Exercises – Long Day

Exercise 1: Linear combinations

We are given the following vectors in \(\mathbb{R}^3\):

\[\begin{split} \mathbf u=\left[ \begin{array}{c} 4\\ -2\\ 0\\ \end{array} \right], \quad \mathbf v=\left[ \begin{array}{c} 7\\ 4\\ 3\\ \end{array} \right], \quad \mathbf w=\left[ \begin{array}{c} 1\\ 7\\ 3\\ \end{array} \right]. \end{split}\]

Question a

Calculate the following linear combinations of the three vectors \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\):

1. \(4\mathbf v\).

2. \(2\mathbf u-3\mathbf w\).

3. \(3\mathbf u-2\mathbf v+2\mathbf w\).

Answer

\[\begin{split}4\mathbf v=\left[ \begin{array}{c} 28\\ 16\\ 12\\ \end{array} \right]. \end{split}\]

\[\begin{split} 2\mathbf u-3\mathbf w=\left[ \begin{array}{c} 5\\ -25\\ -9\\ \end{array} \right]. \end{split}\]

\[\begin{split}3\mathbf u-2\mathbf v+2\mathbf w=\left[ \begin{array}{c} 0\\ 0\\ 0\\ \end{array} \right].\end{split}\]

Question b

Determine whether the vectors \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\) are linearly independent based on your answers to Question a.

Hint

In Definition 7.1.1 in the textbook you can read more about what it means for vectors to be linearly independent/dependent. It may also be useful to have a look at Equations (7.3) and (7.4).

Answer

The vectors \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\) are not linearly independent. This follows from the third item in Question a together with Equation (7.4) in the textbook.

Question c

Show that the vectors \(\mathbf u\) and \(\mathbf v\) are linearly independent.

Hint

First, consider Equation (7.3) in the textbook. If \(c_1,c_2 \in \mathbb{R}\) satisfy

\[\begin{split}c_1 \cdot \mathbf u+c_2 \cdot \mathbf v=\left[ \begin{array}{c} 0\\ 0\\ 0\\ \end{array} \right],\end{split}\]

can it then be shown that \(c_1=c_2=0\)?

Answer

The vectors \(\mathbf u\) and \(\mathbf v\) are linearly independent.

Exercise 2: Linearly independent vectors in \(\mathbb{R}^4\)

We are given the following three vectors in \(\mathbb{R}^4\):

\[\begin{split} \mathbf u=\left[ \begin{array}{c} 1\\ 1\\ 0\\ 0\\ \end{array} \right], \quad \mathbf v=\left[ \begin{array}{c} 0\\ 1\\ 1\\ 0\\ \end{array} \right], \quad \mathbf w=\left[ \begin{array}{c} 0\\ 0\\ 1\\ 1\\ \end{array} \right]. \end{split}\]

Question a

Show that the vectors \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\) are linearly independent.

Hint

Assume that

\[\begin{split} c_1\cdot\left[ \begin{array}{c} 1\\ 1\\ 0\\ 0\\ \end{array} \right]+c_2\cdot\left[ \begin{array}{c} 0\\ 1\\ 1\\ 0\\ \end{array} \right]+c_3\cdot\left[ \begin{array}{c} 0\\ 0\\ 1\\ 1\\ \end{array} \right]=\left[ \begin{array}{c} 0\\ 0\\ 0\\ 0\\ \end{array} \right]. \end{split}\]

What can we now say about \(c_1\), \(c_2\) and \(c_3\)?

Question b

Assume that \(\mathbf c \in \mathbb{R}^4\) is a vector such that \(\mathbf u\), \(\mathbf v\), \(\mathbf w\) and \(\mathbf c\) are linearly independent. Show that \(\mathbf c\) then cannot be written as a linear combination of \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\).

Hint

Using contraposition (Equation (1.21) in the textbook) we see that what we wish to prove is logically equivalent to the proposition:

If \(\mathbf c\) can be written as a linear combination of \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\), then the vectors \(\mathbf u\), \(\mathbf v\), \(\mathbf w\) and \(\mathbf c\) are linearly dependent.

Try proving this proposition instead.

