SOLUTION
[tex]\text{The image matrix at 90}^o\text{ rotation is derived using the formula }[/tex][tex]\begin{gathered} \begin{bmatrix}{\cos \theta} & -\sin \theta & {} \\ \sin \theta & {\cos \theta} & {} \\ {} & {} & {}\end{bmatrix}\text{ where }\theta\text{ is 90}^{} \\ \text{This becomes the matrix of } \\ \begin{bmatrix}{0} & {-1} & {} \\ {} & {} & {} \\ {1} & {0} & {}\end{bmatrix}\text{ }\times\text{ }\begin{bmatrix}{2} & {4} & {6}{}{} \\ {2} & {5} & {1}\end{bmatrix}\text{ = }\begin{bmatrix}{0-2} & {0-5} & {0-1}{}{} \\ {2+0} & {4+0} & {6+0}\end{bmatrix}\text{ = } \\ \\ \begin{bmatrix}{-2} & {-5} & {-1}{}{} \\ {2} & {4} & {6}\end{bmatrix} \\ \\ \text{Therefore, the image matrix is }\begin{bmatrix}{-2} & {-5} & {-1}{}{} \\ {2} & {4} & {6}\end{bmatrix} \end{gathered}[/tex]
The graph is shown below
