Exercises – Short Day

Exercises – Short Day#

Exercise 1: Pivots in a Reduced Row-Echelon Form#

We are given the following matrix:

\[\begin{split}{\mathbf B}= \left[ \begin{array}{cccccccc} 0 & 1 & \sqrt{2} & 0 & 5 & 0 & 0 & 9 \\ 0 & 0 & 0 & 1 & \mathrm e & 0 & 0 & \pi \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 42 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Question a#

Is the matrix \(\mathbf B\) in reduced row-echelon form?

Answer

Yes, since all four requirements from Definition 6.3.1 in the textbook are fulfilled.

Question b#

State the pivot elements in the given matrix \({\mathbf B}\). How many pivots does the matrix \({\mathbf B}\) have and in which columns are they found?

Answer

There are three pivots, found in the second, fourth, and seventh column of \(\mathbf B\), respectively. The three pivot elements are emphasised in bold below:

\[\begin{split}{\mathbf B}= \left[ \begin{array}{cccccccc} 0 & {\mathbf 1} & \sqrt{2} & 0 & 5 & 0 & 0 & 9 \\ 0 & 0 & 0 & {\mathbf 1} & \mathrm e & 0 & 0 & \pi \\ 0 & 0 & 0 & 0 & 0 & 0 & {\mathbf 1} & 42 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Exercise 2: Rank of a Matrix#

The following \(3 \times 4\) matrix with complex elements is given:

\[\begin{split}{\mathbf C}= \left[ \begin{array}{ccc} 1 & i & 1 & 0\\ 3 & 3i & 0 & 0\\ 1+2i & -2+i & -1-i & 0\\ i & -1 & -1 & 0\\ \end{array} \right]. \end{split}\]

What is the rank of the matrix?

Hint

Definition 6.3.2 explains how rank of a matrix is defined and how one can determine the rank directly from the reduced row-echelon form of the matrix. So, first compute the reduced row-echelon form of the matrix.

Hint

The reduced row-echelon form of the matrix is

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

Now use Definition 6.3.1 to determine how many pivots that are found in the reduced row-echelon form of matrix \(\mathbf C\).

Answer

\(\rho({\mathbf C})=2.\)

Exercise 3: Parametric Representation with a Free Parameter#

A system of linear equations over \(\mathbb R\) in the unknowns \(x_1,x_2,x_3\), and \(x_4\) has the following augmented matrix:

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

This matrix and the corresponding linear equation system also appeared in Exercises 4 and 7 on Long Day.

Question a#

What is the linear equation system that corresponds to the given augmented matrix?

Answer

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

or written in a more compact manner:

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

Question b#

Check that the vector

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

is a particular solution to the linear equation system that has the augmented matrix \([{\mathbf A} |{\mathbf b}]\).

Question c#

In Exercise 4 from Long Day the answer stated that the matrix \([{\mathbf A} |{\mathbf b}]\) has a rank of \(3\) and that its reduced row-echelon form is

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

Use the reduced row-echelon form to describe the general solution to the homogeneous linear equation system with coefficient matrix \(\mathbf A\).

Hint

The reduced row-echelon form of coefficient matrix \(\mathbf A\) is achieved by removing the last column from the reduced row-echelon form of the matrix \([{\mathbf A}|{\mathbf b}]\). Hence it is

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

What is the corresponding homogeneous linear equation system? Which columns contain a pivot? Now continue in a similar manner as in Example 6.4.3 in the textbook.

Answer

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ \end{array} \right] = t\cdot \left[ \begin{array}{c} -1\\ -5\\ 9\\ 1\\ \end{array} \right] \quad (t \in \mathbb R). \end{split}\]

It is also perfectly fine to use another symbol for the free parameter \(t\), for instance \(t_1\) or \(s\).

Question d#

What is the general solution to the linear equation system over \(\mathbb R\) that corresponds to the given augmented matrix \([{\mathbf A}|{\mathbf b}]\)? For confirmation, you can compare your answer to the answer from Exercise 7b from Long Day.

Hint

Theorem 6.1.2 in the textbook implies that you can find the answer by combining Question b and the answer to Question c. If necessary, read Example 6.4.4 for an example.

Answer

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ \end{array} \right] = \left[ \begin{array}{c} 0\\ 1\\ 1\\ 0\\ \end{array} \right] + t\cdot \left[ \begin{array}{c} -1\\ -5\\ 9\\ 1\\ \end{array} \right] \quad (t \in \mathbb R). \end{split}\]

This answer should be identical to that to Exercise 7b from Long Day (apart from maybe the answer being written in another form, such as \((-t,1-5t,1+9t,t)\) or the like.

