MATH 2030: ASSIGNMENT 2

Intro to Linear Systems Q.1: pg 69, q 2,4,6. Determine which equations are linear equations in the variables x, y and z, if not, explain why. (1) x2 + y 2 + z 2 = 1 (2) 2x − xy − 5z = 0 √ (3) (cos3)x − 4y + z = z A.1. • This is the equation for a sphere in R3 , due to the powers of two on x, y and z this cannot be linear. • This is almost a linear equation, however as the coordinates x and y are multiplied together in the second term this cannot be linear. • Again, this equation is almost linear, however the square root of the z coordinate immediately invalidates it. Q.2: pg 69, q 28. Find the augmented matrix of the linear system: 2x1 + 3x2 − x3 = 1 x1 + x3 = 0 −x1 + 2x2 − 2x3 = 0.

A.2. The augmented matrix will be   2 3 −1 1  1 0 1 0 . −1 2 −2 0 Q.3: pg 69, pg 32. Find a system of linear matrix  1 −1 0 3 1  1 1 2 1 −1 0 1 0 2 3

equations for the given augmented  2 4 . 0

A.3. Defining the variables x1 , x2 , x3 , x4 , x5 the system of linear equations is then x1 − x2 + 3x4 + x5 = 2, x1 + x2 + 2x3 + x4 − x5 = 4 x2 + 2x4 + 3x5 = 0. 1

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MATH 2030: ASSIGNMENT 2

Direct Methods to Solving Linear Systems of Equations Q.4: pg 85-86, q 14,38. Use elementary tow operations to reduce the given matrix to a) row echelon form and b) reduced row echelon form. •  −2 3 1

6 −9 −3

 −7 10  3

• 

1  6 7

2 5 7

3 4 7

4 3 7

5 2 7

 6 1 . 7

A.4. • Applying the row operations, R1 ↔ R3 , R2 − 3R1 , R3 + 2R1 , R3 − 13R2 gives the matrix in row echelon form   1 −3 3 0 0 1 . 0 0 0 Applying another row operation R3 − 3R2 yields the matrix in reduced row echelon form   1 −3 0 0 0 1 . 0 0 0 • The row operations R3 − R1 − R2 , R2 − 6R1 and − 71 R2 produces the row echelon matrix   1 2 3 4 5 6 0 1 2 3 4 5 . 0 0 0 0 0 0 One last row operation, R1 − 2R2 yields the reduced row echelon form   1 0 −1 −2 −3 −4 0 1 2 3 4 5 . 0 0 0 0 0 0       1 1 2 Q.5: pg 87, q 50. Let p = 2 , u =  1 , and v = 1; describe all points 3 −1 0 Q = (a, b, c) such that the line through Q with direction vector v intersects the line with equations x = p + su. A.5. Writing the vector equation for the line in terms of the vector in standard position with its head at Q and the direction vector. q, x0 = q + tv. The two lines intersect if x = x0 , equating the two vector equations we may simplify the resulting vector equation su − tv = q − p

MATH 2030: ASSIGNMENT 2

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working with components we may determine the system of linear equations in terms of s and t with augmented matrix   1 −2 1 − a  1 −1 b − 2  −1 0 c − 3 The reduced row echelon augmented matrix is found to be     2b − a − 3 1 0 2b − a − 3 1 0 = 0 1   0 1 b−a−1 b−a−1 0 0 c − a + 2b − 3 0 0 c − a + 2b − 6 In order for this system of equations to be consistent the last row must be the zero row vector, thus the coordinates for Q are constrained to a plane with general equation c + 2b − a = 6

Q.6: pg 88, q 56,60. Solve the system of linear equations over the indicated Zp : • 3x + 2y = 1, x + 4y = 1 in Z5 . • 2x + 3y = 4, 4x + 3y = 2 in Z6 . In the latter case p is not prime, one should be careful with solving the linear system; state the complications arising in Z6 if we solve this linear equation with Gaussian elimination. A.6. • The augmented matrix is 

3 2 1 4

1 1

 .

Applying the row operations R2 + 3R1 , 2R1 , 4R2 ,we produce the reduced row echelon matrix   1 4 2 . 0 0 1 • The augmented matrix is 

2 3 4 3

4 2

 .

