Hello, yorkey!
I don't think you can "eyeball" the matrix.
You might have to do a teensy bit of math . . .
$\displaystyle \text{(c) The matrix }\,\begin{pmatrix}0 & 1 \\ \text{}1 & 0 \end{pmatrix}\,\text{ represents a single transformation.}$
$\displaystyle \text{(1) Describe fully this transformation.}$
Let $\displaystyle (a,b)$ be any vector.
Then: .$\displaystyle (a,b)\begin{pmatrix}0 & 1 \\ \text{}1 & 0\end{pmatrix} \;=\;(\text{}b,a)$
It transforms point $\displaystyle P(a,b)$ to the point $\displaystyle Q(\text{}b,a).$
Code:
Q 
* 
:* 
: *  P
: *  *
a: *  * :
: *  * :b
: * * :
+*+
b  a

Point $\displaystyle \,P$ is rotated $\displaystyle 90^o$ counterclockwise about the Origin.
$\displaystyle \text{(ii) Find the coordinates of the image of the point }(5,3).$
. . $\displaystyle (5,3)\begin{pmatrix}0 & 1 \\ \text{}1 & 0\end{pmatrix} \;=\;(\text{}3,5)$