Exercises – Long Day

Exercises – Long Day#

Exercise 1: Properties of Systems of Linear Equations#

We are given the following linear equation system over \(\mathbb R\) in the three unknowns \(x\), \(y\), and \(z\).

\[\begin{split} \left\{ \begin{array}{ccccccl} 3x & - &7y & + &z & = & 1\\ 11x & -&y & +&4z & = & 4. \end{array} \right. \end{split}\]

Question a#

Is the system homogeneous or inhomogeneous?

Hint

If you are in doubt about what the terms “homogeneous” and “inhomogeneous” mean, then you can find an explanation in the textbook just before Theorem 6.1.1.

Answer

The given linear equation system is inhomogeneous since not all constants on the right-hand side of the equations are equal to zero.

Question b#

Is the triple \((-1,-1,-3) \in \mathbb{R}^3\) a solution to the system? What about the triple \((0,0,1)\)?

Answer

The triple \((-1,-1,-3)\) is not a solution to the system (it fulfills the first equation but not the second).

The triple \((0,0,1)\) is a solution to the system.

Question c#

We still consider the linear equation system given in the beginning of the exercise. What is its corresponding homogeneous linear equation system?

Hint

In Theorem 6.1.2 in the textbook the corresponding homogeneous system is described.

Answer

\[\begin{split} \left\{ \begin{array}{ccccccc} 3x & -& 7y & + &z & = & 0\\ 11x & -&y & +&4z & = & 0 \end{array} \right. \end{split}\]

Question d#

We are being informed that the triple \((27,1,-74)\) is a solution to the corresponding homogeneous system from Question c. Now use the result from Question b to find yet another solution to the given inhomogeneous linear equation system.

Hint

If needed, first read Theorem 6.1.2 in the textbook.

Answer

The triple \((27,1,-73)\) is a solution to the inhomogeneous linear equation system.

Exercise 2: Augmented Matrix and Row Operations#

We define the following matrix

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 3 & 4\\ 0 & 4 & -4\\ -2 & 0 & -9\\ \end{array} \right] \end{split}\]

and the following vector

\[\begin{split} {\mathbf b}= \left[ \begin{array}{cccc} 1\\ 0\\ -1\\ \end{array} \right]. \end{split}\]

Question a#

Write out the system of linear equations over \(\mathbb R\) in the unknowns \(x_1\), \(x_2\), and \(x_3\) that has the augmented matrix \([{\mathbf A}|{\mathbf b}]\).

Answer

\[\begin{split} \left\{ \begin{array}{ccccc} x_1 & +&3x_2 & +& 4x_3 & = & 1\\ & &4x_2 & -&4x_3 & = & 0\\ -2x_1 & & & -& 9x_3 & = & -1 \end{array} \right. \end{split}\]

Question b#

Determine the matrix that you get by performing the row operation \(R_3 \leftarrow R_3+2R_1\) on the matrix \([\mathbf{A}|\mathbf{b}]\).

Hint

If you are in doubt about what a row operation like \(R_3 \leftarrow R_3+2R_1\) exactly means, then have a look in the textbook right before Example 6.2.4 as well as in Example 6.2.4 itself, where a similar row operation is being used.

Answer

\[\begin{split} \left[ \begin{array}{cccc} 1 & 3 & 4 & 1\\ 0 & 4 & -4 & 0\\ -2 & 0 & -9 & -1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_3 \leftarrow R_3+2R_1 \\ \end{array} \left[ \begin{array}{cccc} 1 & 3 & 4 & 1\\ 0 & 4 & -4 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \end{split}\]

Question c#

Now consider the matrix that was found in the previous question (after the row operation) and perform the row operation \(R_2 \leftarrow (1/4)\cdot R_2\) followed by the row operations \(R_1 \leftarrow R_1-3R_2\) and \(R_3 \leftarrow R_3-6R_2\). What is the result?

Answer

\[\begin{split} \left[ \begin{array}{cccc} 1 & 3 & 4 & 1\\ 0 & 4 & -4 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_2 \leftarrow (1/4)\cdot R_2\\ \end{array} \left[ \begin{array}{cccc} 1 & 3 & 4 & 1\\ 0 & 1 & -1 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \end{split}\]
\[\begin{split} \left[ \begin{array}{cccc} 1 & 3 & 4 & 1\\ 0 & 1 & -1 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_1 \leftarrow R_1-3\cdot R_2\\ \end{array} \left[ \begin{array}{cccc} 1 & 0 & 7 & 1\\ 0 & 1 & -1 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \end{split}\]
\[\begin{split} \left[ \begin{array}{cccc} 1 & 0 & 7 & 1\\ 0 & 1 & -1 & 0\\ 0 & 6 & -1 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_3 \leftarrow R_3-6\cdot R_2\\ \end{array} \left[ \begin{array}{cccc} 1 & 0 & 7 & 1\\ 0 & 1 & -1 & 0\\ 0 & 0 & 5 & 1\\ \end{array} \right] \end{split}\]

Question d#

Determine the reduced row-echelon form of the matrix \([\mathbf{A}|\mathbf{b}]\).

