Given that M : (x, y) \(\to\) (7x, 3x - y) and N : (x, y) \(\to\) (2x - y; 5x + 3y)
(c) find the coordinates of the point Q whose image is Q(2, 4) under the linear transformation N
(a) The matrix M can be obtained by applying M to the standard basis vectors (1,0) and (0,1):
M(1,0) = (7, -0) = (7,0)
M(0,1) = (0, -1) = (0,-1)
Therefore, the matrix M is:
M =
[7 0;
0 -1]
Similarly, the matrix N can be obtained as follows:
N(1,0) = (2, 5)
N(0,1) = (-1, 3)
Therefore, the matrix N is:
N =
[2 -1;
5 3]
(b) To find the image of P(2, -3) under the linear transformation N followed by M, we need to compute the product MNP, where P is the column vector (2,-3). That is,
MN(2,-3) = M(N(2,-3)) = M(2(-3) - (-1)(-3); 5(2) + 3(-3)) = M(-3, 11) = (77, -11)
Therefore, the image of P(2,-3) under the linear transformation N followed by M is the point (77,-11).
(c) To find the coordinates of the point Q whose image is Q(2,4) under the linear transformation N, we need to solve the equation N(x,y) = (2,4). That is,
2x - y = 2
5x + 3y = 4
Solving for x and y, we get x = 2 and y = -1. Therefore, the point Q is (2,-1).
Now we need to find the preimage of Q under the linear transformation M. That is, we need to solve the equation M(x,y) = (2,-1). That is,
7x = 2
-y = -1
Solving for x and y, we get x = 2/7 and y = 1. Therefore, the preimage of Q(2,4) under the linear transformation N followed by M is the point (2/7,1).