Solutions to the first midterm exam

MIDTERM EXAM 1 SOLUTIONS Directions and rules. The exam will last 70 minutes; the last five minutes of class will be used for collecting the exams. No electronic devices of any kind will be allowed, with one exception: a music player that nobody else can hear, and whose controls you do not use during the exam (just put it on shuffle). Please turn your cell phone off. The exam is closed-book. Each of the ten numbered questions is worth one point. Correct answer earns the point. Wrong answer gets zero points. Check your work carefully. 1. Write the following system of equations in matrix form: x − 3y + z − 5w = 2 x+y−z+w = 1 x − y − z + 6w = 6

      x 2 1 −3 1 −5   1 1 −1 1   y  =  1   z  6 1 −1 −1 6 w Compare exercises 19 and 20, page 26. If you instead omitted the variable names and wrote a single (augmented) matrix, I gave credit, (although technically it’s not correct) because the phrase “matrix form” isn’t emphasized in the textbook and several examples are worked with augmented matrices. 2. Compute the matrix product below. You only have to write the answer.

2 −1 1 5

0 1 3 −1

=

1

−3 3 15 −4

2

MIDTERM EXAM 1 SOLUTIONS

3. Compute the matrix inverse. Just write the answer, do not show your work. −1 −2 3 −2 −3 = −1 −2 1 −2 4. Use Cramer’s rule to solve the following system of equations. You need to show two steps: the result of applying Cramer’s rule (first step) and the result of evaluating those expressions (final solution). 3x − 4y = 1 2x + 3y = 2

x =

y =

1 −4 2 3 11 = 17 3 −4 2 3 3 1 2 2 4 3 −4 = 17 2 3

5. Are the following three vectors linearly independent?       1 1 0  4  ,  5  ,  −3  −1 −3 −1 They are linearly dependent if some linear combination is zero. That is, if we can find a, b, c (not all zero) such that       0 1 1 a  4  + b  5  + c  −3  = 0. −1 −3 −1 That is, b+c = 0 4a + 5b − 3c = 0 −a − 3b − c = 0 Can we solve these equations for a, b, and c, not all zero? Suppose we have a solution. Then the first equation requires b = −c, so the second equation can be written 4a + 8b = 0, which means a = −2b; then the third


3

equation becomes 2b − 3b − (−b) = 0, or 0 · b = 0. This is satisfied for any b; for example, we can take b = 1; then a = −2 and c = −1 will solve the equations (as you can double-check). Since the equations can be solved, the three vectors are linearly dependent, and the answer to the question is “no.” 

     1 1 1 6. Determine if  1  ,  −1 , and  3  span R3 . 0 −1 1 We need to determine if an arbitrary vector (x, y, z) can be expressed as a linear combination of these three vectors. That means, given (x, y, z), can we solve the following equations for (a, b, c)?         1 x 1 1 a  1  + b  −1  + c  3  =  y  0 −1 1 z Those equations will be solvable for arbitrary (x, y, z) if and only if the coefficient matrix on the left has nonzero determinant. That determinant is 1 1 1 1 −1 3 = 0 0 −1 1 Hence the three vectors do not span R3 , so the answer to the question is “no.” An alternate solution that doesn’t use determinants: The equations are a+b+c = x a − b + 3c = y −b + c = c Suppose we have a solution of the equations. Subtracting the first from the second and dividing by 2 we have −b + c = (y − x)/2, but by the third equation that is z. So all we have to do is choose (x, y, z) so that z 6= (y − x)/2, for example, z = 1 and x = y = 0, and we will not be able to solve the equations. That is, (0, 0, 1) is an example of a vector that is not a linear combination of the three given vectors. 7. Are the following polynomials linearly independent? x3 − 1, x2 − 1, x − 1, 1 A linear combination of these polynomials has the form a(x3 − 1) + b(x2 − 1) + c(x − 1) + d = ax3 + bx2 + cx + (d − a − b − c) so it will be the zero polynomial if and only if the coefficients are all zero, i.e. a = b = c = 0 and d − a − b − c = 0, which implies d = 0 too. Hence the polynomials are linearly independent.

4


8. Do the following three vectors form a basis for R3 ? 

     2 1 1  −1  ,  3  ,  −4  0 −1 −1 The question asks two things: are these vectors linearly independent, and do they span R3 ? To see if they are linearly independent, suppose       2 1 1 a  −1  + b  3  + c  −4  = 0 0 −1 −1 That is, 2a + b + c = 0 −a + 3b − 4c = 0 −b − c = 0 The last equation implies b + c = 0, so the first equation implies a = 0, and the second equation then implies (because c = −b) that 7b = 0, so b = 0, so c = 0. Hence the three vectors are linearly independent. Now, do they span R3 ? To show that, we have to show that for any (x, y, z), the equations 2a + b + c = x −a + 3b − 4c = y −b − c = z are solvable. Writing this  2  −1 0

in matrix form we have     1 1 a x     3 −4 b y  = −1 −1 c z

This is solvable for every (x, y, z), because the matrix on the left has nonzero determinant, and hence has an inverse. So the answer to the question is “yes”, the given vectors span R3 . 9. Compute the following determinant. Do not show your work, write only the answer. 1 0 1 1 1 1 =1 1 1 2


5

10. Use Cramer’s rule to solve the following system for x, y, and z: x+y−z = 2 2x − y + z = 3 x − 2y + z = 1 This is Example 1 from page 55 of the textbook. The answer is (x, y, z) = (5/3, 1/3, 0).