SEE ATTACHED FILE - Find bases for the column space of A, row space of A, and null space of...

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SEE ATTACHED FILE - Find bases for the column space of A, row space of A, and null space of a.
	
Document Preview: Find bases for the column space of A, row space of A, and null space of a. verify that rank/nullity theorem holds. ( A is 4x4)A=  [[1,4,-1,1][3,11,-1,4][1,5,2,3][2,8,-2,2]Use cramers rule to find solution. If no solution explain whyX1 +x2 -2x3 = -33*x1-2*x2+2*x3 = 96*x1 -7*x2-x3 = 4Find adj(A) and use it to compute A^(-1)A= [[2,0,1][0,0,2][0,0,1]]Find the characteristic polynomial, the eigenvalues, and a basis for each eigenspace for matrix A.A=[[-1,0,0,0][5,-2,0,0][0,3,1,0][2,0,1,1]]Suppose A is a square matrix with characteristic polynomial (x-3)^3(x-2)^2(x+1)What are dimensions of AWhat are eigenvalues of AIs A invertibleWhat is largest possible dimension for eigenspace of AProve that u cannot be an eigenvector associated with two distinct eigenvalues x1 and x2 of A.Let A be an N x N matrix that is nilpotent of order k = 2.  That is, assume that there exists an integer k (where k = 2) such that Ak = 0 (the N x N matrix consisting of zeros).  Prove that 0 is the only eigenvalue of A.  Find the change of basis matrix from B2 to B1. B1 and B2 contain two column vectors (e.g. B1 has colum vector [2] and [1]  [ 1]         [1]B1= {[[2][1]],  [[1][1]]}B2 = {[[5][7]], [[3][4]]}B1 and B2 contain 3 column vectors with 3 rows. B1= {[[-1][1][-1]], [[1][0][2]], [[-2][5][0]]}  and B2= {[[-2][1][3]], [[2][0][1]], [[4][1][-1]]}.   Find XB1 if XB2= [[-1][1][3]]B2.Fiknd an example that meets given specifications.  A basis B of R3 such that [[3][1][-2]]B = [[1][2][5]].

David · Accepted Answer

ANSWERS: 
1. Find bases for the column space of A, row space of A, and null space of a. verify that 
rank/nullity theorem holds. ( A is 4x4) 
A=  [[1,4,-1,1][3,11,-1,4][1,5,2,3][2,8,-2,2] 
We first reduce the matrix to reduced row echelon form(RREF). 
By the following operations, 
add -4 times the 1st row to the 2nd row 
add 1 times the 1st row to the 3rd row 
add -1 times the 1st row to the 4th row 
multiply the 2nd row by -1 
add -2 times the 2nd row to the 3rd row 
add -1 times the 2nd row to the 4th row 
multiply the 3rd row by 1/5 
add -3 times the 3rd row to the 4th row 
add 1 times the 3rd row to the 2nd row 
add -1 times the 3rd row to the 1st row 
add -3 times the 2nd row to the 1st row 
Rref(A) =[[1,0,0,2][0,1,0,0][0,0,1,0][0,0,0,0]] 
The fourth column is a linear combination of the 1
st
 column. 
So, the basis for the column space are the remaining column vectors in the rref(A) = 
[1,0,0,0] , [0,1,0,0] and [0,0,1,0]. 
Similarly, row space has the basis as: [1,0,0,2] , [0,1,0,0] and [0,0,1,0]. 
For the nullspace, Ax=0 or equivalently, rref(A)x=0, 
If x [x1,x2,x3,x4], we have, 
x = [-2x4,0,0,x4]. Thus, [-2,0,0,1] is the basis for the nullspace. 
Dimension of column space= Dimension of row space =Rank of A=3 
Dimension of null space =1. 
3+1 =4 = Dimension of A .Thus the rank-nullity equation is verified.
2. Use cramers rule to find solution.

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