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MATH 15a: Linear Algebra
Practice Final Exam, Solutions
- (a) (4 points) Complete the definition: Vectors ~v 1 ,... , ~vk are linearly independent if... Answer: ...no vi is in the span of v 1 ,... , vi− 1. OR ...they do not satisfy any nontrivial linear relation. (b) (4 points) Suppose A is an m × n matrix whose columns are linearly independent. What is the nullity of A? Answer: If the columns are linearly independent, the rank of A is equal to n (the dimension of the span of the columns), so the nullity equals zero.
- Let A denote the matrix A =
[
√ 3 12 −^12 2 −
√ 3 2
]
Let T : R^2 → R^2 be the linear transformation given by T (~x) = A~x. (a) (5 points) Describe T geometrically. Answer: For any angle θ, the matrix for counterclockwise rotation by θ is
[cos θ − sin θ sin θ cos θ
]
Since cos(^56 π ) = −
√ 3 2 and sin(^56 π ) =^12 , we see that^ T^ is rotation by^56 π = 150◦. (b) (5 points) Find the characteristic polynomial of A, and use it to find all eigen- values of A or to show that none exist. Explain why your answer makes sense geometrically. Answer: The characteristic polynomial of A is:
det
[
√ 3 2 1 −^ t^ −^12 2 −
√ 3 2 −^ t
]
2 −^ t)
2 +^1
= t^2 + √ 3 t + 1. In the quadratic formula, the quantity b^2 − 4 ac is (√3)^2 − 4 · 1 · 1 = −1, so this polynomial has no real roots, and hence A has no eigenvalues. An eigenvector of A would be a vector ~v such that T (~v) is on the same line as ~v. However, a rotation by an angle other than 0 or π cannot send any line to itself. (c) (5 points) Compute A^2011. (Hint: What power of A is equal to the identity?) Answer: Because T is a 5/12 rotation, doing T twelve times is the same as the identity. Thus A^12 = In. Since 2011 = 167 · 12 + 7, we have A^2011 = A^7. Also, doing T six times is a 1/2 rotation (i.e. rotation by π or 180◦), which is minus the identity. Thus,
A^7 = −A =
[ √ 3
2 12 −^12
√ 3 2
]
- (a) (5 points) Compute the inverse of the matrix
Answer: We apply Gauss-Jordan elimination to [A|I 3 ] until we get [I 3 |A−^1 ]:
0 1 13 0 −^13
0 1 13 0 −^13
0 0 −^13 1 −^83
0 1 13 0 −^13
Thus, A−^1 =
(b) (5 points) Find all solutions to the system of linear equations − 4 x + 5z = − 2 − 3 x − 3 y + 5z = 3 −x + 2y + 2z = − 1 Answer: This system is A~x = ~b, where A is as in the previous part and ~b =
. Hence
x y z
(^) = A−^1 ~b =
- (a) (4 points) Using Gaussian elimination, find all solutions to the following system of linear equations: 2 x 2 + 3x 3 + 4x 4 = 1 x 1 − 3 x 2 + 4x 3 + 5x 4 = 2 − 3 x 1 + 10x 2 − 6 x 3 − 7 x 4 = − 4
(c) (4 points) What is the rank of the linear transformation T : R^4 → R^3 with T (~e 1 ) = ~e 2 − 3 ~e 3 , T (~e 2 ) = 2~e 1 + − 3 ~e 2 + 10~e 3 , T (~e 3 ) = 3~e 1 + 4~e 2 − 6 ~e 3 , and T (~e 4 ) = 4 ~e 1 + 5~e 2 − 7 ~e 3? Answer: The matrix for T is exactly A (given above). Since rref(A) has three pivots, we see that the rank of T is 3. (Hint: The answers to all three parts are related!)
