Math 54 - Ogus, A. - 2005 Spr - Midterm 1 - Solutions

# Math 54 - Ogus, A. - 2005 Spr - Midterm 1 - Solutions -...

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Mathematics 54W Professor A. Ogus Spring, 2005 Midterm Solutions—March 03, 2005 1. (15 pts) Find a matrix X which satisﬁes the given conditions if possi- ble. If not, explain why not. (a) (3 pts) 2 X + ± 3 2 - 1 4 ² = ± 1 0 1 2 ² . 2 X = ± 1 0 1 2 ² - ± 3 2 - 1 4 ² = ± - 2 - 2 2 - 2 ² X = ± - 1 - 1 1 - 1 ² (b) (3 pts) X = ± 1 1 1 - 1 ²± 3 - 2 0 2 5 1 ² . X = ± 5 3 1 1 - 7 - 1 ² . (c) (3 pts) X = ± 3 - 2 0 2 5 1 ²± 1 1 1 - 1 ² . This is undeﬁned because the length of the rows of the ﬁrst factor does not match the length of the columns of the second. (d) (2 pts) X ± 1 2 4 1 1 3 ² = ± 1 2 4 0 - 1 - 1 ² , with X invertible. The elementary matrix X := ± 1 0 - 1 1 ² works here. (e) (2 pts) X ± 1 2 4 1 1 3 ² = ± 1 1 3 1 2 4 ² , with X invertible. The elementary matrix ± 0 1 1 0 ² works. (f) (2 pts) X ± 1 2 4 1 1 3 ² = ± 1 2 4 2 4 8 ² , with X invertible. This is not possible since the rows spaces of the two matrices are not the same—they have diﬀerent dimensions.

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2. (15 pts) Let A := 1 - 2 0 0 1 0 0 0 0 0 0 0 0 0 2 0 0 0 - 1 2 0 0 - 1 1 . Use Gauss elimination in the standard way to: (a) (5 pts) Find a basis for the row space of A .
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## This note was uploaded on 03/08/2010 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.

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Math 54 - Ogus, A. - 2005 Spr - Midterm 1 - Solutions -...

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