Sp06SolPractFinal - M340L Matrices and Matrix Calculations...

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M340L: Matrices and Matrix Calculations Unique Number: 57160 Practice Final sketchy Solution, May , 2006 LAST NAME (PRINT) ................................. First Name ....................................... - Calculators are authorized. Books, hand written notes or other documents are not authorized. - Show all your work. - Explain every answer, either by a computation, or by using a theorem. Cor- rect results with no explanation will not be taken into account. - Freely use both sides of the sheets. - Put your name on any additional or unstapled sheet. - Good Luck. I. Let A = 1 3 2 1 1 0 1 1 0 - 1 - 1 0 1 2 1 0 1 1 2 3 1 2 2 3 . 1) What is the domain of A ? What is the co-domain of A ? (3points) The domain of A is R 6 , number of columns of A . The co-domain of A is R 4 , number of rows of A . Observe that the words source and target are rather used for the linear mapping T (see question 3)), the words domain and co-domain are more usual for the matrix. 2) Give an echelon form for A . Give the reduced echelon form of A . 1
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A = 1 3 2 1 1 0 1 1 0 - 1 - 1 0 1 2 1 0 1 1 2 3 1 2 2 3 1 3 2 1 1 0 0 1 1 1 1 0 0 0 0 0 1 1 0 - 3 - 3 0 0 3 1 3 2 1 1 0 0 1 1 1 1 0 0 0 0 0 1 1 0 0 0 3 0 0 1 3 2 1 1 0 0 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 1 The reduced echelon form is 1 0 - 1 0 0 2 0 1 1 0 0 - 1 0 0 0 1 0 0 0 0 0 0 1 1 . You were expected to show more work than I do. 3) Is the mapping T : x 7→ A x one-to-one? No it is not because there are non pivot columns, so the homogeneous equation A x = 0 has non trivial solutions. Is it onto? (Justify your answers, Yes or No is not acceptable) Yes it is because there is a pivot in each row. 4) Without further calculation give the Rank of A and the dimension of the null space of A . (Justify your answers.) Rank A = 4 is the dimension of Col( A ) (or the dimension of the range of T ) that is the number of pivot columns. From the rank theorem the dimension of the null space of A is 6 - 4 = 2. It is also the number of non pivot columns. 5) Determine a basis for the column space of A . We may take the pivot columns of A , namely: A = ( a 1 = 1 1 1 2 , a 2 = 3 1 2 3 , a 4 = 1 - 1 0 2 , a 5 = 1 - 1 1 2 ) (This is what I am expecting. However since
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