matrix

**samanthaj1001** · December 18th 2008, 12:30 PM

ok, so i know this math problem should be easy, but i really dont understand it. pls help me out!
the following are supposed to be matrices but i do not know how to make them on the computer.

Let X= (1 1
1 1)
Let Y= (1 -1
-1 1)
Calculate X^2, X^3, X^4, Y^2, Y^3, Y^4
By considering integer powers of X and Y, find expressions for Z^n, Y^n, (X+Y)^n

Let A= aX and B=bY where a and b are constants

Use different values of a and b to calculate A^2, A^3, A^4, B^2, B^3, B^4

By considering interger powers of A and B, find expressions for A^n, B^n, (A+B)^n

Now consider M= (a+b a-b
a-b a+b)

Show that M= A+B and that M^2= A^2+B^2

Hence, find the general statement that expresses M^n in terms of aX and bY

Test the validity of your general statement by using different Values of a, b, n

Discuss the scope and limitations of your general statement

Use an algebraic method to explain how you arrived at your general statement.

Thanks so mcuh for your help!

**Soroban** · December 18th 2008, 06:31 PM

Hello, samanthaj1001!

This will take a while . . . I'll answer this in several sessions . . .

1) Let $X\:=\:\begin{bmatrix}1& 1 \\ 1& 1\end{bmatrix}\quad Y \:=\:\begin{bmatrix}1 & \text{-}1\\ \text{-}1 &1 \end{bmatrix}$

(a) Calculate: $X^2,\;X^3,\;X^4,\;Y^2,\;Y^3,\;Y^4$

(b) By considering integer powers of $X$ and $Y$ , find expressions for $X^n,\;Y^n,\;(X+Y)^n$

(a)

$X^2 \;=\;\begin{bmatrix}1&1\\1&1\end{bmatrix}\,\begin{ bmatrix}1&1\\1&1\end{bmatrix} \;=\;\begin{bmatrix}2&2\\2&2\end{bmatrix}$

$X^3 \;=\;X^2\cdot X \;=\; \begin{bmatrix}2&2\\2&2 \end{bmatrix} \,\begin{bmatrix}1&1\\1&1\end{bmatrix} \;=\;\begin{bmatrix}4&4\\4&4\end{bmatrix}$

$X^4 \;=\;X^3\cdot X \;=\;\begin{bmatrix}4&4\\4&4\end{bmatrix}\,\begin{ bmatrix}1&1\\1&1\end{bmatrix} \;=\;\begin{bmatrix}8&8\\8&8\end{bmatrix}$

$Y^2 \;=\;\begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix}\,\begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix} \;=\;\begin{bmatrix}2&\text{-}2\\\text{-}2&2\end{bmatrix}$

$Y^3 \;=\;Y^2\cdot Y \;=\;\begin{bmatrix}2&\text{-}2\\\text{-}2&2\end{bmatrix}\,\begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix} \;=\;\begin{bmatrix}4&\text{-}4\\ \text{-}4&4\end{bmatrix}$

$Y^4 \;=\;Y^3\cdot Y \;=\;\begin{bmatrix}4&\text{-}4\\\text{-}4&4\end{bmatrix}\,\begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix} \;=\;\begin{bmatrix}8&\text{-}8\\\text{-}8&8\end{bmatrix}$

(b)

$X + Y \;=\;\begin{bmatrix}1&1\\1&1\end{bmatrix} + \begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix} \;=\;\begin{bmatrix}2&0\\0&2\end{bmatrix}$

$(X+Y)^2 \;=\;\begin{bmatrix}2&0\\0&2\end{bmatrix}\,\begin{ bmatrix}2&0\\0&2\end{bmatrix} \;=\;\begin{bmatrix}4&0\\0&4\end{bmatrix}$

$(X+Y)^3 \;=\;\begin{bmatrix}4&0\\0&4\end{bmatrix}\,\begin{ bmatrix}2&0\\0&2\end{bmatrix} \;=\;\begin{bmatrix}8&0\\0&8\end{bmatrix}$

. . . $X^n \;=\;\begin{bmatrix}2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{bmatrix} \qquad Y^n \;=\;\begin{bmatrix}2^{n-1} & \text{-}2^{n-1} \\ \text{-}2^{n-1} & 2^{n-1}\end{bmatrix} \qquad (X + Y)^n \;=\;\begin{bmatrix}2^n & 0\\ 0&2^n\end{bmatrix}$

**samanthaj1001** · December 19th 2008, 06:05 AM

ok.. answer it in as many sessions as u want.. what you posted earlier was very helpful! thanks!

