Previous answers holtlinalg1 43003 consider the

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2. 4/4 points | Previous Answers HoltLinAlg1 4.3.003. Consider the following matrices. (To make your job easier, an equivalent echelon form is given for the matrix.) Find a basis for the column space of A . (If a basis does not exist, enter DNE into any cell.) = 1 0 4 7 2 1 13 8 0 1 5 6 1 0 4 7 0 1 5 6 0 0 0 0 1 0 -2 1 0 1 [1, 0; -2, 1; 0, 1] Find a basis for the row space of A . (If a basis does not exist, enter DNE into any cell.) 1 -2 rank( A ) = 2 2 nullity( A ) = 1 1 rank( A ) + nullity( A ) = 3 3 = m , . 1 2 3 3 5 8 , . 1 0 20 0 1 8 A x = 0 , x = s , 20 8 1 . 20 8 1 A = 1 0 4 7 2 1 13 8 0 1 5 6 1 0 4 7 0 1 5 6 0 0 0 0 ~
0 1 -4 13 -7 8 [1, 0; 0, 1; -4, 5; -7, -6] Find a basis for the null space of A . (If a basis does not exist, enter DNE into any cell.) 4 7 -5 6 1 0 0 1 [4, 7; -5, 6; 1, 0; 0, 1] Verify that the Rank-Nullity Theorem holds. (Let m be the number of columns in matrix A .) Solution or Explanation A basis for the column space, determined from the pivot columns 1 and 2, is A basis for the row space is determined from the nonzero rows of the echelon form, We solve to obtain and so our null space basis is rank( A ) = 2 2 nullity( A ) = 2 2 rank( A ) + nullity( A ) = 4 4 = m , . 1 2 0 0 1 1 , . 1 0 4 7 0 1 5 6 A x = 0 , x = s 1 + s 2 , 4 5 1 0 7 6 0 1
3. 4/4 points | Previous Answers HoltLinAlg1 4.3.006. Consider the following matrix. Find a basis for the column space of A . (If a basis does not exist, enter DNE into any cell.)
Find a basis for the row space of A . (If a basis does not exist, enter DNE into any cell.)

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