Step-by-step explanation: To show that T is invertible, we need to show that T is both injective and surjective.
First, to show that T is injective, we need to show that if T(x1, x2) = T(y1, y2), then (x1, x2) = (y1, y2). So:
T(x1, x2) = T(y1, y2)
(6x1 - 8x2, -5x1 + 7x2) = (6y1 - 8y2, -5y1 + 7y2)
This gives us two equations:
6x1 - 8x2 = 6y1 - 8y2
-5x1 + 7x2 = -5y1 + 7y2
We can rewrite these equations as a matrix equation:
[6 -8; -5 7] [x1; x2] = [y1; y2]
This is a linear system of equations with the coefficient matrix [6 -8; -5 7], which we can row-reduce to get:
[1 0; 0 1] [x1; x2] = [(7/29)y1 + (8/29)y2; (5/29)y1 + (6/29)y2]
Since the coefficient matrix reduces to the identity matrix, we can see that the system has a unique solution for any given (y1, y2), which means that T is injective.
Next, to show that T is surjective, we need to show that for any (y1, y2) in R^2, there exists (x1, x2) in R^2 such that T(x1, x2) = (y1, y2). So we need to solve the equation T(x1, x2) = (y1, y2):
(6x1 - 8x2, -5x1 + 7x2) = (y1, y2)
This gives us two equations:
6x1 - 8x2 = y1
-5x1 + 7x2 = y2
We can rewrite these equations as a matrix equation:
[6 -8; -5 7] [x1; x2] = [y1; y2]
Since we showed earlier that the coefficient matrix [6 -8; -5 7] is invertible, we can solve for [x1; x2] using the formula:
[x1; x2] = [6 -8; -5 7]^-1 [y1; y2]
We can compute the inverse of [6 -8; -5 7] as follows:
[6 -8; -5 7]^-1 = (1/74) [7 8; 5 6]
Therefore, we have:
[x1; x2] = (1/74) [7 8; 5 6] [y1; y2]
= [(7/74)y1 + (8/74)y2; (5/74)y1 + (6/74)y2]
This shows that for any (y1, y2) in R^2, there exists (x1, x2) in R^2 such that T(x1, x2) = (y1, y2), which means that T is surjective.
Since T is both injective and surjective, we know that T is invertible. To find a formula for T^-1, we can use the formula for the inverse of a 2x2