3. (a). If x +3y-2z = 0, then x = -3y+2z so that (x,y,z)T = (-3y+2z,y,z)T = y(-3,1,0)T +z (2,0,1)T.
This implies that every vector in H is a linear combinationof 2 linearly independent vectors (-3,1,0)T ,(2,0,1)T. Hence H is a vector space and therefore a subspace of R3.
(b). H = Nul A, where A =
-3 |
1 |
0 |
2 |
0 |
1 |
so H is a subspace of R3.
4. Let A =
1 |
1 |
-2 |
2 |
5 |
-1 |
-3 |
1 |
10 |
It may be observed that the entries in the columns of A are the scalar multiples of 1 and the coefficients of x,x2 in the polynomials in the set S.
The RREF of A is
1 |
0 |
-3 |
0 |
1 |
1 |
0 |
0 |
0 |
This implies that -2+x+10x2 = -3(1+2x-3x2)+( 1+5x+x2) . Thus, the polynomials in the set S are not linearly independent . Hence dim(S) = 2. Since dim(P2) = 3, hence the polynomials in the set S do not span P2.
Further span (S) contains all the linear combinations of the polynomials 1+2x-3x2, 1+5x+x2.
Question (7) Consider the vector space R3 with the regular addition, and scalar aL multiplication. Is The set of all vectors of the form b, subspace of R3 Question (9) a) Let S- {2-x + 3x2, x + x, 1-2x2} be a subset of P2, Is s is abasis for P2? 2 1 3 0 uestion (6) Let A=12 1 a) Compute the determinant of the matrix A via reduction to triangular form. (perform elementary row operations)
Question (7) Consider...
1. Verify that the following linear system does not have an infinite number of solutions for all constants b. 1 +39 - 13 = 1 2x + 2x2 = b 1 + bxg+bary = 1 2. Consider the matrices -=(: -1, -13). C-69--1--| 2 -1 0] 3 and F-10 1 1 [2 03 (a) Show that A, B, C, D and F are invertible matrices. (b) Solve the following equations for the unknown matrix X. (i) AXT = BC (ii)...
how did we get the left null space please use simple
way
6% 0-0, 1:44 AM Fri May 17 , Calc 4 4 Exaimi 3 solutions Math 250B Spring 2019 1. Let A 2 6 5 (a) Find bases for and the dimensions of the four fundamental subspaces. Solution Subtract row onc from row 2, then 8 times row 2 from row 3, then 5 timcs rovw 2 fro row. Finally, divide row1 by 2 to get the row reduced...