![A triangle ABC with vertices A(1, -1), B(3, -1), C(1, 3) is mapped to A'B'C' by transformation T1 = [[-1, 0], [0, 1]]. Triangle A'B'C' is then mapped to A''B''C'' by transformation T2 = [[2, 2], [0, 1](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1777391470675-cc54990c2a032c71.png&w=3840&q=75)
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Understand the transformations. Triangle ABC with vertices A(1, -1), B(3, -1), C(1, 3) is mapped to A'B'C' by transformation T_1 = -1 & 0 \\ 0 & 1 . Triangle A'B'C' is then mapped to A''B''C'' by transformation T_2 = 2 & 2 \\ 0 & 1 . a) Find the coordinates of A'B'C'. To find the coordinates of A'B'C', we multiply the coordinates of A, B, C by the matrix T_1. For A(1, -1): A' = -1 & 0 \\ 0 & 1 1 \\ -1 = (-1)(1) + (0)(-1) \\ (0)(1) + (1)(-1) = -1 \\ -1 So, A'(-1, -1). For B(3, -1): B' = -1 & 0 \\ 0 & 1 3 \\ -1 = (-1)(3) + (0)(-1) \\ (0)(3) + (1)(-1) = -3 \\ -1 So, B'(-3, -1). For C(1, 3): C' = -1 & 0 \\ 0 & 1 1 \\ 3 = (-1)(1) + (0)(3) \\ (0)(1) + (1)(3) = -1 \\ 3 So, C'(-1, 3). The coordinates of A'B'C' are A'(-1, -1), B'(-3, -1), C'(-1, 3). b) Find the matrix which maps A'B'C' onto A''B''C''. The problem statement explicitly gives this matrix as the second transformation. The matrix is 2 & 2 \\ 0 & 1 . c) Determine the area scale factor between A'B'C' and A''B''C''. The transformation from A'B'C' to A''B''C'' is given by the matrix T_2 = 2 & 2 \\ 0 & 1 . The area scale factor is the absolute value of the determinant of the transformation matrix. det(T_2) = (2)(1) - (2)(0) = 2 - 0 = 2 The area scale factor is |2| = 2. d) Find the transformation matrix which maps A''B''C'' onto ABC. This requires finding the inverse of the combined transformation from ABC to A''B''C''. First, find the combined transformation matrix T_total that maps ABC to A''B''C''. This is T_2 · T_1. T_total = T_2 · T_1 = 2 & 2 \\ 0 & 1 -1 & 0 \\ 0 & 1 T_total = (2)(-1) + (2)(0) & (2)(0) + (2)(1) \\ (0)(-1) + (1)(0) & (0)(0) + (1)(1) T_total = -2 & 2 \\ 0 & 1 Now, find the inverse of T_total. For a matrix M = a & b \\ c & d , its inverse is M^-1 = (1)/(ad-bc) d & -b \\ -c & a . Here, a=-2, b=2, c=0, d=1. det(T_total) = (-2)(1) - (2)(0) = -2 - 0 = -2 T_total^-1 = (1)/(-2) 1 & -2 \\ 0 & -2 T_total^-1 = (1)/(-2) & (-2)/(-2) \\ (0)/(-2) & (-2)/(-2) T_total^-1 = -(1)/(2) & 1 \\ 0 & 1 That's 2 down. 3 left today — send the next one.