1. ## matrix

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!

2. Hello, samanthaj1001!

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

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

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

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

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

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

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

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

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

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

(b)

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

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

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

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

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

4. Hello again, Samantha!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

5. 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!

6. Hello again, Samantha!

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

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

(a) Use different values of $\displaystyle a\text{ and }b$ to calculate: .$\displaystyle A^2,\:A^3,\:A^4,\:B^2,\:B^3,\:B^4$
From [1] we have: . $\displaystyle A \:=\:\begin{bmatrix}a&a\\a&a\end{bmatrix}\qquad B \:=\:\begin{bmatrix}b&\text{-}b\\\text{-}b&b\end{bmatrix}$

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

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

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

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

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

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

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

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

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

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

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