Part 1 3 4 4 6 | 1 -3 2 the columns of A. b-| 2 | , and let W-Spanfal,a,,a3], -4
ID: 3185686 • Letter: P
Question
Part 1 3 4 4 6 | 1 -3 2 the columns of A. b-| 2 | , and let W-Spanfal,a,,a3], -4 Let A | 2 1 where ai , a, , a, are , a. Is b in W? Show your work (no calculator) and give a reason for your answer. b. Find the solution, if it exists, to the matrix equation Axb. If the system corresponding to this matrix equation is dependent, write the solution in parametric vector form. Hint: You can use your results from part a c. Is there a non-trivial solution to the homogeneous system corresponding to Ax 0? If so, write the solution in parametric vector form. You may use a calculator or your results from part a d. In words, describe and compare the solutions to Ax - b and Ax0. For example, are the solutions lines? planes? Is one a translation of the other? Be specific. e. If b is in W, write b as a linear combination of the columns of A. Hint: Pick a value for your free variable. Answers may vary f. Are the columns of A linearly independent? Why or why not? Do the columns of A span R? Why or why not? g. Using A and b write 1) The matrix equation Ax-b 2) The equivalent system of equations 3) The equivalent vector equation 4) The augmented matrix |A b . Circle the pivot positions.Explanation / Answer
a. Let M = [A|b] =
3
4
4
6
2
1
6
2
-1
-3
2
-4
To determine, whether b is in W, we will reduce M to its RREF as under:
Multiply the 1st row by 1/3
Add -2 times the 1st row to the 2nd row
Add 1 times the 1st row to the 3rd row
Multiply the 2nd row by -3/5
Add 5/3 times the 2nd row to the 3rd row
Add -4/3 times the 2nd row to the 1st row
Then the RREF of M is
1
0
4
2/5
0
1
-2
6/5
0
0
0
0
Thus, b = (2/5)a1 +(6/5)a2. Hence b is in W.
b. If X = (x,y,z)T, then the equation AX = b is equivalent to x+4z = 2/5 or, x = 2/5 -4z and y-2z = 6/5 or, y = 6/5 +2z. Then, X = (2/5-4z,6/5+2z,z)T = (2/5,6/5,0)T +z(-4,2,1)T =(2/5,6/5,0)T +t(-4,2,1)T where t = z is an arbitrary real number.
c. If b = (0,0,0)T, then the equation AX = 0 is equivalent to x+4z =0 or, x = -4z and y-2z =0 or, y = 2z, so that X = (-4z,2z,z) = z(-4,2,1)T= t(-4,2,1)T where t = z is an arbitrary real number.
d. The solution to AX = b and AX = 0 are both lines. In the former case, the line passes through the point (2/5,6/5,0)T and has the direction of the vector (-4,2,1)T whereas in case of the homogeneous equation AX = 0, the line passes through the origin and has the direction of the vector (-4,2,1)T.
e. As in part a. above, b = (2/5)a1 +(6/5)a2.
e. It is apparent from the RREF of M that a3 = 4a1-2a2 so that the columns of A are not linearly independent. Further, dim(R3 ) = 3 so that the columns of A do not span R3.
Please post the remaining part again .
3
4
4
6
2
1
6
2
-1
-3
2
-4