MATH 240 – Spring – Exam 2 Solutions
ATH 240 – Spring 2011 – Exam 2 Solutions 1. (20 points) Given the matrix A = 0 1 1 2 3 0 2 2 4 6 0 −1 0 −1 0 0 4 2 6 6 , (a) (5 pts.) Put A into reduced row echelon form. SOLUTION. A = 0 1 1 2 3 0 2 2 4 6 0 −1 0 −1 0 0 4 2 6 6 → 0 1 1 2 3 0 0 0 0 0 0 0 1 1 3 0 0 −2 −2 −6 → 0 1 1 2 3 0 0 0 0 0 0 0 1 1 3 0 0 0 0 0 → 0 1 1 2 3 0 0 1 1 3 0 0 0 0 0 0 0 0 0 0 → 0 1 0 1 0 0 0 1 1 3 0 0 0 0 0 0 0 0 0 0 = R. (b) Then find a basis for each of the following: i. (4 pts.) Row(A), the row space of A. ii. (4 pts.) Col(A), the column space of A. iii. (7 pts.) Nul(A), the null space of A. SOLUTION. The pivot positions are (1, 2) and (2, 3) (these are the (i, j) such that R(i, j) has a leading entry). Therefore, (i) Rows 1 and 2 of R are a basis for Row(A). (ii) Columns 2 and 3 of A are a basis of Col(A). (iii) For the equation Ax = 0, the free variables are x1, x4, x5 (because 1, 4, 5 are the columns without pivot positions). Therefore the following three vectors form a basis for Nul(A): 1 0 0 0 0 , 0 0 −3 0 1 , 0 −1 −1 1 0 . (Each vector is the solution to Rx = 0 determined by setting one free variable equal to 1 and the others equal to zero. The equations Rx = 0 and Ax = 0 have the same solutions.)
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- 5 de enero de 2023
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