Vaz_MAT_343_ONLINE_B_Fall_2019.hourogne.Section_1.5_Elementary_Matrices.pdf - Hassan Ouro-gneni Vaz MAT 343 ONLINE B Fall 2019 Assignment Section 1.5

# Vaz_MAT_343_ONLINE_B_Fall_2019.hourogne.Section_1.5_Elementary_Matrices.pdf

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Hassan Ouro-gneni Vaz MAT 343 ONLINE B Fall 2019 Assignment Section 1.5 Elementary Matrices due 10/21/2019 at 11:59pm MST 1. (1 point) Suppose that A is a 3 × 3 matrix. For each of the question (a)-(d), find the elementary matrix E such that EA gives the matrix resulting from A after the given row operation is performed. Then give E - 1 . a. R 2 1 5 R 2 E = , E - 1 = b. R 3 ← - 3 R 3 E = , E - 1 = c. R 1 R 3 E = , E - 1 = d. R 2 R 2 + 6 R 3 E = , E - 1 = Solution: a. E 1 is obtained by multiplying by 1 2 the first row of the identity matrix I = 1 0 0 1 . This gives E 1 = 1 2 0 0 1 The inverse of E 1 is obtained by multiplying the first row of the identity by 2. This gives E - 1 1 = 2 0 0 1 . b. E 2 is obtained by multiplying by - 6 the second row of the identity matrix I = 1 0 0 1 . This gives E 2 = 1 0 0 - 6 The inverse of E 2 is obtained by dividing the first row of the identity by - 6. This gives E - 1 2 = 1 0 0 1 - 6 . c. E 3 is obtained by switching the first and second rows of the identity matrix I = 1 0 0 1 . This gives E 3 = 0 1 1 0 The inverse of E 3 is obtained by switching the first and second rows of the identity . This gives E - 1 3 = 0 1 1 0 . Note that E 3 = E - 1 3 . This is always the case for permutation matrices, that is, matrices that are obtained by permuting the rows of the identity matrix. d. E 4 is obtained by performing the row operation r 2 r 2 + 7 r 1 on the identity matrix I = 1 0 0 1 . This gives E 4 = 1 0 7 1 The inverse of E 4 is obtained by performing the row op- eration r 2 r 2 - 7 r 1 on the identity matrix. This gives E - 1 4 = 1 0 - 7 1 . e. E 5 is obtained by multiplying by 1 5 the second row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 5 = 1 0 0 0 1 5 0 0 0 1 The inverse of E 5 is obtained by multiplying the second row of the identity by 5. This gives E - 1 5 = 1 0 0 0 5 0 0 0 1 . f. E 6 is obtained by multiplying by - 3 the third row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 6 = 1 0 0 0 1 0 0 0 - 3 The inverse of E 6 is obtained by dividing the third row of the identity by - 3. This gives E - 1 6 = 1 0 0 0 1 0 0 0 1 - 3 . g. E 7 is obtained by switching the fist and the third row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 7 = 0 0 1 0 1 0 1 0 0 The inverse of E 7 is obtained by switching the first and third row of the identity . This gives E - 1 7 = 0 0 1 0 1 0 1 0 0 . h. E 8 is obtained by performing the row operation r 2 r 2 + 6 r 3 on the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 8 = 1 0 0 0 1 6 0 0 1 The inverse of E 8 is obtained by performing the row operation r 2 r 2 - 6 r 3 the identity . This gives E - 1 8 = 1 0 0 0 1 - 6 0 0 1 . Answer(s) submitted: 1 0 0 0 1/5 1

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0 0 0 1 1 0 0 0 5 0 0 0 1 1 0 0 0 1 0 0 0 -3 1 0 0 0 1 0 0 0 -1/3 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 0 0 1 6 0 0 1 1 0 0 0 1 -6 0 0 1
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