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### Quiz6Solutions

Course: MATH 247, Spring 2003
School: Stanford
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Math Newberger 247 Spring 03 Quiz 2.1-2.3 100 1. Find the inverse of A = 5 1 0 . 022 name: 1 100100 100 5 1 0 0 1 0 0 1 0 5 022001 022 0 100 1 0 1 0 5 0 0 2 10 Thus A1 1 00 1 0 . = 5 1 5 1 2 00 1 0 01 00 1 0 2 1 2. Suppose A is invertible and AB = 0, where 0 denotes the zero matrix. Show that B = 0. Since A is invertible, we have the matrix A1 to work with. Multiplying AB = 0 by A1 , we get...

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Math Newberger 247 Spring 03 Quiz 2.1-2.3 100 1. Find the inverse of A = 5 1 0 . 022 name: 1 100100 100 5 1 0 0 1 0 0 1 0 5 022001 022 0 100 1 0 1 0 5 0 0 2 10 Thus A1 1 00 1 0 . = 5 1 5 1 2 00 1 0 01 00 1 0 2 1 2. Suppose A is invertible and AB = 0, where 0 denotes the zero matrix. Show that B = 0. Since A is invertible, we have the matrix A1 to work with. Multiplying AB = 0 by A1 , we get A1 (AB ) = A1 0. Using the associative law, this becomes (A1 A)B = A1 0. But A1 A = I and the right hand side is 0, so we get B = 0. 3. Suppose B = P AP 1 , where A P and are invertible n n matrices. Find B 1 . The inverse of the product of invertible matrices is the product of the inverses in reverse order. B 1 = (P AP 1 )1 = (P 1 )1 A1 P 1 = P A1 P 1 4. Let A and B be invertible n n matrices. Explain why the columns of A1 B span all of Rn . Since A is invertible, so is A1 (its inverse is A). Now since A1 and B are invertible, so is A1 B (since the product of invertible matrices is invertible). Now by the invertible matrix theorem, A1 B has n pivot positions so it has a pivot in every row, and the columns of A1 B span all of Rn .
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Stanford - MATH - 247
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Stanford - MATH - 247
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Stanford - MATH - 247
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