The matrix operation needed to transform the original matrix (pre-image) is represented by option C. ( [tex]T' = \left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]\cdot \left[\begin{array}{cccc}-5&-8&-5&-3\\-9&-7&-3&-6\end{array}\right][/tex]. )
The operation used to obtain the image in dash lines is a rotation around the origin, which is defined by following vectorial expression:
[tex]T' = R\cdot T[/tex] (1)
Where:
- [tex]T[/tex] - Original matrix (pre-image).
- [tex]R[/tex] - Rotation matrix.
- [tex]T'[/tex] - Resulting matrix (image).
Where the original matrix is defined by the following matrix:
[tex]T = \left[\begin{array}{cccc}x_{1}&x_{2}&...&x_{n}\\y_{1}&y_{2}&...&y_{n}\end{array}\right][/tex] (2)
And the rotation matrix is:
[tex]R = \left[\begin{array}{cc}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{array}\right][/tex] (3)
Where [tex]\theta[/tex] is the rotation angle, in sexagesimal degrees.
If we know that [tex]\theta = 180^{\circ}[/tex], [tex](x_{1}, y_{1}) = (-5, -9)[/tex], [tex](x_{2},y_{2}) = (-8,-7)[/tex], [tex](x_{3},y_{3}) = (-5, -3)[/tex] and [tex](x_{4}, y_{4}) = (-3, -6)[/tex], then the resulting matrix is represented by this entity:
[tex]T' = \left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]\cdot \left[\begin{array}{cccc}-5&-8&-5&-3\\-9&-7&-3&-6\end{array}\right][/tex]
The matrix operation [tex]T' = \left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]\cdot \left[\begin{array}{cccc}-5&-8&-5&-3\\-9&-7&-3&-6\end{array}\right][/tex] is represented in the graph.
We kindly invite to see this question on matrix rotation: https://brainly.com/question/22363471
