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# Concepts

## Multiplication by rows

$Ax$ comes from dot products, each row times the column $x$:

$Ax = \begin{pmatrix}(row 1) \cdot x \\ (row 2) \cdot x \\ (row 3) \cdot x \end{pmatrix}$

## Multiplication by columns

$Ax$ is a combination of column vectors:

$Ax = x(\text{column 1}) + y(\text{column 2}) + z(\text{column 3})$

# Questions

## Question 1

Write $2x+3y+z+5t=8$ as a matrix $A$ (how many rows?) multiplying the column vector $x = (x, y, z, t)$ to produce $b$. The solutions $x$ fill a plane or “hyperplane” in 4-dimensional space. The plane is 3-dimensional with no 4D volume.

## Question 2

Find the matrix $P$ that multiplies $(x, y, z)$ to give $(y, z, x)$. Find the Matrix $Q$ that multiplies $(y, z, x)$ to bring back $(x, y, z)$.

## Question 3

(a) What 2 by 2 matrix $R$ rotates every vector by $90\,^{\circ}$? $R$ times $\begin{pmatrix}x \\ y\end{pmatrix}$ is $\begin{pmatrix}y \\ -x\end{pmatrix}$ (b) What 2 by 2 matrix $R^{2}$ rotates every vector by $180\,^{\circ}$?

$Ax = \begin{pmatrix}2 & 3 & 1 & 5\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ t \\ \end{pmatrix}$

how many rows?

one row.

Find the matrix $P$ that multiplies $(x, y, z)$ to give $(y, z, x)$.

$\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}y \\ z \\ x\end{pmatrix}$ $\therefore P = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$

Find the Matrix $Q$ that multiplies $(y, z, x)$ to bring back $(x, y, z)$.

$\begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix}y \\ z \\ x\end{pmatrix}$ $\therefore Q = \begin{pmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$

(a)

$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}y \\ -x\end{pmatrix}$ $\therefore R = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$

(b)

Intuitively, if a $90\,^{\circ}$ rotation occurs when R is applied once, wouldn’t a $180\,^{\circ}$ rotation occur when R is applied twice?

$\therefore \text{ Let's apply R to the answer obtained from (a) again.}$ $\begin{pmatrix}y \\ -x\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} = \begin{pmatrix}-x \\ -y\end{pmatrix}$

$\text{By the way, It is the same as:}$ $\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-x \\ -y\end{pmatrix}$

$\text{Here, }\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix} = -I$ $\therefore R^{2} = -I$

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