of Department Economics Ryerson University ECN230-Mathematics For Economics Problem Set #6 - Solutions
EXERCISE 5.1 Problem 5: (a) Interchange row 2 and row 3 in A to get a matrix A1 . In A1 keep row 1 as is, but add row 1 to row 2, to get a matrix A2 . In A2 , divide row 2 by 5. Then multiply the new row 2 by -3, and ass the result to row 3. 5 Finally, multiply row 3 by 42 . The resulting echelon matrix: 1 5 1 A3 = 0 1 1 5 0 0 1
contains three nonzero rows. Hence, r (A) = 3. (b) Interchange row 1 and row 3 in B to get a matrix B1 . In B1 , divide row 1 by 6. Then multiply the new row 1 by -3, and add the result to row 2, to get B2 . In B2 , multiply row 2 by 2, then add the new row 2 to row 3. The resulting echelon matrix: 1 1 0 6 B3 = 0 1 4 0 0 0 has two nonzero rows. Hence, r (B) = 2. There is linear dependence in B: 0 0 0 v1 = v3 - 2v2 . Matrix B is singular. (c) Interchange row 2 and row 3 of matrix C, to get matrix C1 . In C1 , divide row 1 by 7. Then, multiply the new row 1 by -8, and add the result to row 2, to get C2 . In C2 , multiply row 2 by -7/48. Then, multiply the 1
contains two nonzero rows. Hence r (D) = 2. Again, the question of nonsingularity is not relevant here because matrix D is not square.
There are three nonzero rows in C4 . Hence, r (C4 ) = 3. The question of singularity is not relevant here because matrix C is not square. (d) Interchange row 1 and row 2 in matrix D to get matrix D1 . In D1 , multiply row 1 by -2 and add the result to row 2 to get matrix D2 . Since the last two rows of D2 are identical, linear dependence is obvious. To produce the echelon matrix, divide row 2 in D2 by 5 and then add (-5) times the new row 2 to row 3. The resulting echelon matrix: 1 1 0 1 D3 = 0 1 9 - 3 5 5 0 0 0 0
new row 2 by -1 and add the result to row 3 to get C3 . In C3 , multiply row 3 by 2/3 to get the echelon 3 matrix: 1 6 3 7 7 7 C4 = 0 1 1 - 2 . 2 3 0 0 1 10 9
Problem 6: The link is provided by the third elementary row operation. If, for instance, row 1 of a given matrix is equal to row 2 minus k times row 3 (showing a specific pattern of linear combination), then by adding (-1) times row 2 and k times row 3 to row 1, we can produce a zero row. This process involves the third elementary row operation. The usefulness of the echelon matrix transformation lies in its systematic approach to force out zero-rows if they exist. EXERCISE 5.3 Problem 5: In Part (d), the rank is 3. In Parts (a), (b) and (c), the rank is less than 3. EXERCISE 6.2 Problem 1: (a) 4 (x + x)2 + 9 - 4x2 + 9 y = = 8x + 4x x x 2
(b) dy = f 0 (x) = 8x dx (c) f 0 (3) = 24 and f 0 (4) = 32 Problem 2: (a) y = 10x + 5x - 4 x (b) dy = f 0 (x) = 10x - 4 dx (c) f 0 (2) = 16 and f 0 (3) = 26 Problem 3: (a) y =5 x This is a constant function. (b) It does not make any difference whether x is large or small. The limit of the difference quotient as x approaches zero is dy/dx = 5. EXERCISE 6.4 Problem 1: Left-side limit = right-side limit = 15. Function q has a limit as v approaches 7. The limit is 15. Problem 2: The function q can be rewritten as: i h 3 (v + 2)3 - 8 v + 6v 2 + 12v + 8 - 8 = = q = v v v3 + 6v 2 + 12v = v2 + 6v + 12. = v 3
(a) lim q = 12.
v0
(b) lim q = 28. (c) lim q = a2 + 6a + 12.
va v2
Problem 3: (a) lim q = 5.
v+
(b) lim q = 5.
v-
EXERCISE 6.6 Problem 1: (a)
v0
lim q = 7 - 0 + 0 = 7
(b)
v3
lim q = 7 - 27 + 9 = -11
(c)
v-1
lim q = 7 + 9 + 1 = 17
Problem 2: (a)
v-1
lim q = lim (v + 2) lim (v - 3) = 1 (-4) = -4
v-1 v-1
(b)
v0
lim q = 2 (-3) = -6
(c)
v5
lim q = 7 (2) = 14
4
Problem 3: (a) lim q =
v0
lim (3v + 5)
v0
v0
lim (v + 2)
=
5 1 =2 2 2
(b) lim q = 20 6 (15 + 5) = =2 (5 + 2) 7 7
v5
(c) lim q = (-3 + 5) =2 (-1 + 2)
v-1
5