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


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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]].



Answered Same DayDec 29, 2021

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

David answered on Dec 29 2021
129 Votes
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. If no solution explain why
X1 +x2 -2x3 = -3
3*x1-2*x2+2*x3 = 9
6*x1 -7*x2-x3 = 4
Evaluating the determinants we get
Δ= 49
Δx= 79
Δy=22
Δz= 124
Thus, x= Δx /Δ=79/49
y= Δy /Δ=22/49
z= Δz /Δ=124/49
3. Find adj(A) and use it to compute...
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