MATLAB to go through all $2^{16}$ possible $4\times 4$ matrices over $\mathbb{Z}_2$.
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#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #12 on: December 24, 2018, 02:36:37 pm »
+1
\text{Let}\\ \begin{align*}A &= \begin{pmatrix} 1/3 & 1/4 & 1/4\\ 1/3 & 1/2 & 1/4\\ 1/3 & 1/4 & 1/2\end{pmatrix}\\ Q &= \begin{pmatrix} 3 & 0 & -2\\ 4 & -1 & 1\\ 4 & 1 & 1\end{pmatrix}\\ D &= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1/4 & 0\\ 0& 0 & 1/12 \end{pmatrix}\end{align*}
$\text{1. Verify that }QDQ^{-1} = A.\\ \text{2. Hence find }\lim_{k\to \infty}A^k.$
[First Year Uni]
3. Explain in one sentence why the result in Q2 is unsurprising.

#### lzxnl

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #13 on: December 24, 2018, 05:28:24 pm »
+2
I'm not going to spoil RuiAce's question so I'm going to let it be.

The solution to these questions are well within the confines of a VCE Specialist Mathematics 3/4 student or the HSC equivalent (highest level of mathematics offered in high school), even if it doesn't look simple.

Evaluate the integral
$
\int_0^{\frac{\pi}{2}} \frac{\sin^{\pi}(x)}{\sin^{\pi}(x) + \cos^{\pi}(x)}\,dx
$

If you think this one is too easy, try
$
\int_0^{\pi/2} \log(\sec(x))\,dx
$
« Last Edit: December 24, 2018, 07:00:54 pm by lzxnl »
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#### RuiAce

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##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #14 on: December 24, 2018, 05:44:39 pm »
0
If you think this one is too easy, try
$
\int_0^\pi \log(\sec(x))\,dx
$

Shouldn't the upper boundary be $\frac\pi2$ here if you're trying to allude to a classic?

(In any case it can't be $\pi$ because "$\sec$ is negative in the second quadrant", so $\ln (\sec x)$ would be non-real)