ANSWERS TO TERM TEST 3b
1. (4 marks) Find the standard matrices for the following linear transformations:
(a) T1: R2 -> R3
w1 = 2x1 - 5x2
w2 = -x1 + 3x2 [T1] =
w3 = 2x1 - 7x2
(b) T2: R3 -> R3
w1 = x1 + 2x2 + 3x3 [T2] =
w2 = - 4x2 + 6x3
w3 = 3x1 + 5x2 + 7x3
2. (4 marks) Determine whether the linear transformation in question 1(a) is one-to-one.
det[T2] = 1(-28-30) - 2(-8) + 3(12) = 14
Since this determinant is non-zero the transformation is one-to-one.
3. (5 marks)
(a) Find the composition T2 o T1 of the two linear transformations in questions 1(a) and (b).
[T2oT1] = [T2][T1] =
(b) Is the composition in part (a) commutative? Justify your answer.
It is not commutative since the matrix product [T1][T2] is not defined.
4. (2 marks) Identify the linear transformations represented by the following standard matrices:
(a) Reflection about the x-axis
(b) Projection into the xz-plane
5. (7 marks) Find the standard matrix for the linear transformation T:R3 -> R3 which rotates a vector about the positive x-axis clockwise through an angle pi/4 and then reflects it about the xz-plane by considering the effect of this transformation on the standard basis e1, e2, e3.
The linear transformations fave the following effect on the standard basis:
rotation about x-axis reflection about xz-plane
e1 = (1, 0, 0) -> (1, 0, 0) -> (1, 0, 0) = T(e1)
e2 = (0, 1, 0) -> = T(e2)
e3 = (0, 0, 1) -> = T(e3)
Thus the standard matrix is [T] = [T(e1), T(e2), T(e3)] =
The following are the axioms which define a vector space V where u, v and w are vectors in V and k and m are scalars:
(1) If u and v are in V then u + v is in V.
(2) u + v = v + u
(3) u + (v + w) = (u + v) + w
(4) There is a zero vector 0 in V such that u + 0 = 0 + u = u for all u in V.
(5) For each u in V there is a negative -u such that u + (-u) = (-u) + u = 0
(6) For any u in V and any scalar k, ku is in V.
(7) k(u + v) = ku + kv
(8) (k + m)u = ku + mu
(9) k(mu) = (km)u
(10) 1u = u
6. (3 marks) Consider the set of all vectors of the form (x, y) with the operation + denoted by
(x, y) + (u, v) = (xu, yv) and the usual definition of scalar multiplication.
(a) What is the zero vector defined in axiom 4) for this definition of 'addition'?
Since u + 0 = (x, y) + (01, 02) = (x01, y02) = (x, y) = u this means that 01 = 1 and 02 = 1 i.e. the zero vector is (1, 1).
(b) Whaaat is the negative vector defined in axiom 5) for this definition of addition? Does it exist for all vectors (x, y)?
Since u + (-u) = 0 or (x, y) + (-x, -y) = (x(-x), y(-y)) = (1, 1) then -x = 1/x and -y = 1/y, i.e. -u = (1/x, 1/y). This vector does not exist if x and/or y is zero.
7. (5 marks) Is the set of 2 by 2 matrices of the form
a subspace of M22? Justify your answer.
We must show that axioms 1 and 6 hold.
A = and B =
A + B = which is of the required form.
kA = which is of the required form.
Hence the set is a subspace.
8. (5 marks) Are the vectors (3, -3, 1), (2, -1, 0) and (1, 4, -2) linearly independent? Justify you answer.
The three vectors are linearly independent if
a(3, -3, 1) + b(2, -1, 0) + c(1, 4, -2) = (0, 0, 0)
has the only the trivial
solution a = b = c = 0.
This is true ifhas
a non-zero determinant. The determinant has the value 3(2) - 2(6-4)
+ 1(1) = 3 so the vectors are linearly independent.
This is true ifhas a non-zero determinant. The determinant has the value 3(2) - 2(6-4) + 1(1) = 3 so the vectors are linearly independent.
9. (2 marks) Are the polynomials 1 + x2, 2 - x a basis for P2? Justify your answer.
A basis for P2 must have three basis vectors. Hence the given polynomials aaaer NOT a basis.
10. (7 marks) Find the coordinates of (2, 3, -5) relative to the basis (1, 0, -1), (2, -1, 0) and (0, 1, 2) for R3.
We must find the coefficients a, b and c such that (2, 3, -5) = a(1, 0, -1) + b(2, -1, 0) + c(0, 1, 2)
or a + 2b = 2, -b + c = 3, -a + 2c = -5. This has the solution a = 13/2, b = -9/4 and c = 3/4 (the method of solution is not important) so that the co-ordinates are (13/2, -9/4, 3/4).
11. (4 marks) Prove that an nxn invertible matrix has rank n.
If A is invertible then Ax = 0 has only the trivial solution which means the nullity of A is 0. But rank + nullity = n, the number of rows of A, so that the rank = n.
12. (7 marks)
(a) For what value of s are the vectors (1, s, 3) and (2, -2, s) orthogonal?
For the vectors to be orthogonal their dot product must be zero, i.e. 2 - 2s + 3s = 0 or s = -2.
(b) Find a vector which is orthogonal to the two vectors in part (a).
Let the vector be (x, y, z). Then x + sy + 3z = 0 and 2x - 2y + sz = 0 with s = -2. Solving this system yields z = t, y = 4t, x = 5t so the vectors is t(5, 4 1).
13. (10 marks) Use the Gram-Schmidt method to transform the vectors (1, 3) and (2, 2)} into an orthonormal basis for R2.
Let u1 = (1, 3) and u2 = (2, 2).
Then take v1 = (1, 3) and v2 = u2 - (u2 . v1)/(v2 . v1) v1 = (2, 2) - (8/10)(1, 3) = 6/5, -2/5)
(b) Find the coordinates of (2, -2) with respect to the basis in part (a).
Let the vector be v. Then the coordinates are
Thus the coordinates of v with respect to this basis are