Chapter 8.PS, Problem 1PS

To determine

(a)

To find:

The matrices $AT$ and $ATT$, sketch of the original and the two images of the triangle. Also find the transformation represented by $A$.

Answer to Problem 1PS

Solution:

The matrices $AT$ and $ATT$ are $\left[\begin{array}{ccc}-1& -4& -2\\ 1& 2& 3\end{array}\right]$ and $\left[\begin{array}{ccc}-1& -2& -3\\ -1& -4& -2\end{array}\right]$ respectively and the sketch of the original and the two images of the triangle is shown below in figure $\left(1\right)$, figure $\left(2\right)$, and figure $\left(3\right)$ and transformation represented by $A$ is the rotation by $90°$ counterclockwise.

Explanation of Solution

Given:

Matrix $A$ is,

$A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$

Matrix $T$ is,

$T=\left[\begin{array}{ccc}1& 2& 3\\ 1& 4& 2\end{array}\right]$

Approach:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$.

Here $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\mathrm{...}+a_{in}{b}_{nj}$.

The following condition prevails for the matrix multiplication.

$\begin{array}{l}A\times B=AB\\ m\times nn\times pm\times p\end{array}$

Calculation:

Multiply matrix $A$ and $T$.

$\begin{array}{rll}AT& =& \left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 1& 4& 2\end{array}\right]\\ & =& \left[\begin{array}{ccc}0\left(1\right)-1\left(1\right)& 0\left(2\right)-1\left(4\right)& 0\left(3\right)-1\left(2\right)\\ 1\left(1\right)+0\left(1\right)& 1\left(2\right)+0\left(4\right)& 1\left(3\right)+0\left(2\right)\end{array}\right]\\ & =& \left[\begin{array}{ccc}-1& -4& -2\\ 1& 2& 3\end{array}\right]\end{array}$

From the above calculation, the matrix $AT$ is $\left[\begin{array}{ccc}-1& -4& -2\\ 1& 2& 3\end{array}\right]$.

Multiply matrix $A$ and $AT$.

$\begin{array}{rll}AAT& =& \left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& -4& -2\\ 1& 2& 3\end{array}\right]\\ & =& \left[\begin{array}{ccc}0\left(-1\right)-1\left(1\right)& 0\left(-4\right)-1\left(2\right)& 0\left(-2\right)-1\left(3\right)\\ 1\left(-1\right)+0\left(1\right)& 1\left(-4\right)+0\left(2\right)& 1\left(-2\right)+0\left(3\right)\end{array}\right]\\ & =& \left[\begin{array}{ccc}-1& -2& -3\\ -1& -4& -2\end{array}\right]\end{array}$

From the above calculation, the matrix $AAT$ is $\left[\begin{array}{ccc}-1& -2& -3\\ -1& -4& -2\end{array}\right]$.

The graph for the original triangle is shown below.

Figure $\left(1\right)$

The graph for the $AT$ triangle is shown below.

Figure $\left(2\right)$

The graph for the $AAT$ triangle is shown below.

Figure $\left(3\right)$

The transformation $A$ interchanges the coordinates of $x$ and $y$ and then takes the negative of the $x$ coordinate. $A$ represents a rotation by $90°$ counterclockwise.

Conclusion:

Hence, the matrices $AT$ and $ATT$ are $\left[\begin{array}{ccc}-1& -4& -2\\ 1& 2& 3\end{array}\right]$ and $\left[\begin{array}{ccc}-1& -2& -3\\ -1& -4& -2\end{array}\right]$ respectively and the sketch of the original and the two images of the triangle is shown above in figure $\left(1\right)$, figure $\left(2\right)$, and figure $\left(3\right)$ and transformation represented by $A$ is the rotation by $90°$ counterclockwise.

To determine

(b)

To find:

The transformation that produces the triangle determined by $AT$ and $T$.

Answer to Problem 1PS

Solution:

The transformation that produces the triangle determined by $AT$ is the rotation of $AAT$ by $90°$ clockwise and the transformation that produces the triangle determined by $T$ is the rotation of $AT$ by $90°$ clockwise.

Explanation of Solution

The matrix $AAT$ is rotated $90°$ in the clockwise to obtain $AT$. Then, $AT$ is rotated $90°$ clockwise to obtain $T$.

Conclusion:

Hence, the transformation that produces the triangle determined by $AT$ is the rotation of $AAT$ by $90°$ clockwise and the transformation that produces the triangle determined by $T$ is the rotation of $AT$ by $90°$ clockwise.