1 2 1 s r 3 3 are the vectors 1 1 2 2 1 1 and 1 2 3

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- 1 2 1 0 : s R 3
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3. Are the vectors 1 0 1 2 , 2 1 0 - 1 and - 1 - 2 3 10 linearly independent or linearly dependent. If they are linearly dependent exhibit one linear dependency. Solution. We solve the equation x 1 ~ V 1 + x 2 ~ V 2 + x 3 ~ V 3 = ~ 0 for non-zero vector ~ X = ( x 1 , x 2 , x 3 ), where ~ V 1 , ~ V 2 and ~ V 3 are the three given vectors. 1 2 - 1 0 1 - 2 1 0 3 2 - 1 10 0 0 0 0 - 1 1 - 2 1 1 2 - 1 0 1 - 2 0 - 2 4 0 - 5 12 0 0 0 0 - 2 1 2 1 5 1 1 0 3 0 1 - 2 0 0 0 0 0 2 0 0 0 0 1 0 3 0 1 - 2 0 0 2 0 0 0 0 0 0 0 1 2 1 0 3 0 1 - 2 0 0 1 0 0 0 0 0 0 0 Thus we see that the only solution is x 1 = x 2 = x 3 = 0 . The vectors are therefore linearly independent. 4
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4. Consider the matrix A = 1 1 2 1 y 3 2 3 5 with unspecified entry y . (a) Find the row reduced echelon form for A ? Does it depend on the value of y ? (b) Show that there is a value for y such that the columns of A are linearly dependent (c) Show that there is a value for y such that the columns of A are linearly independent Solution. We find the row reduced echelon form by the row reduction tableau method. 1 1 2 1 y 3 2 3 5 - 1 1 - 2 1 1 1 2 0 y - 1 1 0 1 1 1 1 2 0 1 1 0 y - 1 1 - 1 1 1 - y 1 1 0 1 0 1 1 0 0 2 - y Thus the row reduced echelon form of A is 1 0 0 0 1 0 0 0 1 if y 6 = 2 . and is 1 0 1 0 1 1 0 0 0 if y = 2 . Furthermore we see that when y is chosen to have value 2, then the columns of A are linearly dependent and if y is chosen to be any other value, then the columns are linearly independent.
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