# LINEAR ALGEBRA TRANSFORMATION

May 29th, 2015
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The function t : R3 −→ R3 is given by the rule (x, y, z) 0−→ (y − z, x + z, x + y). (a) Use Strategy 1.1 in Unit LA4 to show that t is a linear transformation. (b) Write down the matrix of t with respect to the standard basis for R3. (c) Determine the

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Def Consider the square matrix A The matrix A is invertable if and only if there is a matrix B such that AxB = BxA (Identity Matrix)[1 0][0 1] = I[1 0 0][0 1 0] = I Identity matrix[0 0 1]We use the notation A^-1To find A^-1Use elementary row operationsI|A^-1Problem Find A^-1[2 3][7 8] = A[2 3 | 1 0] x (1/2)[7 8 | 0 1][1 3/2 | 0][7 8 | 0 1 ] x 7[1 3/2 | 0][0 -5/2| -7/2 1] x (-2/5)[1 3/2 0 ][0 1 7/5 2/5] x (-3/2)[1 0 -1/10 3/5][0 1 7/5 -2/5][-16/10 3/5][7/5 -2/5 ] = A^-1Determinants[a b][c d] = ad-bcA^-1 = 1/( ad bc) [d b] [-c a]A= [2 3] A^-1 = 1/(16-21)[8 -3] [7 8] [-7 2]= -1/5[8 -3][ -8/5 3/5 ] [-7 2] = [7/5 -2/5]Let a = [cos(theta) sin(theta)][-sin(theta) cos(theta)]A^-1 = 1/(cos^2(theta) (-sin^2(theta) [cos(theta) -sin(theta)] = [sin(theta) cos(theta)][cos(theta) -sin(theta)][sin(theta) cos(theta)]3x3 Determinant Evaluate determinant A[1 2 0][4 1 5] = A[1 1 -1]1 2 0 | 1 24 1 5 | 4 11 1 -1| 1 1= [(1)(1)(-1) + 2(5)(1) + 0(4)(1)] [ (2)(4)(-1) + (1)(5)(1) + (0)(1)(1)]= -1 +10 + 0 +8 -5 = 12Important fact A square matrix is invertible if and only if determinant =/= 0Det A = 0 -> singularDet A =/= -> non-singularExample let A =[1 -1 1][-1 1 1][1 1 -1]Find A^-11 -1 1| 1 0 0-1 1 1| 0 1 01 1 -1 | 0 0 1---------------1 -1 1| 1 0 00 0 2 | 1 1 00 2 -2| -1 0 11 -1 1 | 1 0 00 2 -2 | -1 0 10 0 2 | 1 1 0__

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