Linear Algebra 1 Exam 1 Solutions 6/12/3

Transcription

1 Linear Algebra 1 Exam 1 Solutions 6/12/3 Question 1 Consider the linear system in the variables (x, y, z, t, u), given by the following matrix, in echelon form: Reduce the system to reduced echelon form and give the general solution. We reduce as follows: Here we did the row operations: R 1 R 1 2R 2, R 2 R 2 3R 3 and R 1 R 1 + 3R 3. Written out, the reduced system is as follows: x 3z + 9u = 3, y + z 7u = 5 and t + 2u = 3. Put z = p and u = q. Then x = 3z 9u + 3 = 3p 9q + 3, y = z + 7u 5 = p + 7q 5 and t = 3 2u = 3 2q. So, with p and q arbitrary real numbers, the complete solution is: [x, y, z, t, u] = [3p 9q + 3, p + 7q 5, p, 3 2q, q] = [3, 5, 0, 3, 0] + p[3, 1, 1, 0, 0] + q[ 9, 7, 0, 2, 1].

2 Write your answer in the form of the sum of a particular solution and the general solution of the associated homogeneous system. Reading off from our general solution, we see that a particular solution is: [x, y, z, t, u] = [3, 5, 0, 3, 0]. Also the general solution of the associated homogeneous system is: [x, y, z, t, u] = p[3, 1, 1, 0, 0] + q[ 9, 7, 0, 2, 1], where p and q are arbitrary scalars. How many free variables are there? There are two free variables, the variables z and u. Which are the pivot variables? There are three pivot variables, the variables x, y and t. Is there a solution with y = z = t = 1 and if so is what is the solution and is it unique? We have shown that the general solution is: [x, y, z, t, u] = [3p 9q + 3, p + 7q 5, p, 3 2q, q]. In particular, we have: y = p + 7q 5, z = p and t = 3 2q. If now z = 1, then we must have p = 1. If also t = 1, then we must have 3 2q = 1, so q = 1. If p = q = 1, then the solution is [x, y, z, t, u] = [ 3, 1, 1, 1, 1]. This has y = z = t = 1. So the required solution exists and is the unique solution: [x, y, z, t, u] = [ 3, 1, 1, 1, 1].

3 Question 2 Consider the following linear system: x 3y + 2z = 8 3x 8y 5z = 11 2x 4y 18z = 10 Find the general solution of the system. We reduce the augmented matrix for this system as follows: Here we did the row operations: R 2 R 2 3R 1, R 3 R 3 2R 1, R 3 R 3 2R 2 and R 1 R 1 + 3R 2. The system is now in reduced echelon form. x and y are the pivot variables and the variable z is the sole free variable. The reduced equations are x 31z = 31 and y 11z = 13, so, putting z = s, we get: x = 31z 31 = 31s 31 and y = 11z 13 = 11s 13. So the complete solution is [x, y, z] = [ 31, 13, 0] + s[31, 11, 1], with s an arbitrary real number. Give a geometric interpretation of the system and your solution. Geometrically each of the given equations represents a plane. The solutions to the system represent the intersection of the three planes and form a line in three dimensions. So the three planes share a common line: the line through the point [ 31, 13, 0] with direction vector [31, 11, 1]. Is there a solution of the system with x = 0? If so, find it. We have shown that the complete solution is [x, y, z] = [ 31, 13, 0] + s[31, 11, 1], with s an arbitrary real number. If now x = 0, then we have 31s 31 = 0, so s = 1. Then z = 1 and y = = 2. So there is a unique solution with x = 0, namely [x, y, z] = [0, 2, 1].

4 Question 3 Everyplace.com has three levels of employee, levels A, B and C. Last year level A employees each received 10, 000 stock options, level B employees each recieved 5, 000 stock options and level C employees 2, 500 stock options. Bonuses for a record year were paid out at $20, 000 for levels A and B and $10, 000 for level C. Base salaries were $120, 000 for level A, $80, 000 for level B and $50, 000 for level C. Last year a total of 300, 000 stock options were given out, total bonuses of $1, 000, 000 and total base salaries of $5, 000, 000. How many employees does Everyplace.com have? The linear system for the numbers A, B and C of each level of employee has the following augmented matrix: 10, 000 5, 000 2, , , , , 000 1, 000, , , , 000 5, 000, 000 We divide the first row by 2, 500 and other two rows each by 10, 000 and reduce: Here we did the row operations: R 1 R 1 2R 2, R 3 R 3 6R 1, R 3 R 3 2R 1, R 2 R 2 + R 1,R R 2, R 1 R 1 + R 3, R R 1 and finally R 1 R 2. From the reduced matrix, we read off that there are 10 employees at levels A and B and 60 at level C, so there are a total of 80 employees at Everyplace.com.

5 Question 4 Give an example of each of the following or explain why no such example can exist: An inconsistent linear system in three variables, with a coefficient matrix of rank two. Here we need three equations not all proportional, that are incompatible: so for example the following system will do: x + y + z = 4 y + z = 5 x + 2y + 2z = 10. This system is inconsistent, since the sum of the first two equations gives x + 2y + 2z = 9, which is incompatible with the third equation. The coefficient matrix is not rank zero, since it is not zero. It is not rank one, since otherwise all the left-hand sides of the equations would be proportional. It is not rank three, since otherwise the coefficient matrix would be invertible and the system would have a solution. Therefore it is rank two, as required. A consistent linear system with three equations and two unknowns, with a coefficient matrix of rank one. The following system will do: x + y = 4 2x + 2y = 8 3x + 3y = 12. The equations are all proportional, so the coefficient matrix is rank one. The system is consistent, since x = y = 2 solves it.

