1 answer

Question 5 True of False part II: 5 problems, 2 points each. (6). Let w be...

Question:

Question 5 True of False part II: 5 problems, 2 points each. (6). Let w be the x-y plain of R3, then wlis any line that is or
Question 1 True or False parti: 5 problems, 3 points each. (1). If each column of A has a pivot, then A = 7 has only one solu
Question 5 True of False part II: 5 problems, 2 points each. (6). Let w be the x-y plain of R3, then wlis any line that is orthogonal to w. (Select) (7). Let A be a 3 x 3 non-invertible matrix. If Ahas eigenvalues 1 and 2, then A is diagonalizable. Sele (8). If an x n matrix A is diagonalizable, then n eigenvectors of A form a basis of " [Select] (9). Letzbean x 1 vector. Then all matrices A s.t. Az z form a subspace. [Select] (10). Let A be an x n matrix, then Null(A) C Null(A²), Col(A²) Col(A). [Select]
Question 1 True or False parti: 5 problems, 3 points each. (1). If each column of A has a pivot, then A = 7 has only one solution. (Select] (2). det(A? - B) = det(A - B) det(A + B). Because A2 - B2 = (A - B)(A + B). [Select] (3). If A and A + I are both invertible, then (A+1)-1 = A-1 + 1. (Select] (4). Because (Null(A))+ = Col(AT), so (Null(A))+ + Col(A). (Select] . (5). If A is similar to B, then A3 + 2A2 + 51 is equal to B3 + 2B2 + 51. Select]

Answers

(...) True (2) Since multiplication of matrices does not satisfy Commutative Low If we choose two maprices A and B such that(8) (8) → True. 19 Let w={A : AX*=* } Set ă= then oz = o te 1 - The zero matrise 0& W. So wishof a subspace. e false (00لا lo

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