[4014]
[4400]
[4004]
[4104]
Correct answer is C
P:(x,y)→(2x+y,−2y)
p[xy]=[2xy0−2y]
∴
\therefore p^2 = \begin{bmatrix} 2&1\\0&-2\end{bmatrix} \begin{bmatrix} 2&1\\0&-2\end{bmatrix} = \begin{bmatrix} 4&0\\0&4\end{bmatrix}
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