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Quiz 2: Linear Algebra and Differential Equations (Summer 2022)

Question 1

You must take this quiz completely alone. Showing it to or discussing it with anyone else is forbidden. Reproducing or sharing this material on any other websites or platforms is also forbidden. You are permitted to consult your notes, the textbook, any materials handed out in class, and any materials on the bCourses site. You may not consult any other resources (for instance public websites, Wolfram Alpha, and calculators).

I have read and understood these instructions. I agree that I will follow both the instructions and the Berkeley Honor Code.

I will not follow these instructions or the Berkeley Honor Code.

Question 2

True or false: It is impossible for a list of 5 vectors in to be linearly independent.

True

False

If you form the associated matrix, it can't possibly have a pivot in every column because there aren't enough rows!

Question 3

True or false: A linear transformation can be injective.

True

False

You can reason this in two ways: First, the transformation has domain and codomain , so the domain is "bigger" than the codomain. This means that the transformation can not be injective.

Second, we can find a matrix such that . Then, the transformation is injective if and only if has a pivot in each column. However, since only has three rows, it has at most three pivots, so it can't have a pivot in each column. This means that the transformation cannot be injective.

Question 4

Alice has a matrix that has pivots. Which of the following statements are true?

The system is consistent.

The system has a nontrivial solution.

The parametric vector form of the solution set of has 2 parameters.

The system has a solution for every in . First, the system is always consistent, with the solution, which gives us one of the answers. The interesting question is if has a nontrivial solution, which happens when has at least one free variable. Since the number of pivots is less than the number of columns, we know that at least one column doesn't have a pivot, meaning that there is a free variable. Thus, the system does have a nontrivial solution.

Next, we know that the number of parameters in the parametric vector form of the solution set is equal to the number of free variables, which is 1. This doesn't match the answer, explaining why it wasn't chosen to be correct.

Finally, we know that the system has a solution for every in when each row has a pivot. Since the number of pivots is less than the number of rows, we have that the statement is false.

Question 5

Select which of the following sets of vectors are linearly independent. Hint: you should be able to determine each by inspection.

A set of vectors in linearly independent if and only if the matrix with the vectors as columns has a pivot in each column. Using this fact, you should be able to determine if each of the sets is linearly independent.

Some general facts to know:

A set of vectors that includes the zero vector is always linearly dependent.

A set of two nonzero vectors is linearly dependent when the two vectors are multiples of one another. Thus, checking linear independence amounts to making sure they are not multiples.

A set of vectors that has more vectors than components in each vector (e.g. 17 vectors in) is always linearly dependent because you can't have a pivot in each column when there are more columns than rows.

Question 6

Bob has a linear transformation . Bob knows that the matrix associated to the linear transformation has 3 pivots. Bob wants to know if is injective but not surjective, surjective but not injective, neither injective nor surjective, or bijective (meaning injective and surjective). Which is it?

is injective but not surjective. 

is surjective but not injective.

is neither injective nor surjective.

is bijective (i.e., it is injective and surjective).

The matrix has a pivot in every column if and only if is injective. It has a pivot in every row if and only if is surjective.

It has a pivot in every column and every row if and only if is bijective. Using this information lets us tell Bob which of the four options it is!

Question 7

In the following matrix, how many columns are linear combinations of the columns preceding it?

(That is, for how many is column a linear combination of columns through?)

We row reduce the matrix to count the number of columns without pivots:

A column without a pivot is a linear combination of the preceding columns.

Question 8

Consider the matrix

Suppose is the linear transformation defined by the formula. Determine if is injective but not surjective, surjective but not injective, neither injective nor surjective, or bijective (meaning injective and surjective).

is injective but not surjective.

is surjective but not injective.

is neither injective nor surjective.

is bijective (i.e., it is injective and surjective).

We need to row reduce the given matrix and decide from its RREF what the correct answer is. If its RREF has a pivot in every column then is injective. If it has a pivot in every row then is surjective. If it has a pivot in every column and every row then is bijective. The reduced form of the given matrix is and looking at the number of pivots in the rows and columns, we get the answer.

Question 9

Consider the matrix

What is the smallest number of parameters needed to describe all solutions of in parametric vector form?

The number of parameters in the parametric vector form equals the number of free variables in our matrix . To determine this, we row reduce the matrix to count the number of columns without pivots:

We count 0 column(s) without pivots, so that is our answer.

Question 10

Consider the set of vectors

For what value of is the set of vectors linearly dependent?

-4

To solve this, we create a matrix with the vectors as columns:

The vectors are linearly independent if and only if each column has a pivot, so we want to find a value for which one of the columns doesn't have a pivot. To do this, we first row reduce the matrix, getting

We choose the value of so that the last column doesn't have a pivot, which is when.

Question 11

Suppose is a linear transformation such that and

What is the first entry of ?

1

3 (with margin: 0)

We want to find some constants and such that because if we apply to both sides and apply the properties of linearity of , we get and then substituting in the given values for , we have

Finding and amounts to solving a linear system. Let's row reduce an augmented matrix:

Hence, and .

Therefore, the first entry of is