Inverses of Linear Transformations

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Department
Mathematics
Course
MATH 2331
Professor
Rita Jimenez Rolland
Semester
Spring

Description
Linear Algebra Notes 2.4 Linear Transformation Inverses January 30, 2014 n n T : R ! R is said to be invertible if the transformax) = x = y has a unique solution. If rref(A) =nI and rk(A) = n, then A is invertible. 0 1 1 1 1 @ A Ex.: Is the matrix A = 2 3 2 invertible? 3 8 2 x1 + x2 + x3 = y1 2x1 + 3x2 + 2x3 = y2 (▯2I) 3x + 8x + 2x = y (▯3I) 1 2 3 3 x1 + x2 + x3 = y1 (▯II) x2 = ▯2y1 + y2 5x ▯ x = ▯3y + y (▯5II) 2 3 1 3 x1 + x3 = 31 ▯ y2 (+III) x2 = ▯2y1 + y2 ▯x 3 = 71 ▯ 5y2 + y3 x1 = 10y1 ▯ 6y2 + y3 x2 = ▯2y1 + y2 x3 = ▯7y1 + 5y2 ▯ y3 Because the system of equations had a single unique solution, the matrix is invertible. n n The inverse should satisfy the following: fx) : R ! R , for a giy, Ty) = x such thatx) = y 0 1 10 ▯6 1 T ▯1(y) [email protected]▯2 1 0A y ▯7 5 ▯1 0 1 10 ▯6 1 A ▯1= @ ▯2 1 0 A ▯7 5 ▯1 If you are givenx) = x = y... h . i h . i Take rref A . I ▯ > rrefA . B n Does rref(A) = n ? Yes: B = A▯1 No: A is not invertible. 1 Example: 0 1 1 0 0 a.) A = @ 0 1 0A . Is this invertible? 0 0 1 h i ▯1 . Yes. To ▯nd A , we need to ▯nd rrefA . I3 2 3 . h i 61 0 0 . 1 0 07 h i . 6 . 7 . A . I3 = 40 1 0 . 0 1 05 rref A . 3 . 0 0 1 . 0 0 1 ▯1 (I3) = I3 0 1 0 0 1 ▯1 b.) A = @ 0 1 0A . Find A . 1 0 0 rk(A) = 3 2 . 3 2 . 3 0 0 1 . 1 0 0
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