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# hw4sol - Solutions Assignment 4 3.3.20 Find the redundant...

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Solutions: Assignment 4 3.3.20 Find the redundant column vectors of the given matrix A “by inspection”. Then find a basis of the image of A and a basis of the kernel of A . A = 1 0 5 3 - 3 0 0 0 1 3 0 0 0 0 0 0 0 0 0 0 The second and third columns are mutliples of the first. And the fifth column is 3 times the third column minus 12 times the first. So the second, third, and fifth columns are redundant. And a basis for the image is just 1 0 0 0 and 3 1 0 0 . To get a basis for the kernel we look at A x = 0. This tells us that relationship between entries of x are just x 1 = - 5 x 3 - 3 x 4 +3 x 5 and x 4 = - 3 x 5 . So this gives us a basis of 3 elements: x 2 = 1 x 3 = 0 x 5 = 0 x 2 = 0 x 3 = 1 x 5 = 0 x 2 = 0 x 3 = 0 x 5 = 1 0 1 0 0 0 - 5 0 1 0 0 3 0 0 - 3 1 3.3.22 Find the reduced row-echelon form of the given matrix A . Then find a basis of the image of A and a basis for the kernel of A . A = 2 4 8 4 5 1 7 9 3 2 4 8 4 5 1 7 9 3 1 2 4 4 5 1 7 9 3 1 2 4 0 - 3 - 15 0 - 5 - 25 1 2 4 0 1 5 0 - 5 - 25 1 0 - 6 0 1 5 0 0 0 There are leading ones in the first two columns of rref ( A ), So a basis for the image is 2 4 7 and 4 5 9 . We know that x is in the kernel is equiv- alent to A x = 0. That’s equivalent to rref ( A ) x = 0. Which is equivalent 1

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to x 1 = 6 x 3 and x 2 = - 5 x 3 . Which is equivalent to x = x 3 6 - 5 1 . So 6 - 5 1 is a basis for the kernel. 3.3.33 A subspace V of R n is called a hyperplane if V is defined by the homoge- neous equation c 1 x 1 + c 2 x 2 + . . . + c n x n = 0 where at least one of the coefficients c i is nonzero. What is the dimen- sion of a hyperplane in R n ? Justify your answer carefully. What is a hyperplane in R 3 ? What is it in R 2 .
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hw4sol - Solutions Assignment 4 3.3.20 Find the redundant...

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