**Soroban** · December 19th 2008, 07:25 AM

Hello again, Samantha!

$X\:=\:\begin{bmatrix}1&1\\1&1\end{bmatrix}\quad Y\:=\:\begin{bmatrix}1&\text{-}1 \\ \text{-}1 & 1\end{bmatrix}$

Let: . $A\:=\:aX,\;\;B\:=\:bY\text{ where }a\text{ and }b\text{ are constants.}$ .[1]

Now consider: . $M\:=\:\begin{bmatrix}a+b& a-b \\ a-b & a+b \end{bmatrix}$

(a) Show that: . $M\:=\: A+B$

From [1], we have: . $A \:=\:\begin{bmatrix}a&a\\a&a\end{bmatrix} \quad B \:=\:\begin{bmatrix}b & \text{-}b \\ \text{-}b & b \end{bmatrix}$

Hence: . $A + B \:=\:\begin{bmatrix}a&a\\a&a\end{bmatrix} + \begin{bmatrix}b & \text{-}b \\\text{-}b&b\end{bmatrix} \;=\;\begin{bmatrix}a+b & a-b \\ a-b & a+b\end{bmatrix} \;=\;M$

(b) Show that: . $M^2 \:=\:A^2+B^2$

$M^2 \;=\;\begin{bmatrix}a+b&a-b\\a-b&a+b\end{bmatrix}\, \begin{bmatrix}a+b&a-b\\a-b&a+b\end{bmatrix}$

. . $= \;\begin{bmatrix}(a+b)^2+(a-b)^2 & (a+b)(a-b)+(a-b)(a+b) \\ (a-b)(a+b) + (a+b)(a-c) & (a-b)^2 + (a+b)^2 \end{bmatrix}$

. . $= \;\begin{bmatrix}a^2+2ab + b^2 + a^2 - 2ab + b^2 & a^2 - b^2 + a^2 + b^2 \\ a^2 - b^2 + a^2 - b^2 & a^2 - 2ab + b^2 + a^2 + 2ab + b^2 \end{bmatrix}$

. . $= \;\begin{bmatrix}2a^2+2b^2 & 2a^2-2b^2 \\ 2a^2-2b^2 & 2a^2+2b^2\end{bmatrix}$ .[2]

$A^2 \:=\:\begin{bmatrix}a&a\\a&a\end{bmatrix}\,\begin{ bmatrix}a&a\\a&a\end{bmatrix} \;=\;\begin{bmatrix}2a^2 &2a^2\\2a^2 & 2a^2 \end{bmatrix}$ . . . $B^2 \;=\;\begin{bmatrix}b&\text{-}b\\\text{-}b&b\end{bmatrix}\,\begin{bmatrix}b&\text{-}b\\\text{-}b&b\end{bmatrix} \;=\;\begin{bmatrix}2b^2&\text{-}2b^2\\\text{-}2b^2&2b^2\end{bmatrix}$

$A^2 + B^2 \;=\;\begin{bmatrix}2a^2&2a^2\\2a^2&2a^2\end{bmatr ix} + \begin{bmatrix}2b^2&\text{-}2b^2\\\text{-}2b^2&2b^2\end{bmatrix} \;= \;\begin{bmatrix}2a^2+2b^2 & 2a^2-2b^2\\2a^2-2b^2 & 2a^2+2b^2\end{bmatrix}$ .[3]

Since [2] = [3]: . $M^2 \:=\:A^2+B^2$

(c)Hence, find the general statement that expresses $M^n$ in terms of $A$ and $B$

Using a lot of scrap paper, I found that the pattern holds . . .

. . $\begin{array}{ccc}M &=& A + B \\ M^2 &=& A^2 + B^2 \\ M^3 &=& A^3 + B^3 \\ M^4 &=& A^4+B^4 \\ \vdots & & \vdots \end{array}$

Therefore: . $M^n \;=\;A^n + B^n$

**samanthaj1001** · December 19th 2008, 09:22 AM

wow this is all soo helpful.. thanks so much! the rest of it seems to be the trickier part tho, but i know ull be able to help me! thanks!

**Soroban** · December 19th 2008, 11:01 AM

Hello again, Samantha!

