This is an linear algebra question. You are given S = {vector V_1, ..., vector V
ID: 3038150 • Letter: T
Question
This is an linear algebra question.
You are given S = {vector V_1, ..., vector V_5} and b (all from R^4) a) [2 -3 4 -5 1 -2 2 -3 1 2 2 1 5 -3 7 -6 6 7 3 7| 11 15 1 18] b) Find rref of (a) c) Explain why vector b elementof span(s) d) Express vector b as a linear of vector V_1, ... vector V_5 in the simplest way possible. Ex vector b = ? vector v + > vector v_4 e) Write down all the solu to C_1 vector V_1 + C_2 vector V_2 + ... + C_5 vector V_5 = vector b f) Can we write vector b as a linear vector V_1, vector V_2, and vector V_5? How? C_1 = 0, C_2 = 0 g) Write down all the solution to the test equ: X_1 vector V_1 + X_2 vector V_2 + ... + X_5 vector V_5 = vector h) Is S dependent or indep? Explain. I) Use (g) to find a dependence equ vector V_1, vector V_2, vector V_4, vector V_5 (vector) j) Same as (i) butt vector V_1, vector V_2, and vector V_3Explanation / Answer
(b). The matrix A ( as described in (a) ) can be reduced to its RREF as under:
Multiply the 1st row by ½
Add 3 times the 1st row to the 2nd row
Add -4 times the 1st row to the 3rd row
Add 5 times the 1st row to the 4th row
Multiply the 2nd row by -2
Add 1/2 times the 2nd row to the 4th row
Multiply the 3rd row by -1/3
Add -2 times the 3rd row to the 4th row
Add 9 times the 3rd row to the 2nd row
Add -5/2 times the 3rd row to the 1st row
Add -1/2 times the 2nd row to the 1st row
Then the RREF of A is
1
0
4
0
-2
-12
0
1
-7
0
-5
0
0
0
0
1
3
7
0
0
0
0
0
0
(c). It is apparent from the RREF of A above that b = -12v1+7v3. Since b is a linear combination of vectors in S, therefore, b is in span S.
(d). b = = -12v1+7v3.
(e). The required solution is same as the solutions to AX = b, and if X = (x1,x2,x3,x4,x5)T, then we have x1 +4x3-2x5= -12, x2-7x3-5x5 = 0 and x4 +3x5 = 7 so that x1 = -12-4x3+2x5, x2= 7x3+5x5 and x4 = 7-3x5 so that X = (-12-4x3+2x5, 7x3+5x5 ,x3, 7-3x5 , x5)T = (-12,0,0,7,0)T +x3( -4,7,1,0,0)T+x5( 2,5,0,-3,1)T. Here, x3 and x5 are arbitrary. Hence, there are infinite solutions.
(f ). From theRREF of A, it is apparent that b can be expressed as a linear combination of v1,v2,v3,v4 and v5 as b = = -12v1+0 v2+ 7v3 +0v4+0v5.
(g). Please see (e ) above.
(h). Since v3 is a linear combination of v1 and v2 and v5 is a linear combination of v1 ,v2 and v4, therefore S is linearly dependent.
(i) v5+2v1+5v2-3v4 = 0
(j) v3-4v1+7v2 = 0.
1
0
4
0
-2
-12
0
1
-7
0
-5
0
0
0
0
1
3
7
0
0
0
0
0
0
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