Solution. Area(OABC) = Area(OAB) + Area(OBC) = 1 2 det( [ Question 2. Let A = (a) Calculate the nullspace of the matrix A.


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1 Solutions to Math 30 Takehome prelim Question. Find the area of the quadrilateral OABC on the figure below, coordinates given in brackets. [See pp of the book.] y C(, 4) B(, ) A(5, ) O x Area(OABC) = Area(OAB) + Area(OBC) = det( [ 5 ] [ ] ) + det( ) = = 7. 4 Question. Let A = (a) Calculate the nullspace of the matrix A. (b) Let B = A T. Find the rank of B. (c) Find a basis for the column space of B. [ 3 ] (a) The nullspace of A is the set of solutions x = GaussJordan reduction to solve it. The reduced row echelon form of A is [ ] x x x 3 x 4 to the linear system Ax = 0. Use Setting x = r, x 4 = s, we have x = r + 3s, x 3 = s, or equivalently r + 3s 3 x = r s = 0 r + 0 s, where r and s are free. s 0
2 (b) rank B = rank A T = rank A = since the rows of A are linearly independent. (c) Since the columns of matrix B = are linearly independent, they form a basis of it s column space. Question 3. Let 3 A = 4 3 (a) Find the reduced row echelon form of A. (b) Do the rows of A span R 3? Explain your answer. (c) Do the columns of A span R 3? Explain your answer. (d) Your friend Bob claims that there exist bases S = {v, v, v 3 } and T = {w, w, w 3 } of R 3 such that [x] S = A[x] T for all x in R 3. Explain why this cannot possibly be true. (a) The reduced row echelon form is R = (b) The row space of A equals the row space of R, which is dimensional. Hence rows of A do not span R 3. (c) No, because the dimension of the column space of A is equal to that of the row space, which is. (d) Suppose, this is true. Since rank A = < 3, A is singular. Therefore there is a nonzero 3vector u = u u u 3 such that Au = 0. Let x = u w + u w + u 3 w 3. Then x 0 (because T is linearly independent and u 0) and [x] T = u. We have [x] S = A[x] T = Au = 0, which means x = 0. Contradiction. Question 4. Let A be an n n matrix with integer entries. (a) If det(a) =, show that A has integer entries. (b) Suppose A has integer entries. What are the possibilities for det(a)? Explain.
3 (a) By the formula of the inverse matrix, (i, j) entry of A A is ji. If det(a) =, this det(a) entry equals A ji, which is an integer. Indeed, up to sign, the cofactor A ji is the determinant of a submatrix of A, and the determinant of a matrix with integer entries is an integer. (b) det(a) det(a ) = det(aa ) = det(i n ) =. So, the product of integers det(a) and det(a ) equals, which implies det(a) = ±. These possibilities are actually realized, e. g. A = I n (det(a) = ) or the diagonal matrix with entries,,,..., on the diagonal (det(a) = ). Question 5. Find out whether the matrices [ ] [ ] 4 u =, u 3 4 =, u 3 3 = form a basis in the space of all matrices. [ ] 3 4, u 4 = [ ] 3 4 Check linear independence. The equation x u + x u + x 3 u 3 + x 4 u 4 = 0 is equivalent to a homogeneous linear system Ax = 0, where 4 3 A = Zero out below (, ) entry using elementary row transformations of the third type: 4 3 B = We have det(a) = det(b). By the cofactor expansion in the first column of B, we have det(b) = B = Then A is invertible and the only solution is x = x = x 3 = x 4 = 0. Thus, matrices u, u, u 3, u 4 are linearly independent. Since their number (four) equals the dimension of the space of matrices, they form a basis in that space. Question 6. Find all vectors in R 3 of length with integer entries. Which of them are orthogonal to the vector v =? x Denote the set of integer vectors of length by S. The vector y belongs to S if and z only if x + y + z = 4. 3
4 If x, y, z are integers, their absolute values can be equal to 0, or (otherwise the length will be bigger than ). All the vectors in S satisfying 0 x y z are , 0,,, 0. 0 The rest of S is obtained from these by permutating entries and multiplying some of them by. [Because any integer vector in S can be transformed to one with 0 x y z by such operations.] In total, S has 33 elements. Orthogonality to v means x + y + z = 0. All vectors in S satisfying to this relation are 0 0,,, 0 0 0,. Question 7. The population of sapsuckers in Sapsucker Woods is described by the following model. Let c k denote the number of chicks in year k, let j k denote the number of juveniles in year k, and let a k denote the number of adults in year k. Then c k j k+ = a k c k j k a k Let A be the matrix A = (a) A vector v in R 3 is called a steadystate vector of A if Av = v. Explain what this means in terms of the model. (b) Find all steadystate vectors for A. (c) After heavy logging in Sapsucker woods, biologists find that the model is no longer accurate. Instead, a more suitable model is c k c k j k+ = j k a k a k Under this new model, what do you think will happen to the population of sapsuckers in the long term? Explain your answer. (a) This means that the population is in balance: c k = c, j k = j, a k = a are constants (do not depend on k). 4
5 (b) The equation Av = v is equivalent to the linear system (I 3 A)v = 0. The reduced row echelon form of I 3 A = is c Setting a = 5r, we have v = j = r, where r is free. a 5 (c) Under the new model, the population will be extinct. Indeed, set t k = c k + j k + a k (the total of all species in year k). Then 0 t k+ = c k+ + j k+ + a k+ = 0.a k + 0.5c k + 0.5j k 0.5(a k + c k + j k ) = 0.5t k. Thus 0 t k+ 0.5t k 0.5 t k 0.5 k t. Since the limit of 0.5 k t as k, is 0, so is the limit of t k. Question 8. Let A = Calculate det(a).. Is A invertible? Explain your answer. 3. Calculate det(aa T ). (a) = 0 0 = Here, the first identity is obtained by subtracting row 3 from row, the second identity follows from the fact that the determinant of a matrix having two equal rows is 0. (b) No, A is singular since det(a) = 0. (c) det(aa T ) = det(a) det(a T ) = 0. 5
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