1. [25 points] For the matrix A below, find a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A, and the nullity of A. The reduced row echelon form of A is the matrix R given below. A =    2 2 6 6 −2 −18 −5 3 9 1 37 5 0 −3 −9 −6 −12 15 1 5 15 11 15 −29    R =    1 0 0 1 −5 −4 0 1 3 2 4 −5 0 0 0 0 0 0 0 0 0 0 0 0    Solution: To find a basis for the null space, you need to solve the system of linear equations A~x = ~0, or equivalently, R~x = ~0. Parameterizing the solutions to this equation, and writing the answer in expanded form, produces        x1 x2 x3 x4 x5 x6        = α1 ·        0 −3 1 0 0 0        + α2 ·        −1 −2 0 1 0 0        + α3 ·        5 −4 0 0 1 0        + α4 ·        4 +5 0 0 0 1        so           4 5 0 0 0 1        ,        5 −4 0 0 1 0        ,        −1 −2 0 1 0 0        ,        0 −3 1 0 0 0           is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 4. A basis for the row space can be found by taking the nonzero rows of R: {[ 1, 0, 0, 1, −5, −4 ] , [ 0, 1, 3, 2, 4, −5 ]} A basis for the column space can be found by taking the columns of A which have pivots in them, so       2 3 −3 5    ,    2 −5 0 1       is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space) of A, so the rank of A is 2. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a basis for the column space, the nullity, and the rank. Grading for common mistakes: −3 points for forgetting a variable in the parameterization; −3 points for choosing columns of R for the column space of A; −3 points for choosing rows from A for the row space of A; −3 points for choosing the non-pivot columns of A for the null space of A.