midterm_2_solutions

midterm_2_solutions - MA 35100 MIDTERM EXAM #2 SOLUTIONS...

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Unformatted text preview: MA 35100 MIDTERM EXAM #2 SOLUTIONS Problem. Consider the matrix A = 1 1- 1- 1 . a. Compute its reduced row-echelon form. Show all of your work. b. Find a basis for ker( A ). What is the nullity of A ? c. Find a basis for im( A ). What is the rank of A ? Solution: (a) To compute the reduced row-echelon form for A , add the first row to the second: rref( A ) = 1 1 0 0 . (b) The reduced row-echelon form above is the coefficient matrix for the system x 1 + x 2 = 0; the kernel of A consists of solutions of this linear system. We see that x 2 = t is a free variable, so the general solution is in the form x 1 x 2 = t- 1 1 i.e.,- 1 1 is a basis for ker( A ). The nullity of A is the dimension of the kernel of A , so null( A ) = 1. (c) According to Theorem 3.3.5 on page 128 of the text, a basis for im( A ) corresponds to the pivot columns of rref( A ). This is the first column of A , so 1- 1 is a basis for im( A ). The rank is the number of elements in a basis, so rank( A ) = 1. Problem. Consider the matrix A = 1- 2- 3 6 . a. Compute its reduced row-echelon form. Show all of your work. b. Find a basis for ker( A ). What is the nullity of A ? c. Find a basis for im( A ). What is the rank of A ? Solution: (a) To compute the reduced row-echelon form for A , add thrice the first row to the second: rref( A ) = 1- 2 . 1 (b) The reduced row-echelon form above is the coefficient matrix for the system x 1- 2 x 2 = 0; the kernel of A consists of solutions of this linear system. We see that x 2 = t is a free variable, so the general solution is in the form x 1 x 2 = t 2 1 i.e., 2 1 is a basis for ker( A ). The nullity of A is the dimension of the kernel of A , so null( A ) = 1....
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midterm_2_solutions - MA 35100 MIDTERM EXAM #2 SOLUTIONS...

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