- Let A denote the matrix (^)
(a) (4 points) Find the eigenvalues of A. Answer: The characteristic polynomial of A is
det
1 − t 0 − 2 0 5 − t 0 − 2 0 4 − t
(^) = (1 − t)(5 − t)(4 − t) − (−2)(5 − t)(−2)
= (5 − t)(t^2 − 5 t + 4 − 4) = −t(t − 5)^2 So the eigenvalues are 0 (with multiplicity 1) and 5 (with multiplicity 2). (Note: Even if I don’t ask explicitly, you should always give the algebraic multiplicities of eigenvalues.) (b) (6 points) Find an orthonormal basis of R^3 consisting of eigenvectors for A. Answer: For the eigenvalue 0, we row-reduce A − 0 I 3 = A:
The solution to the system of equations is x 1 = 2x 3 x 2 = 0 x 3 = free
so the kernel is spanned by
. We normalize to get a unit vector:
~v 1 = √^15
For the eigenvalue 5, we row-reduce A − 5 I 3 :
The solution to the system of equations is x 1 = −^12 x 3 x 2 = free x 3 = free
So the kernel is spanned by the vectors
(^) and
. Normalizing, we get:
~v 2 = √^15
(^) and ~v 3 =
(It happens that these vectors are already orthogonal to each other. If they weren’t, we would have to use Gram-Schmidt to find an orthonormal basis for E 5 .) Combining the bases for E 0 and E 5 , we get an orthonormal basis for R^3. (c) (3 points) Find a 3 × 3 orthogonal matrix S and a 3 × 3 diagonal matrix D such that A = SDST^. Answer: S is gotten by putting the three basis vectors together in a matrix:
S =
√^25 − √^15
√ 5 √^25
Since S is orthogonal, S−^1 = ST^. D is gotten by listing the eigenvalues down the diagonal (in the same order as we wrote the corresponding eigenvectors in S):
D =
(If you had put the eigenvectors in a different order, you would need a different D). (d) (4 points) For any integer t, write an explicit formula for At.
The solution to the system is x 1 = −x 3 x 2 = − 4 x 3 x 4 = free
so any vector in V ⊥^ is of the form
−t − 4 t t
. Since the norm of this must be 1,
we get t^2 + 16t^2 + t^2 = 18t^2 = 1, so t = ± √^118 = ± 3 √^12 = ±
√ 2
- Thus, there are two possible answers. Let us choose t =
√ 2 6 , so that
~v 3 =
(b) (4 points) Let T : R^3 → R^3 denote the linear transformation that interchanges ~v 1 and ~v 3 and has ~v 2 as an eigenvector with eigenvalue −5. Write down [T ]B, the matrix of T with respect to B. Answer: The matrix [T ]B is gotten by writing down T (~v 1 ), T (~v 2 ), and T (~v 3 ) in B coordinates and putting them as the columns of a matrix. You do not know what ~v 1 , ~v 2 , and ~v 3 are. Specifically, since T (~v 1 ) = ~v 3 T (~v 2 ) = − 5 ~v 2 T (~v 3 ) = ~v 1 , we have [T ]B =
(c) (3 points) Write down a product of matrices that equals the standard matrix of T. (Note: Your answer will depend on your choice of ~v 3 in part (a).) Answer: The change-of-basis matrix SB is
SB =
Since SB is orthogonal, we have S B−^1 = SB. The standard matrix for T is then given by
A = SB[T ]BS B−^1 =
(d) (3 points) What is the determinant of the standard matrix of T? Answer: Since A and [T ]B are similar, they have the same determinant: det A = det[T ]B = −(−5) = 5.
- Compute the determinant of each of the following matrices. Indicate clearly the method being used. (a) (5 points) (^)
Answer:
det A = −7 det
(along the second row)
= (−7)(4) det
(^) (along the fourth row)
= (−7)(4)(2) det
[ 2
]
(along the first row) = (−7)(4)(2)(−2) = 56 (b) (5 points) (^)
1 b b^2 b b^2 b^3 b^2 b^3 b^4
(where b is any real number). Answer: We can do row operations: subtract b times the first row from the second row and b^2 times the first row from the second row. This gives
1 b b^2 0 0 0 0 0 0
Thus, the original matrix is not invertible, so its determinant is zero.
- Determine whether each statement is true or false, and provide a justification or a counterexample.
