Question

1. f:R2 to R2 is a linear transformation so that f(1,2)=(16,4) and f(8,1)=(38,-43). Find f(4,-2).

2. f:R² to R² is a linear transformation so that f(1,4)=(32,15) and f(7,1)=(35,-3). Find the determinant of the matrix of f.

Answer 1

1) Let us consider a relation a(1,2)+b(8,1) = (4,-2) where a,b are real.

Then, a+8b = 4

2a+b = -2

Solving we get, a = -4/3 and b = 2/3.

Then we have, (4,-2) = (-4/3)*(1,2)+(2/3)*(8,1)

Now, f(4,-2) = (-4/3)*f(1,2)+(2/3)*f(8,1)

i.e., f(4,-2) = (-4/3)*(16,4)+(2/3)*(38,-43)

i.e., f(4,-2) = (4,-34)

Therefore, f(4,-2) = (4,-34).

2) Let the matrix be : A = .

Here, f(1,4) = (32,15) and f(7,1) = (35,-3).

Then we have,

=

i.e., a+4b = 32...............(i)

c+4d = 15...............(ii)

And, =

i.e., 7a+b = 35..............(iii)

7c+d = -3................(iv)

Solving (i) and (iii) we get, a = 4 and b = 7.

Solving (ii) and (iv) we get, c = -1 and d = 4.

Then, A = .

And, det(A) = = 4*4-7*(-1) = 16+7 = 23

Therefore, the determinant of the matrix of f is 23.