Hint

If \(\mathbf c=c_1\cdot \mathbf u+c_2\cdot \mathbf v+c_3\cdot \mathbf w\) for some \(c_1,c_2,c_3 \in \mathbb{R}\), then

\[\begin{split}\mathbf c-c_1\cdot \mathbf u-c_2\cdot \mathbf v-c_3\cdot \mathbf w=\left[ \begin{array}{c} 0\\ 0\\ 0\\ 0\\ \end{array} \right].\end{split}\]

Question c

Find a vector \(\mathbf b \in \mathbb R^4\) such that the vectors \(\mathbf u\), \(\mathbf v\), \(\mathbf w\) and \(\mathbf b\) are linearly independent.

Hint

What we learned in Question b can be used here: try to find a vector \(\mathbf b \in \mathbb R^4\) that cannot be written as a linear combination of \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\).

Hint

Can the vector

\[\begin{split}\left[ \begin{array}{c} 1\\ 0\\ 0\\ 0\\ \end{array} \right]\end{split}\]

be written as a linear combination of \(\mathbf u\), \(\mathbf v\) and \(\mathbf w\)?

Exercise 3: Linear independence and reduced row-echelon form

We are given the following three vectors in \(\mathbb R^3\):

\[\begin{split} \mathbf v_1=\left[ \begin{array}{c} 1\\ 0\\ 4\\ \end{array} \right], \quad \mathbf v_2=\left[ \begin{array}{c} 1\\ 1\\ 7\\ \end{array} \right], \quad \mathbf v_3=\left[ \begin{array}{c} 1\\ 2\\ 3\\ \end{array} \right]. \end{split}\]

Question a

Let \(\mathbf A\) be the \(3 \times 3\) matrix that has \(\mathbf v_1\), \(\mathbf v_2\) and \(\mathbf v_3\) as columns. Calculate the reduced row-echelon form of the matrix and use this form to determine the rank of \(\mathbf A\).

Answer

The reduced row-echelon form of \(\mathbf A\) is

\[\begin{split} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]. \end{split}\]

In other words, the reduced row-echelon form of \(\mathbf A\) is \({\mathbf I}_3\), the \(3 \times 3\) identity matrix. It contains three pivot elements, so \(\rho(\mathbf A)=3.\)

Question b

Are the given vectors \(\mathbf v_1\), \(\mathbf v_2\) and \(\mathbf v_3\) linearly independent?

Hint

Theorem 7.1.3 in the textbook is useful here.

Answer

Using Theorem 7.1.3 we conclude that the three given vectors are linearly independent.

Question c

Can four vectors \(\mathbf w_1\), \(\mathbf w_2\), \(\mathbf w_3\) and \(\mathbf w_4\) in \(\mathbb R^3\) be linearly independent?

Hint

Let \(\mathbf B\) be the \(3 \times 4\) matrix whose columns are \(\mathbf w_1\), \(\mathbf w_2\), \(\mathbf w_3\) and \(\mathbf w_4\). Consider why the rank of \(\mathbf B\) cannot be greater than three.

Answer

Since \(\mathbf B\) has three rows, its reduced row-echelon form can have at most three pivots. Hence, \(\rho(\mathbf B) \le 3\). Using Theorem 7.1.3 we conclude that four vectors \(\mathbf w_1\), \(\mathbf w_2\), \(\mathbf w_3\) and \(\mathbf w_4\) in \(\mathbb R^3\) will always be linearly dependent. In conclusion, the answer is no.

Exercise 4: Matrix products

We are given the following three matrices:

\[\begin{split} \mathbf A=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right], \quad \mathbf B=\left[ \begin{array}{cc} 1 & 4\\ 2 & 1\\ 3 & 0\\ \end{array} \right], \quad \mathbf C=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 2 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Question a

Which of the nine matrix products \(\mathbf A\cdot\mathbf A\), \(\mathbf A\cdot\mathbf B\), \(\mathbf A\cdot\mathbf C\), \(\mathbf B\cdot\mathbf A\), \(\mathbf B\cdot\mathbf B\), \(\mathbf B\cdot\mathbf C\), \(\mathbf C\cdot\mathbf A\), \(\mathbf C\cdot\mathbf B\) and \(\mathbf C\cdot\mathbf C\) are defined?