Let $X\:=\:\begin{bmatrix}1&1 \\1&1 \end{bmatrix} \qquad Y \:=\:\begin{bmatrix}1&\text{-}1\\\text{-}1&1\end{bmatrix}$

Let: . $A\:=\:aX\:\text{ and }\:B\:=\:bY\:\text{ where }a\text{ and }b\text{ are constants.}$ . [1]

(a) Use different values of $a\text{ and }b$ to calculate: . $A^2,\:A^3,\:A^4,\:B^2,\:B^3,\:B^4$

From [1] we have: . $A \:=\:\begin{bmatrix}a&a\\a&a\end{bmatrix}\qquad B \:=\:\begin{bmatrix}b&\text{-}b\\\text{-}b&b\end{bmatrix}$

$A^2 \;=\;\begin{bmatrix}a&a\\a&a\end{bmatrix}\,\begin{ bmatrix}a&a\\a&a\end{bmatrix} \;=\; \begin{bmatrix}2a^2&2a^2\\2a^2&2a^2\end{bmatrix}$

$A^3 \;=\;A^2\cdot A \;=\;\begin{bmatrix}2a^2&2a^2\\2a^2&2a^2\end{bmatr ix}\,\begin{bmatrix}a&a\\a&a\end{bmatrix} \;=\;\begin{bmatrix}4a^3&4a^3\\4a^3&4a^3 \end{bmatrix}$

$A^4 \;=\;A^3\cdot A \;=\;\begin{bmatrix}4a^3&4a^3\\4a^3&4a^3\end{bmatr ix}\,\begin{bmatrix}a&a\\a&a\end{bmatrix} \;=\; \begin{bmatrix}8a^4&8a^4\\8a^4&8a^4\end{bmatrix}$

$B^2 \:=\:\begin{bmatrix}b&\text{-}b\\ \text{-}b&b\end{bmatrix}\,\begin{bmatrix}b&\text{-}b\\ \text{-}b&b^2\end{bmatrix} \;=\;\begin{bmatrix}2b^2 & \text{-}2b^2\\\text{-}2b^2&2b^2\end{bmatrix}$

$B^3 \;=\;B^2\cdot B \;=\;\begin{bmatrix}2b^2&\text{-}2b^2\\ \text{-}2b^2&2b^2\end{bmatrix}\,\begin{bmatrix}b&\text{-}b\\ \text{-}b&b\end{bmatrix} \;=\; \begin{bmatrix}4a^3&\text{-}4b^3 \\ \text{-}4b^3&4b^3\end{bmatrix}$

$B^4 \;=\;B^3\cdot B \;=\;\begin{bmatrix}4b^3&\text{-}4b^3\\ \text{-}4b^2 & 4b^3 \end{bmatrix}\,\begin{bmatrix}b&\text{-}b\\ \text{-}b&b\end{bmatrix} \;=\;\begin{bmatrix}8b^4&\text{-}8b^4 \\ \text{-}8b^4&8b^4\end{bmatrix}$

By considering interger powers of $A\text{ and }B$ , find expressions for: . $A^n,\:B^n,\: (A+B)^n$

$A^n \;=\;\begin{bmatrix}2^{n-1}a^n & 2^{n-1}a^n \\ 2^{n-1}a^n & 2^{n-1}a^n \end{bmatrix} \;=\;2^{n-1}a^nX$

$B^n \;=\;\begin{bmatrix}2^{n-1}b^n & \text{-}2^{n-1}b^n \\ \text{-}2^{n-1}b^n & 2^{n-1}b^n\end{bmatrix} \;=\;2^{n-1}b^nY$

We have already worked out the powers of: $M \:=\:A + B$ in a previous segment.

I'll leave the "Test, Discuss, and Explain" portions for you . . .

**gacaggabs** · January 7th 2009, 01:10 PM

I was wondering how an algebraic method can be used to explain how to arrive at the general statement (M^n=A^n+B^n) or (M^n=aX^n+bY^n)...same thing.

**gacaggabs** · January 8th 2009, 01:49 PM

by the way, i just want to point out that I'm not asking for the answer to my question...it's not that I want the exact answer on how to do it. I just want some hints on how to approach that question or some clarification and guidance on what exactly it is that it is asking. I've done the whole IB internal assesment on my own (unlike some people who are too lazy and just wanted the straight answer so they could copy/paste and hand it in...at least the answer given here helped me check my own work) but I'm stuck on how to work that last one. I would really appreciate some help.

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