Math 313 (Linear Algebra) Exam 2 - Practice Exam

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Instructions:

• For questions which require a written answer, show all your work. Full credit will be given only if the necessary work is shown justifying your answer. • Simplify your answers. • Scientiﬁc calculators are allowed. • Should you have need for more space than is allocated to answer a question, use the back of the page the problem is on and indicate this fact. • Please do not talk about the test with other students until after the last day to take the exam.

For Instructor use only. Question MC FB 10 11 12 13 14 15 16 Total Points 100 Earned Part I: Multiple Choice Questions: Mark all answers which are correct for each question.

0 1 2 1. Let A be a 3 × 3 matrix such that the columns of A sum to the zero vector. Let B = 0 1 2. 0 1 2 Which of the following could be the product matrix AB?

0 0 0 0 1 2 a) 1 1 1 b) 0 1 2 2 2 2 0 1 2

0 1 1 0 0 0 c) 0 1 1 d) 0 0 0. 0 1 1 0 0 0

Solution: d

2. Which of the following rules for determinants concerning n × n matrices A and B are correct.

a) det(AB) = det(A) det(B).

b) det(A + B) = det(A) + det(B).

c) det(AB) = det(BA).

d) det(AT ) = det(A).

e) Swapping two rows in A results in a change of the sign of the corresponding determinant.

f) Subtracting column number 2 from column number 1 does not alter the value of the determinant.

Solution: True are a) c) d) e) f) 3. Which of the following statements is true? (Mark all which are true)

a) det(AT B) = det(BT A).

b) In a determinant of a 3 × 3-matrix A one may swap the ﬁrst row and the ﬁrst column without changing the value of the determinant.

c) Adding to the ﬁrst column of a determinant a linear combination of the remaining columns does not aﬀect its value.

d) A determinant is linear as a function of each of its vector arguments.

e) If the determinant of a 3 × 3-matrix is equal to the product of its elements in the main diagonal then it is either an upper triangular or a lower triangular matrix.

f) For a 4 × 4-matrix A one always has det(2A) = 16 det(A).

Solution: True are a) c) d) f). A counterexample to b) is

0 1 0 0 0 1

0 0 1 = 1 1 0 1 = 2

1 2 0 0 2 0

A counterexample to e) is

0 0 0

1 0 1

0 0 0 Each correct choice may count half a point.

4. Which of the following are vector spaces? (Addition and scalar multiplication are deﬁned normally.)

a) The set of all polynomials with even coeﬃcients.

b) The set of all polynomials with odd coeﬃcients.

a + 1 3 c) The set of all vectors in R with the form a + b where a, b, c are real numbers. a + c

 q + t  3 d) The set of all vectors in R with the form 2r − t where q, r, s, t are real numbers. 3s + t

x e) The set of all vectors in 2 of the form where y = x + 1. R y

f) The set of all real-valued functions f on R such that f(1) = 1. Solution: 3 c) and d) are true. c) because one obtains all of R . The argument that a subspace must contain 0 kills b), e) and f). 5. Which of the following sets form a basis for P2? a) {1, t, t2}

b) {t2 + t − 2, −t + 2}

c) {t2 + 7, 2t − 3, t + 5, t2 + t + 1}

d) {2, t2 + t, 2t2 + 2t + 3}

e) {1, t, t − 1}

f) {t2 + 2t + 3, 7t, t2 + 1}

Solution: a), c), f) Part II: Fill in the blank with the best possible answer.

6. Let A and B be n×n matrices and express them in terms of their column vectors: A = a1 a2 ··· an and B = b1 b2 ··· bn . Then the products AB and BA can also be expressed in terms of their column vectors:

AB =

BA = . Solution:

AB = [Ab1,..., Abn], the answer for BA has letters A/a interchanged with B/b

7. Let A be a 2 × 2 matrix. If det(A) = −3, then we know that A is . Solution: invertible/one to one/onto. Certainly invertible is the best answer.

0 1 8. Suppose that S is a region in the plane with area 15. Let T (x) = Ax, where A = . Then the 3 1 area of T (S) is . Solution: 45, just the Area times the absolute value of the determinant. Part III: Justify your answer and show all work for full credit.

9. A matrix C is idempotent if C2 = C. Let A be an m × n matrix and B be an n × m matrix. Prove that if BA = In, then AB is idempotent. Solution: 2 2 (AB) = ABAB = A(BA)B = AInB = AB that is, C = C for the matrix C := AB. Hence AB is an idempotent.

10. Consider the matrix 0 0 2 0 1 0 0 1 A =   . 0 −1 3 0 2 1 5 −3

Show that the matrix A is invertible by ﬁnding the inverse of A using row reductions. Verify that the matrix that you ﬁnd is the actual inverse of the matrix A. Solution: By verifying errors can be detected.

11. Find the determinant of  1 0 1 2 2  0 1 2 1 −1   A =  2 1 1 0 −3 .    0 1 0 2 1 −2 2 −1 1 0

Solution: −12

12. Suppose that x, y, and z satisfy

2x − 3y + 3z = 1 2x − y + z = 1 . x + 2y + z = −2

Use Cramer’s rule to ﬁnd the value of y. Solution: det(a , b, a ) 5 Cramer’s rule ﬁnds y = 1 3 = − det(A) 6

13. Let 1 −3 4 A = 1 −3 3 . 2 −2 2 Find the adjugate of A, and use it to compute A−1. Solution:

One should test the inverse, i.e., AB = I3 should hold for the result B.

