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exam1problemsetsolutions-1 - Topic 3: Null Space. Q6: A= 10...

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Unformatted text preview: Topic 3: Null Space. Q6: A= 10 0 12 16 we want Ax = 0, x = (x1 , x2 , x3 )T , which is just x1 10 1 2 x2 = 0 016 x3 Thus, we have 10x1 + x2 + 2x3 = 0, (1) x2 + 6x3 = 0, (2) Look at the second equation, we get x2 = −6x3 Plug it into the first equation, we get 10x1 − 4x3 = 0 Therefore, x1 and x3 are dependent, I could now pick x1 = 1, then x3 = 5/2, again use x2 = −6x3 we get x3 = −15, thus the null space of A could be spanned by x = (1, −15, 5/2)T , in other words, x = (1, −15, 5/2)T is a basis of the null space. Notice rank (A) = 2, the x has 3 variables, so the dimension of your null space is 3 − 2 = 1, which coincides with the answer. 1 Q7: A= 10 0 123 168 x1 x2 x3 = 0 x4 we want Ax = 0, x = (x1 , x2 , x3 , x4 )T , which is just 10 1 2 3 0 1 6 −8 Thus, we have Look at the second equation, we get 10x1 + x2 + 2x3 + 3x4 = 0, (1) x2 + 6x3 − 8x4 = 0, (2) x2 = −6x3 + 8x4 Plug it into the first equation, we get 10x1 − 4x3 + 11x4 = 0 Now x1 , x3 and x4 are dependent, we can choose any values for two of them to determine the third one. Here, I will choose x1 = 1, x3 = 0 first, then x4 = −10/11, again use x2 = −6x3 + 8x4 we get x2 = −80/11, so x(1) = (1, −80/11, 0, −10/11)T . We are not done yet, remember rank (A) = 2, and the x has 4 variables, so the dimension of the null space should be 4 − 2 = 2, so you still need one more vector. OK, now I choose x1 = 0, x3 = 1, then x4 = 4/11, again use x2 = −6x3 + 8x4 we get x2 = −34/11, so x(2) = (0, −34/11, 1, 4/11)T . Now we are done. The null space of A could be spanned by x(1) and x(2) , in other words, x(1) and x(2) is the basis of the A’s null space. 2 ...
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This note was uploaded on 12/01/2009 for the course MATH 322 taught by Professor Staff during the Spring '08 term at Arizona.

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exam1problemsetsolutions-1 - Topic 3: Null Space. Q6: A= 10...

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