P5: The matrix 1 2 0 0 3 0 -1 1 2 1 0 20-3 A= 1201114 1 2 1 0 2 1 1 3 6 2 1 5 2
ID: 3115332 • Letter: P
Question
P5: The matrix 1 2 0 0 3 0 -1 1 2 1 0 20-3 A= 1201114 1 2 1 0 2 1 1 3 6 2 1 5 2 2 has reduced row echelon form 1 2 0 0 3 0-1 0 0 1 0-1 0 -2 0 0 01 -2 0 1 0 0 0 0 0 4 (a) Without a calculation, give a basis for Ro(A) using row vectors from RREF(A) (b) Without a calculation, give a basis for Col(A) using column vectors from A (c) Determine the general solution of the system A0, giving a basis for the solution space (d) Verify the rank of A plus the dimension of the solution space is the number of variablesExplanation / Answer
(a) The linearly independent rows of A constitute a basis for row (A). Hence row(A) = { (1,2,0,0,3,0,1),(0,0,1,0,-1,0,-2),(0,0,0,1,-2,0,1),(0,0,0,0,0,1,4)}. Since we do not know whether there were any row interchanges in deriving the RREF of A, we have chosen the linearly independent rows from the RREF of A.
(b) A basis for col(A) is { (1,0,0,0,0)T,(0,1,0,0,0)T,(0,0,1,0,0)T,(0,0,0,1,0)T}.
(c)If X = (x1,x2,x3,x4,x5,x6,x7)T, then the equation AX =0 is equivalent to x1+2x2+3x5-x7= 0…(1), x3-x5-2x7 = 0…(2) , x4-2x5+x7 = 0…(3) and x6+4x = 0…(4). Now, let x2 =r, x5= s and x7 = t. Then, we have x1 = -2r-3s+t, x3= s+2t, x4= 2s-t , x6 = -4t so that X = (-2r-3s+t, r, s+2t, 2s-t, s,t)T = r(-2,1,0,0,0,0)T+ s(-3,0,1,2,1,0)T + t(1,0,2,-1,0,1)T, where r,s,t are arbitrary real numbers. This is the general solution.
(d)The rank of A equals the number of non-zero rows in its RREF so the rank of A is 4. As per the rank-nullity theorem, the rank + nullity of A is equal to the number of columns in A. Hence the dimension of the solution space of the equation AX = 0 (which is the nullity of A) is 7-4 = 3. Even otherwise, a basis for the solution space of the equation AX = 0 is {(-2,1,0,0,0,0)T,(-3,0,1,2,1,0)T, (1,0,2,-1,0,1)T} so that the dimension of the solution space of the equation AX = 0 is 3.
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