# ch3 - Solutions to Homework 7 Math 110 Fall 2006 Prob 3.1.3...

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Solutions to Homework 7. Math 110, Fall 2006. Prob 3.1.3. To find the inverses of the elementary matrices, we must simply undo the corresponding elementary operations: ( a ) 0 0 1 0 1 0 1 0 0 , ( b ) 1 0 0 0 1 / 3 0 0 0 1 , ( c ) 1 0 0 0 1 0 2 0 1 . Prob 3.1.5. If E is an elementary matrix of type I corresponding to the swap or rows i and J , then E has 1 in positions ( i, j ) and ( j, i ), zeros in positions ( i, i ) and ( j, j ), and the other entries exactly as in the identity matrix of the appropriate size. But then E is symmetric, i.e., its transpose is E itself. Likewise, if E is of type II, then its transpose is itself. If E is of type III and corresponds to adding row j multiplied by c to row i , then it looks like the identity matrix except for the entry c in position ( i, j ). Then its transpose has c in position ( j, i ), which corresponds to adding row i multiplied by c to row j . So, in all three cases E is an elementary matrix if and only if so is E t . Prob 3.2.2. (a) 2, (b) 3, (c) 2, (d) 1, (e) 3, (f) 3, (g) 1. Prob 3.2.6. (a) T is invertible. Writing the matrix of T in the standard basis for P 2 (IR) and inverting it, we get T - 1 ( ax 2 + bx + c ) = - ax 2 - (4 a + b ) x - (10 a + 2 b + c ) . (b) T is not invertible, since it has a nontrivial kernel, for example, T 1 = 0.

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• Spring '08
• Matrices, Invertible matrix, Row echelon form, Carpenter, Aj Bj

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