Answer :
Answer:
[tex]((-\frac{1}{2}(x+y), -\frac{1}{2} (x-y) )[/tex]
Step-by-step explanation:
T(x , y) = (-x -y, -x +y)
T(1, 0) = (-1, -1) = -1(1 , 0) -1(0 , 1)
T(0, 1) = (-1, 1) = -1(1 , 0) +1(0 , 1)
Therefore,
[tex]T = \left[\begin{array}{ccc}-1&-1\\-1&1\end{array}\right][/tex]
|T| = [-1 -1] = -2
T is invertible
[tex]T^-^1 = -\frac{1}{2} \left[\begin{array}{ccc}1&+1\\+1&-1\end{array}\right] \\\\=\left[\begin{array}{ccc}-1/2&-1/2\\-1/2&1/2\end{array}\right][/tex]
Therefore,
[tex]T^-^1(x,y)=(-\frac{1}{2}x -\frac{1}{2} y,-\frac{1}{2} x+\frac{1}{2} y)[/tex]
[tex]((-\frac{1}{2}(x+y), -\frac{1}{2} (x-y) )[/tex]