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**PRACTICE** MiprERM 1 SOLUTIONS MAT223 - FALL 2021


1 (5 points)

Find all complex number solutions z to the equation92 z5 +i = 一1. Express your solutions in polar form. Also express at least one of the solutions in a +bi form.

First, we put this in zn = a bi form, as z5 = 1/92 i/92.

Now, z = reiθ and 1/92 i/92 = e5π/4i

Thus, r5 = 1 ÷ r = 1.

And, 5θ = 5π/4 +2πk ÷ θ = π/4 +2kπ/5 = (5 +8k)π/20.

Naturally, k = 0, 1, 2, 3, 4 are the values that give unique solutions, and plugging them in successively, we reach our answers:

z = e5π/20 = eπ/4 = 1/92 +i/92, e13π/20, e21π/20, e29π/20, e37π/20 .


2

(5 points)

Let A = '

'

1

2

0

c2

2c2

2

c

8(1) ''. For what values of c (if any) does the matrix A have rank 2? Justify your

answer and show all steps in your work.

┌  1    c2          c      ┐

' 0     0     1 2c  '.

From here, we can see that there will be a row of zeroes if 1 一 2c = 0 is in the bottom row. With that 0, you can confirm by row reducing a little more (or just by pointing out that there will be two leading ones in the other rows) that it has rank 2 (so c = 1/2). If c 1/2, again a little more row

reducing or a short argument show that the rank will be 3.

So our final answer is just c = 1/2.

(4 points)

If possible, find a 3 × 3 matrix M which has the following properties:

• The system Mx = 0 has exactly one basic solution;

•  M is not in RREF form.

Justify why your matrix has these properties; or, if it is not possible to construct such a matrix, explain why.


Lots of solutions exist here. To make one quickly, make something upper triangular (say), so that

┌  1    1    1  ┐

'                ',

┌  1    0    0  ┐

which has RREF '', and clearly has rank 2 (so one basic solution for Ax = 0).


4

(9 points total)

Determine if the statements below are true or false.

Make sure to justify your answers! You will receive no credit for simply selecting “true" or “false", or providing little explanation.

4.1 (3 points) True or False:

If A is a 3 × 3 skew-symmetric matrix (i.e. At = A), then the system Ax = 0 has non-trivial solutions.

4.2 (3 points) True or False:

If A is invertible, and A3 = A, then Af 1 = A.

4.3 (3 points) True or False:

For any square matrix A, if R is the reduced row echelon form (RREF) of A, and det(A= det(R, then A can be row reduced to R with only Type 3 row operations (adding a multiple of one row to another.)

4.1   True.   det(A)  =  det(At)  =  det(一A)  =  (一1)3 det(A)  = 一det(A), so det(A)  = 一det(A)  and thus det(A)= 0.

4.2   True.  Notice that A3  = A implies that A2AAf 1  = AAf 1  = In  so A2  = In  and thus AA = In  so AAAf 1 = Af 1 and finally A = Af 1 . (Note that we cancelled AAf 1 in a couple steps there without comment.)

┌ 2    0    0  ┐

0    0    0                has det(0) as does R, but it needs to have row1 divided by 2 (it’s clear enough without further justification that Type 3 row operations are useless here, but you could mention it.)

(3 points)

Read the following definition carefully, and then answer the question below.

True or False:

For any rtoeqinueqsible matrix A, det(A)= 士1.

2                                                                     Midterm 1

This is true.

Notice that AAt = I, by superinvertibility, so det(A)2 = 1 ÷ det(A= 91 = 士1.

Read the following carefully, then answer the question that follows.