I. (8 points) Find the value or values of a (if any) that make the vectors vi =
ID: 3168510 • Letter: I
Question
I. (8 points) Find the value or values of a (if any) that make the vectors vi = (1 ,(1,1), t 2 = (0,2, a) and U3 = (-1, a, 2) a linearly independent collection in IR3 2, (5-1 5= 10 points) Let T : R4 R3 be given by Find (a) a basis for Ker(T), (b) a basis for R(T). ints) Let B-(i, 2} be the basis of R2 given by the vectors vi- N 3. (10po and t 2 = | | (in standard coordinates). Let Let D-(01, u, us} be the basis of R3 given by 03 and4 (in standard coordinates). Let T : R2 R3 be the linear map whose representation with -1 2 respect to bases B and D is given by the matrix ADB= 1-1 21 . Find the matrix A that represents T in the standard coordinates of IR2 and R3 (continue on back page)Explanation / Answer
1
0
-1
a
2
a
1
a
2
To determine whether v1,v2,v3 are linearly independent, we will reduce A to its RREF as under:
Add –a times the 1st row to the 2nd row
Add -1 time the 1st row to the 3rd row
Multiply the 2nd row by ½
Add–a times the 2nd row to the 3rd row
Then the RREF of A is
1
0
-1
0
1
a
0
0
3-a2
Now, it is apparent that v1,v2,v3 will be linearly independent if 3-a2 0 i.e. if a ±3.
2. We have T(x1,x2,x3,x4) =(x1+2x2+5x3-2x4,2x1+x2+4x3-x4,3x1-x2+x3+x4).Then T(e1)=(1,2,3) ,T(e2)= (2,1,-1) ,T(e3) =(5,4,1) and T(e4) = (-2,-1,1). Thus, the standard matrix of T is A =
1
2
5
-2
2
1
4
-1
3
-1
1
1
Now, Ker (T), being same as ker(A), is the set of solutions to the equation AX = 0. The RREF of A is
1
0
1
0
0
1
2
-1
0
0
0
0
Thus, if X = X = (x,y,z,w)T, then the equation AX = 0 is equivalent to x+z = 0 or, x = -z and y+2z-w = 0 or, y = -2z+w. Then X = (-z,-2z+w,z,w)T = z(-1,-2,1,0)T+w(0,1,0,1)T. Hence {(-1,-2,1,0)T, (0,1,0,1)T
Also, it is apparent from the RREF of A that { (1,0,1,0),(0,1,2,-1)} is a basis for R(A) or, R(T).
Please upload the 3rd question again separately.
1
0
-1
a
2
a
1
a
2