14. Let H be the subspace of P3 made up of only those polynomials p(x) such that p(0) = 0,

H = {p(x) ∈ P3 | p(0) = 0}. Find a basis for H. Solution: A basis is {x, x, x3}.

15. If the n × n matrix A is invertible, show that

n −1 (a) For any b in R , x = A b is a solution for the matrix equation Ax = b. (b) The solution x = A−1b is unique.

Solution: −1 −1 For a) put we check A(A b) = (AA )b = Inb = b. b) Let c be any solution. Then Ac = b implies by left multiplication with A−1 that

A−1Ac = A−1b

Observing that the l.h.s. equals b the desired claim follows.

16. Let Mn×n be the vector space of all n × n matrices, and let C ∈ Mn×n be a ﬁxed n × n matrix. Let H be the set of all n × n matrices that commute with A. Is H a subspace of Mn×n? Justify your answer. Solution: If XA = AX and YA = AY then (X +Y )A = XA+YA = AX +AY = A(X+Y ), so with X and Y also X+Y belongs to the subspace. Similarly (cX)A = c(XA) = c(AX) = (cA)X = (Ac)X = A(cX) shows that cX belongs there. Finally, IMPORTANT, since In is in that subset, it is not the empty set, hence we have a subspace.

n 17. Let A be an m × n matrix. Prove that the solution set of Ax = 0 is a subspace of R . Solution: Ax = 0 and Ay = 0 implies that A(x + y) = Ax + Ay = 0 + 0 = 0, so x + y belongs to the subset H. Similarly A(cx) = (Ac)x = (cA)x = c(Ax) = c0 = 0 shows that cx is in H. That H is not the empty set follows from 0 ∈ H.

18. Let V be a vector space, with zero vector 0 ∈ V . Using the vector space axioms to justify each step (see the attached sheet), prove that c · 0 = 0 for any scalar c ∈ R. Solution: c0 = c(0 + 0) = c0 + c0. Add on both sides the inverse −(c0) then

0 = (c0 + c0) + (−(c0)) = c0 + (c0 + (−(c0)) = c0 + 0 = c0

The ends of this chain of equations yields the desired equality.

19. Let T : V → W be a linear transformation from a vector space V into a vector space W . Prove that the range of T is a subspace of W . Solution: Let w and w0 be elements in the range of T . Then there are v and v0 in V such that w = T (v) and w0 = T (v0). Therefore

w + w0 = T(v) + T(v0) = T(v + v0)

shows that w + w0 is in the range of T since it is the image of v + v0 under T . Similarly cw = cT(w) = T(cv) implies that cw is in the range. Finally, since 0 = T(0) belongs to the range, the latter is not an empty set.

m 20. The column space of an m × m matrix A is all of R if and only if the equation Ax = b has a m solution for each b in R . Solution: The column space is the set of linear combinations of the columns of m A. Thus the columns span R iﬀ every b is such a linear combination. m X However, it is a linear combination b = ciai iﬀ Ac = b where c is i=1 the column vector with entries ci. 21. Indicate whether the statement is always true or sometimes false. Provide a complete detailed proof for statements which are always true, or a counterexample for statements which can be false.

(a) If A and B are invertible n × n matrices then A(B−1) is also invertible. (b) Let A and B be invertible n × n matrices. i.( AB)T is also invertible. ii.(( AB)T )−1 = (A−1)T (B−1)T . (c) Let A and B be n × n matrices. i. If AB is invertible, then A is invertible. ii. If AB = 0, then A = 0 or B = 0. (d) If A and B are n × n matrices, det(A) + det(B) = det(A + B). (e) Given a 3 × 3 matrix with columns A = a1 a2 a3 , then det(A) = det a2 a3 a1 . Solution: a) The matrix A is invertible iﬀ det(A) 6= 0. Since det(AB−1) = det(A) det(A) det(B−1) = 6= 0 the claim holds. det(B) b) (i) det((AB)T )) = det(AB) = det(A) det(B) 6= 0. Hence (AB)T is invertible. (ii) It suﬃces to show that

T −1 T −1 T ((AB) )(A ) (B ) = In

This follows from observing ﬁrst

(A−1)T (B−1)T = (B−1A−1)T

and then plugging this expression into the l.h.s. of the to be proven equality:

T −1 −1 T −1 −1 T T (AB) (B A ) = ((AB)((B A ))) = In = In

as claimed.

c) Let C denote the inverse of AB. Then In = (AB)C = A(BC) and since the matrices are quadratic, we found that A−1 = BC. d) False. Take two by two matrices with 0s everywhere except one 0 in the main diagonal. Their sum is I2 with determinant 1 while each of the summands has determinant 0. e) This fact is true. Proof: Since det(AT ) = det(A) one may use column operations. Then

det a1 a2 a3 = − det a2 a1 a3 = det a2 a3 a1

My suggestion is to put 4 pts for correct answer and 4 pts for the proof.

END OF EXAM