Hint

According to Definition 7.2.2 in the textbook two matrices can only be multiplied together if the number of columns in the first matrix equals the number of rows in the second matrix.

Answer

Only these matrix products are defined: \(\mathbf A\cdot\mathbf B\), \(\mathbf A\cdot\mathbf C\), \(\mathbf B\cdot\mathbf A\), \(\mathbf C\cdot\mathbf B\) and \(\mathbf C\cdot\mathbf C\).

Question b

Calculate the matrix products \(\mathbf A\cdot\mathbf B\) and \(\mathbf B\cdot\mathbf A\). Does it hold that \(\mathbf A\cdot\mathbf B=\mathbf B\cdot\mathbf A\)?

Answer

\[\begin{split}\mathbf A\cdot\mathbf B=\left[ \begin{array}{cc} 1 & 4 \\ 2 & 1 \\ \end{array} \right] \quad \text{and} \quad \mathbf B\cdot\mathbf A=\left[ \begin{array}{ccc} 1 & 4 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 0 \\ \end{array} \right].\end{split}\]

It is clear that \(\mathbf A\cdot\mathbf B \neq \mathbf B\cdot\mathbf A\). The matrices \(\mathbf A\cdot\mathbf B\) and \(\mathbf B\cdot\mathbf A\) do not even have the same size!

Question c

Calculate the product of the scalar \(-1/2\) with the matrix \(\mathbf C\). In other words, calculate \((-1/2) \cdot \mathbf C\).

Answer

\[\begin{split}(-1/2) \cdot \mathbf C=\left[ \begin{array}{ccc} -1/2 & 0 & 0 \\ 0 & 1/2 & 0 \\ -1 & 0 & 0 \\ \end{array} \right]\end{split}\]

Question d

Determine whether the following sums of matrices are defined, and calculate those that are: \((-1/2) \cdot \mathbf C+\mathbf A \cdot \mathbf B\) and \((-1/2) \cdot \mathbf C+\mathbf B \cdot \mathbf A.\)

Answer

The sum \((-1/2) \cdot \mathbf C+\mathbf A \cdot \mathbf B\) is not defined since \((-1/2) \cdot \mathbf C\) is a \(3 \times 3\) matrix while \(\mathbf A \cdot \mathbf B\) is a \(2 \times 2\) matrix.

The sum \((-1/2) \cdot \mathbf C+\mathbf B \cdot \mathbf A\) is defined since the matrices \((-1/2) \cdot \mathbf C\) and \(\mathbf B \cdot \mathbf A\) have the same size. The result is:

\[\begin{split} (-1/2) \cdot \mathbf C+\mathbf B \cdot \mathbf A=\left[ \begin{array}{ccc} 1/2 & 4 & 0 \\ 2 & 3/2 & 0 \\ 2 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Exercise 5: Matrix-vector product and systems of linear equations

We are given the following system of linear equations:

\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_4 & = & 0\\ -2x_1+x_2 +3x_4 & = & 1\\ 5x_1 +x_3 -4x_4 & = & 1\\ 4x_1 +x_2 +x_3 & = & 2\\ \end{array} \right. \end{split}\]

Question a

Write the system in the same form as Equation (7.5) in the textbook.

Answer

\[\begin{split} \left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -2 & 1 & 0 & 3 \\ 5 & 0 & 1 & -4 \\ 4 & 1 & 1 & 0 \\ \end{array} \right] \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{array} \right]= \left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ \end{array} \right]. \end{split}\]

Question b

We define

\[\begin{split}\mathbf A=\left[ \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -2 & 1 & 0 & 3 \\ 5 & 0 & 1 & -4 \\ 4 & 1 & 1 & 0 \\ \end{array} \right], \quad \mathbf b=\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ \end{array} \right].\end{split}\]

Check that the given system of linear equations has the augmented matrix \([\mathbf A|\mathbf b]\). In Exercise 3b from Short Day in Week 6 we checked that the system of linear equations whose augmented matrix is \([\mathbf A|\mathbf b]\) has the following particular solution:

\[\begin{split} \mathbf v=\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \\ \end{array} \right]. \end{split}\]

Check this again, but this time by calculating the product \(\mathbf A\cdot\mathbf v\).

