Find the matrix that represents a rotation clockwise around the origin by 30∘ followed by a magnification by a factor of 4.
My attempt:
I multiplied the magnification matrix $\left[ {\begin{array}{cc} 4 & 0 \\ 0 & 1 \\ \end{array} } \right]$ by the rotation matrix of $-30$ degree, $\left[ {\begin{array}{cc} \cos(t) &-\sin(t) \\ \sin(t) & \cos(t \\ \end{array} } \right]$ where $t=\theta$.
On
Another way to do this is to find out where the standard basis vectors $\langle 1, 0 \rangle$ and $\langle 0, 1 \rangle$ map to.
In general, if we have a linear transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $T(\langle 1, 0 \rangle) = \langle a, b \rangle$ and $T(\langle 0, 1 \rangle) = \langle c, d \rangle$, then $T$ can be represented by the matrix $\displaystyle \begin{bmatrix} a & c \\ b & d \\ \end{bmatrix}$.
The problem with this is that $[\{4\;0\};\{0\;1\}]$ is not a scaling matrix, it is a "stretching" matrix, only stretching vectors in the x direction. What you are looking for is $[\{4\;0\};\{0\;4\}]$. Upon multiplying these together, we get $[\{4\cos(-30^{\circ})\;\;-4\sin(-30^{\circ})\};\{4\sin(-30^{\circ})\;\;4\cos(-30^{\circ})\}]$, or $[\{2\sqrt{3}\;\;-2\};\{2\;\;2\sqrt{3}\}].$