Hint

You do not have to start all over. You can continue with the matrix that was the answer to Question c.

Answer

After the row operation \(R_3 \leftarrow (1/5)\cdot R_3\) followed by \(R_2 \leftarrow R_2+R_3\) and then \(R_1 \leftarrow R_1-7R_3\) we arrive at the matrix

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

This matrix is in the wanted reduced row-echelon form.

Question e#

Describe and solve the linear equation system over \(\mathbb R\) that has the reduced row-echelon form found in Question d as its augmented matrix. Check that the solution you find is a solution to the original linear equation system that had the augmented matrix \([{\mathbf A}|{\mathbf b}].\)

Answer

The equation system corresponding to the found reduced row-echelon form is

\[\begin{split} \left\{ \begin{array}{rrrrr} x_1 & & & = & -2/5&\\ & x_2 & & = & 1/5&\\ & & x_3 & = & 1/5&. \end{array} \right. \end{split}\]

The solution is the triple \((-2/5,1/5,1/5).\) By insertion we see that the triple is a solution to the original system. This is just as it should be since Theorem 6.2.1 in the textbook predicts that the solution set to a linear equation system does not change when elementary row operations are performed.

Exercise 3: Reduced Row-Echelon Forms#

The following matrices are defined

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 3 & 4\\ 0 & 1 & 5\\ 0 & 0 & 1\\ \end{array} \right], \quad {\mathbf B}=\left[ \begin{array}{ccc} 2 & 0 & 4\\ 0 & 4 & 0\\ \end{array} \right], \quad {\mathbf C}=\left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \right], \quad {\mathbf D}=\left[ \begin{array}{ccc} 1 & 4 & 0 & 9\\ 0 & 0 & 1 & 2\\ \end{array} \right]. \end{split}\]

Question a#

Which of these matrices are in row-echelon form? Which are in reduced row-echelon form?

Hint

The definition of a matrix in reduced row-echelon form is given in Definition 6.3.2 in the textbook. Right after this definition it is described what a matrix in row-echelon form fulfills.

Answer

The matrices \(\mathbf A,\mathbf B,\) and \(\mathbf D\) are in row-echelon form. \(\mathbf C\) is not. Amongt hose, only matrix \({\mathbf D}\) is in reduced row-echelon form.

Question b#

Determine the reduced row-echelon form of each of the given matrices \({\mathbf A}\), \({\mathbf B}\), \({\mathbf C}\), and \({\mathbf D}\).

Answer

The following sequences of row operations lead to the reduced row-echelon forms of the matrices:

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 3 & 4\\ 0 & 1 & 5\\ 0 & 0 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_1 \leftarrow R_1-3\cdot R_2\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & -11\\ 0 & 1 & 5\\ 0 & 0 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_1 \leftarrow R_1+11\cdot R_3\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 5\\ 0 & 0 & 1\\ \end{array} \right] \end{split}\]
\[\begin{split} \begin{array}{c} \longrightarrow \\ \scriptstyle R_2 \leftarrow R_2-5\cdot R_3\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \right]. \end{split}\]
\[\begin{split} {\mathbf B}=\left[ \begin{array}{ccc} 2 & 0 & 4\\ 0 & 4 & 0\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_1 \leftarrow (1/2)\cdot R_1\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 2\\ 0 & 4 & 0\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_2 \leftarrow (1/4)\cdot R_2\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 2\\ 0 & 1 & 0\\ \end{array} \right]. \end{split}\]
\[\begin{split} {\mathbf C}=\left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_2 \leftrightarrow R_3\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\\ \end{array} \right] \begin{array}{c} \longrightarrow \\ \scriptstyle R_3 \leftrightarrow R_4\\ \end{array} \left[ \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{array} \right]. \end{split}\]

As we concluded in the previous Question, matrix \(\mathbf D\) is already in reduced row-echelon form.

Exercise 4: Reduced Row-Echelon Form of \(4\times 5\) Matrix#

We define the matrix

\[\begin{split}{\mathbf A}= \left[ \begin{array}{ccc} 1 & 0 & 0 & 1 & 0\\ -2 & 1 & 0 & 3 & 1\\ 5 & 0 & 1 & -4 & 1\\ 4 & 1 & 1 & 0 & 2\\ \end{array} \right]. \end{split}\]

How many non-zero rows does the reduced row-echelon form of \(\mathbf A\) have?