6 A consistent linear system with three equations and two unknowns, with a coefficient matrix of rank larger than one. The following system will do: x + y = 4 x + 2y = 5 x + 3y = 6. The coefficient matrix is not rank one, since the left-hand sides of the equations are not proportional and is clearly not rank zero either. So the rank is larger than one, as required. In fact we know that the rank r is always less than or equal to the number of unknowns (proved in class), so 0 r 2, so, since r 0 and r 1, we have r = 2. Also the system is consistent with a solution (x, y) = (3, 1). A linear system of two equations in three unknowns, with an invertible coefficient matrix. This is impossible: the coefficient matrix is 2 by 3, so is not square. But we proved in class that the only invertible matrices are square. Alternatively, the homogeneous system with the given coefficient matrix has rank at most two, so at least one degree of freedom in its solution, whereas if the coefficient matrix were invertible, there would be a unique such solution. A linear system in three variables, whose geometrical interpretation is three planes intersecting in a line. The system of question two above was shown to have the required property: x 3y + 2z = 8 3x 8y 5z = 11 2x 4y 18z = 10 A simpler version would be the following: x = 0, y = 0, x + y = 0. Each of these equations represents a plane through the z-axis, which is their common intersection.

7 Question 5 Let A be the following matrix: Compute the matrices A 2, AA T and A 1. By matrix multiplication, we get: A 2 = AA T = = = For A 1, we first compute the determinant of A: det(a) = 1( 8) 3( 2) = = 2. Since det(a) 0, we can use a known formula for the inverse of A: we interchange the diagonal elements, change the sign of the off diagonal elements and divide by det(a): A 1 = ( 1 2 ) = We may check by matrix multiplication that AA 1 = A 1 A = I, as required. Find numbers p and q, such that A 2 = pa + qi, where I is the 2 2 identity matrix. We want A 2 = pa+qi, or A 2 pa qi = 0, so we write out A 2 pa qi: A 2 pa qi = p 3p 2p 8p q 0 0 q = 5 p q 21 3p p p q We want this to be the zero matrix, giving in particular the equations p = 0 and 0 = 5 p q, so p = 7 and q = 5 p = 2. With p = 7 and q = 2, we check that every entry is zero, so we have (p, q) = ( 7, 2) as the unique solution. So we have A 2 = 7A + 2I.

8 Write A and A T as a product of elementary matrices. We reduce A to the identity matrix as follows: Here we did the row operations E 1 : R 2 R 2 + 2R 1, E 2 : R R 2 and E 3 := R 1 R 1 3R 2. The corresponding elementary matrices and their inverses, and transposes, which are also elementary matrices, are as follows: E 1 = , E 1 1 = , ET 1 = , (E 1 1 ) T = , E 2 = E2 T = , E2 1 = (E2 1 ) T = , 2 E 3 = , E 1 3 = , ET 3 = , (E 1 3 ) T = Then we have: I = E 3 E 2 E 1 A, so A 1 = E 3 E 2 E 1 gives an expression for A 1 as a product of elementary matrices. Taking the inverse, we find: A = (A 1 ) 1 = E1 1 E2 1 E3 1 giving an expression for A as a product of elementary matrices. Finally, taking the transpose, we find: A T = (E3 1 ) T (E2 1 ) T (E1 1 ) T giving an expression for A T as a product of elementary matrices. Let B = A ti, where t is a scalar. For which values of t is B not invertible? We first compute det(b): det(b) = det(a ti) = det 1 t t = (1 t)( 8 t) (3)( 2) = t2 +7t 2. Then B is not invertible if and only if det(b) = 0, which gives the quadratic equation t 2 + 7t 2 = 0, with the solutions: t = 7 ± 49 4(1)( 2) 2 = 7 ± So for t = 1 2 ( 7 ± 57) the matrix B is not invertible and for all other t-values, B is invertible.

9 Question 6 Calculate the following determinant (using suitable properties of the determinant to simplify the calculation): det x y z Also give the geometrical interpretation of the vanishing of the determinant. We have: = (x 1) det = det det = det x y z x 1 y + 2 z x 1 y + 2 z (y + 2) det (z 3) det = (x 1)( 8) + (y + 2)(6) (z 3)( 1) = 8x 8 + 6y z 3 = 8x + 6y + z Here we first did the three row operations R 1 R 1 R 2, R 3 R 3 R 2 and R 4 R 4 R 2, then did a Laplace expansion down the fourth column, finishing with the standard Laplace expansion for a three by three determinant. The vanishing of the determinant gives the equation of the plane in three dimensions, through the points [1, 2, 3], [2, 3, 1] and [4, 6, 3], since it is easily seen that the determinant vanishes at these points.

10 Question 7 Let A and B be the following matrices: A = B = By row reducing a suitable matrix, solve the equation AX = B. Is the matrix A invertible? Explain your answer. We reduce the following augmented matrix: [A B] = Here we did the row operations: R 2 R 2 3R 1, R 3 R 3 2R 1, R 1 R 1 + 2R 3, R 2 R 2 5R 3, R 3 R 3 R 2, R 3 R 3 and finally R 2 R 3. The reduced system is of the form [I C] and has the same solutions as the original system. But the new system is IX = C, so the unique solution to our original solution is X = C, the augmented part of our reduced system. So the required solution is: X = We may check by matrix multiplication that the equation AX = B holds. Since this solution is unique, the matrix A must be invertible, for if it were not invertible, the system would either be inconsistent or have an infinity of solutions.