Exercise 6: Finding inverse matrices

We are given the following square matrix:

\[\begin{split}\mathbf A=\left[ \begin{array}{ccc} 1 & 1 & 0 \\ -1 & 1 & 0 \\ 5 & 0 & 1 \\ \end{array} \right].\end{split}\]

Determine whether the matrix \(\mathbf A^{-1}\) exists, and if so, find it.

Hint

The procedure explained right before Example 7.3.1 in the textbook can be followed. Examples 7.3.1 and 7.3.2 can be useful as well if you wish to see some examples.

Answer

The reduced row-echelon form of the matrix \([\mathbf A|{\mathbf I}_3]\) is

\[\begin{split}\left[ \begin{array}{cccccc} 1 & 0 & 0 & 1/2 & -1/2 & 0 \\ 0 & 1 & 0 & 1/2 & 1/2 & 0 \\ 0 & 0 & 1 & -5/2 & 5/2 & 1 \\ \end{array} \right]. \end{split}\]

Hence, \(\mathbf A^{-1}\) exists, and we have

\[\begin{split}\mathbf A^{-1}=\left[ \begin{array}{ccc} 1/2 & -1/2 & 0 \\ 1/2 & 1/2 & 0 \\ -5/2 & 5/2 & 1 \\ \end{array} \right].\end{split}\]

Exercise 7: Inverse matrices and systems of linear equations

Let \(n\) be a natural number and \(\mathbf A \in \mathbb{C}^{n \times n}\) an invertible square matrix. Furthermore, let \(\mathbf b \in \mathbb{C}^n\) be a vector.

Question a

Show that the vector \(\mathbf A^{-1}\cdot \mathbf b \in \mathbb{C}^n\) is a solution to the system of linear equations whose augmented matrix is \([\mathbf A|\mathbf b].\)

Hint

First, write the system in matrix form as in Equation (7.5) in the textbook.

Hint

In matrix form the system will look as follows:

\[\begin{split} \mathbf A \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]=\mathbf b.\end{split}\]

Is this equation satisfied if

\[\begin{split} \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]=\mathbf A^{-1} \cdot \mathbf b?\end{split}\]

Question b

Show that the vector \(\mathbf A^{-1} \cdot \mathbf b\) is the only solution to the system of linear equations whose augmented matrix is \([\mathbf A|\mathbf b].\)

Hint

Multiply both sides of the equation

\[\begin{split} \mathbf A \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]=\mathbf b\end{split}\]

from the left by \(\mathbf A^{-1}\).

Exercise 8: Systems with variable coefficients

For each real number \(a\), we are given the following system of linear equations:

\[\begin{align*} ax_1 + x_2 + x_3 &= 1\\ x_1 + ax_2 + x_3 &= 1\\ x_1 + x_2 + ax_3 &= 1 \end{align*}\]

Question a

State the augmented matrix of the system.

Question b

What is the solution set to the system for \(a=-2\)?

Question c

Solve the system of linear equations for all \(a\in \Bbb R\).

Hint

Along the way in your calculations you may risk dividing by zero. Make a note of these special cases and treat them separately afterwards.

Answer

For \(a\in\mathbb R\setminus\{-2,1\}\) the solution is

\[(x_1,x_2,x_3)=\left(\frac1{a+2},\frac1{a+2},\frac1{a+2}\right),\]

and for \(a=1\) the general solution is

\[(x_1,x_2,x_3)=(1,0,0)+t_1(-1,1,0)+t_2(-1,0,1)\quad,t_1,t_2\in\mathbb R.\]

There is no solution for \(a=-2\).

Thematic Python Module 1

The Jupyter Notebook for module 1 will be released today at 15:30 on the course’s DTU Learn module.

Next Wednesday at 15:30, a 1.5-hour Python test on module 1 will take place in your study area. Work through today’s Notebook to prepare.

Have you finished all the exercises and the Jupyter Notebook?

Are you done with the exercises of the day and just waiting for the Python module to be released, then click here for some challenging optional extra exercises.