Answer

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

\[\begin{split} \left[ \begin{array}{ccc} 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 5 & 1\\ 0 & 0 & 1 & -9 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{array} \right]. \end{split}\]

It has three non-zero rows. This number is by the way called the rank of the matrix (see Definition 6.3.3).

Exercise 5: Relation between Row-Echelon Form and Reduced Row-Echelon Form#

We define the following matrix, which is already is in row-echelon form:

\[\begin{split}{\mathbf B}= \left[ \begin{array}{ccc} 1 & 6 & 10 & 1 & 0\\ 0 & 0 & 1 & 3 & 1\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{array} \right]. \end{split}\]

Question a#

Without carrying out any computations, justify that \(\mathbf B\) will have a reduced row-echelon form of this type:

\[\begin{split} \left[ \begin{array}{ccc} 1 & a & 0 & b & 0\\ 0 & 0 & 1 & c & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ \end{array} \right] \end{split}\]

for certain numbers \(a,b,c \in \mathbb{R}\).

Question b#

Consider why it more generally applies that the number of non-zero rows in a matrix in row-echelon form is equal to the number of non-zero rows in its reduced row-echelon form.

Exercise 6: Examples of Systems of Linear Equations#

Question a#

Provide an example of an inhomogeneous linear equation system over \(\mathbb R\) in the three unknowns \(x_1\), \(x_2\), and \(x_3\) that does not have any solutions.

Hint

Corollary 6.4.3 describes when an inhomogeneous linear equation system has no solutions.

Question b#

Can a homogeneous linear equation system have no solutions?

Question c#

Provide an example of a homogeneous linear equation system over \(\mathbb R\) that contains at least two equations and has at least two solutions.

Exercise 7: A System of Four Equations in Four Unknowns#

We again consider the matrix

\[\begin{split} \left[ \begin{array}{ccc} 1 & 0 & 0 & 1 & 0\\ -2 & 1 & 0 & 3 & 1\\ 5 & 0 & 1 & -4 & 1\\ 4 & 1 & 1 & 0 & 2\\ \end{array} \right] \end{split}\]

from Exercise 4. We are informed that this matrix is the augmented matrix of a linear equation system over \(\mathbb R\) in the unknowns \(x_1,x_2,x_3\), and \(x_4\).

Question a#

Find a solution to the system that fulfills \(x_4=0\). Also, find a solution to the system that fulfills \(x_4=1\).

Question b#

We are given a real number \(t\). Find a solution to the system that fulfills \(x_4=t\).

Exercise 8: Square Matrix and corresponding System of Linear Equations#

A matrix \(\mathbf A\) is called a square matrix if it contains just as many rows as columns. In other words, the matrix \(\mathbf A\) is square precisely if \({\mathbf A} \in \mathbb{F}^{n \times n}\) for a natural number \(n\).

Question a#

Given a square matrix \({\mathbf A} \in \mathbb{R}^{4 \times 4}\) and a vector \(\mathbf b\in\mathbb R^4\). We are informed that the reduced row-echelon form of \(\mathbf A\) is:

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

Show that a system of linear equations over \(\mathbb R\) with the augmented matrix \({[{\mathbf A}|{\mathbf b}]} \in \mathbb{R}^{4 \times 5}\) has exactly one solution.

Hint

Try to use the given information that the reduced row-echelon form of \({\mathbf A}\) is

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

to realise that the reduced row-echelon form of the augmented matrix \({[{\mathbf A}|{\mathbf b}]}\) looks as follows:

\[\begin{split} \left[ \begin{array}{ccc} 1 & 0 & 0 & 0 & c_1\\ 0 & 1 & 0 & 0 & c_2\\ 0 & 0 & 1 & 0 & c_3\\ 0 & 0 & 0 & 1 & c_4\\ \end{array} \right] \end{split}\]

for some \(c_1,c_2,c_3,c_4 \in \mathbb{R}.\)

“Exercise 9”: Thematic Python Module#

Today at 15:30 the first of a series of thematic Python modules will be released. It will be published via DTU Learn as a Jupyter Notebook file containing a set of Python-focused exercises. The purpose is to introduce the use of Python - in particular the Python package SymPy - for mathematical computations on the topics we have covered this far in the course. The intention is that you finish the exercises of the day before you start working on the Python problems. Next Long Day (in semester week 7, after the Autumn break) a test on this module will be carried out in Möbius. Details about this test will be found in the description of week 7.

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