examf - MAT 242 A 1 1 3 6 -1 -5 -2 -4 1 3 0 1. (15) Given...

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MAT 242 A 1 1. (15) Given that A = 3 6 - 1 - 5 - 2 - 4 1 3 3 6 - 2 - 4 4 8 - 2 - 6 is row-equivalent to R = 1 2 0 - 2 0 0 1 - 1 0 0 0 0 0 0 0 0 : (a) Find all solutions to the homogeneous system of linear equations A x = 0 . (b) What is the dimension of the solution space of the system in (a)? (c) Find a basis for Col( A ).
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Final Exam 2 2. (16) If A = ± 1 2 k 4 ² , find all values of k for which. .. (a) A is invertible. (b) det( A ) = 0. (c) A is row-equivalent to ± 1 0 0 1 ² . (d) A x = b has a solution for every b .
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MAT 242 A 3 3. (18) If V is a subspace of R 3 , find its dimension. If not, show why not. (a) V = { ( x,y,z ) | xy + yz + zx = 0 } . (b) V = { ( x,y,z ) | x + y + z = 0 } . (c) V = { ( x,y,z ) | x = 0 , y + z = 0 , and y - 2 z = 0 } .
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Final Exam 4 4. (20) Consider the following elements of R 3 : u = 1 2 2 , v = 0 - 1 - 2 , w = 1 1 0 , a = 0 2 4 , b = - 9 1 2 .
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This note was uploaded on 06/30/2008 for the course MAT 242 taught by Professor Kaliszewski during the Fall '02 term at ASU.

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examf - MAT 242 A 1 1 3 6 -1 -5 -2 -4 1 3 0 1. (15) Given...

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