Applying the row operations R2 + R1 ,then we produce the reduced row echelon matrix   2 3 4 . 0 0 0 The solution to the equation 2x+3y = 2 mod 6 has has solutions: (2, 0), (2, 2), (2, 4), (5, 0), (5, 2), (5, 4) this may be seen by multiplying the above equation by 2 and 3 respectively to produce 4x = 2 mod6 and 3x = 0 mod 6.

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MATH 2030: ASSIGNMENT 2

Spanning Sets and Linear Independence   3 Q.7: pg 103 q 4. Determine if v =  1  is a linear combination of −2     1 0 u1 = 1 , u2 = 1 . 0 1

A.7. Form the augmented matrix with coefficient matrix u1 and u2 as column vectors and v as the constant vector   1 0 3  1 1 1 . 0 1 −2 We apply the row operations R1 − R2 and R3 − R2 to produced the reduced row echelon matrix   1 0 3  0 1 −2  . 0 0 0 Thus we see that v = 3u1 − 2u2 . Q.8: pg 104, q 26,30. In both cases, state whether the sets of vectors are linearly independent or dependent, by direct calculation using an appropriate theorem or by inspection. If dependent, find the dependence relation between the vectors. • 

       2 −5 4 3 −3 ,  1  , 3 , 1 7 1 0 5 •   0 0  , 0 1

  0 0  , 2 1

  0 3  , 2 1

  4 3   2 1

A.8. • Applying Theorem 0.26 from the May 17th notes, we see that there are m = 4 vectors and n = 3 dimensions, hence one of these must be a linear combination of the others. By forming the augmented matrix with these vectors and the zero vector, and simplifying to reduced row echelon form we find the constants to be   100 0 1 0 0 151  0 1 0 55 0  . 151 132 0 0 1 151 0

MATH 2030: ASSIGNMENT 2

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• Combining the vectors into an augmented matrix as before, we note that a reversal of the order of the rows in this matrix produces a new matrix in upper-triangular form. By dividing the second row of this new matrix by 2 and the third row by 3, we find a matrix in row echelon form.   1 0 0 0 0  0 1 0 0 0     0 0 1 0 0 . 0 0 0 0 1 Q.9 pg 105, q 44. In R2 , prove that two vectors are linearly dependent if and only if one is a scalar multiple of the other (Consider the case where one vector is the 0 vector separately.). A.9. Given u and v, we assume they are linearly dependent so that there exists c1 and c2 - not both zero - such that c1 u + c2 v = 0 If u = 0, the zero vector, we may set c1 6= 0 and c2 = 0, this proves v is a scalar multiple of 0. Assuming u, v 6= 0, c1 and c2 must be non-zero. Without loss of generality we may divide by c1 to find c2 u=− v c1 implying that u is a scalar multiple of v. To show the other direction, suppose u = cv, then c1 = 1 and c2 = −c proves the two vectors are linearly dependent. Application: Allocation of Resources Q.10: pg 120, q 4. (1) In your pocket you have some nickels, dimes and quarters. There are 20 coins altogether and exactly twice as many dimes as nickels. The total value of the coins is $3.00. Find the number of coins of each type. (2) Find all possible combinations of 20 coins (nickels, dimes, and quarters) that will make exactly $3.00. A.10. (1) Denoting x as the number of nickels, y as the number of dimes and z as the number of quarters, we have the following equations x + y + z = 20, y = 2x, 5x + 10y + 25z = 300, with augmented matrix   1 1 1 20  −2 1 0 0  5 10 25 300 the equivalent reduced row  1  0 0

echelon matrix is then  0 0 4 1 0 8 , 0 1 8

Thus we need 4 nickels, 8 dimes and 8 quarters.

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MATH 2030: ASSIGNMENT 2

(2) Dropping the constraint y − 2z = 0, we have a new augmented matrix   1 1 1 20 5 10 25 300 This has the following reduced row echelon matrix   1 0 −3 20 0 1 4 40 As there is no such thing as negative number of change in one’s pocket x, y and z must be non-negative. This gives the following two constraints on z 3z ≥ 20, 40 ≥ 4z Solving these two inequalities, we find 6 < z ≤ 10. Thus the possible combinations of change for this are defined as x = 3z − 20, y = 40 − 4z, z ∈ [7, 10] References [1] D. Poole, Linear Algebra: A modern introduction - 3rd Edition, Brooks/Cole (2012).