Exercise 4: Parametric Representation with Multiple Free Parameters#

We again consider matrix \({\mathbf B}\) from Exercise 1:

\[\begin{split}{\mathbf B}= \left[ \begin{array}{cccccccc} 0 & 1 & \sqrt{2} & 0 & 5 & 0 & 0 & 9 \\ 0 & 0 & 0 & 1 & \mathrm e & 0 & 0 & \pi \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 42 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Question a#

Matrix \({\mathbf B}\) can be considered as an augmented matrix of a linear equation system over \(\mathbb C\). Is the system homogeneous or inhomogeneous? How many unknowns/variables are a part of this system?

Answer

The system is inhomogeneous. It contains seven unknowns.

Question b#

What is the linear equation system corresponding to the given augmented matrix \(\mathbf B\)?

Answer

If the seven unknowns are symbolised as usual by \(x_1,x_2,x_3,x_4,x_5,x_6\), and \(x_7\), then we have:

\[\begin{split} \left\{ \begin{array}{cccccccc} & x_2 & +\sqrt{2}\cdot x_3 & & +5\cdot x_5 & & & = 9 \\ & & & x_4 & +\mathrm e\cdot x_5 & & & = \pi \\ & & & & & & x_7 & = 42 \\ & & & & & & 0 & = 0 \\ \end{array} \right. \end{split}\]

Question c#

Compute a particular solution to the system from Question b.

Hint

We know from Exercise 1 that matrix \(\mathbf B\) has three pivots located within the second, fourth, and seventh columns of \(\mathbf B\). Hence, we can find a particular solution by following the same strategy as in Theorem 6.4.2 in the textbook. More precisely: if one lets \(x_1,x_3,x_5,\) and \(x_6\) all be zero, then one can afterwards compute the values of \(x_2,x_4,\) and \(x_7\) from the equation system.

Answer

If you follow the above hint, then you will arrive at the following particular solution:

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ x_7\\ \end{array} \right] = \left[ \begin{array}{c} 0\\ 9\\ 0\\ \pi\\ 0\\ 0\\ 42\\ \end{array} \right] \end{split}\]

Question d#

Compute the general solution to the system from Question b.

Answer

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ x_7\\ \end{array} \right] = \left[ \begin{array}{c} 0\\ 9\\ 0\\ \pi\\ 0\\ 0\\ 42\\ \end{array} \right] + t_1 \cdot \left[ \begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array} \right] + t_2 \cdot \left[ \begin{array}{c} 0\\ -\sqrt{2}\\ 1\\ 0\\ 0\\ 0\\ 0\\ \end{array} \right] + t_3 \cdot \left[ \begin{array}{c} 0\\ -5\\ 0\\ -\mathrm e\\ 1\\ 0\\ 0\\ \end{array} \right] + t_4 \cdot \left[ \begin{array}{c} 0\\ 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ \end{array} \right] \quad (t_1,t_2,t_3,t_4 \in \mathbb C). \end{split}\]

Exercise 5: An Inhomogeneous System#

Determine whether the following linear equation system over \(\mathbb R\) in the unknowns \(x_1,x_2,\) and \(x_3\) has a solution:

\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_3 & = & 0\\ x_1+x_2 +3x_3 & = & 0\\ 10x_1 +3x_2+16x_3 & = & 1\\ \end{array} \right. \end{split}\]

Hint

Corollary 6.4.3 in the textbook can be used to determine whether the system has a solution or not. It is thus a good idea to determine the reduced row-echelon form of the system’s augmented matrix first.

Answer

The system has no solutions.

Exercise 6: Three Examples on Systems of Linear Equations#

Compute the general solutions to the following three systems of linear equations over \(\mathbb R\):

\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_2+ x_3 & = & 1\\ x_1+ 2x_2 +4x_3 & = & 1\\ x_1 +3x_2+9x_3 & = & 1\\ \end{array} \right. \end{split}\]
\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_2+ x_3 & = & 1\\ x_1+ 2x_2 +4x_3 & = & 1\\ x_1 +3x_2+7x_3 & = & 1\\ \end{array} \right. \end{split}\]
\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_2+ x_3 & = & 1\\ x_1+ 2x_2 +4x_3 & = & 1\\ x_1 +3x_2+7x_3 & = & 0\\ \end{array} \right. \end{split}\]

Answer

  1. The general solution is

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array} \right] = \left[ \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right]. \end{split}\]

  1. The general solution is

\[\begin{split} \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ \end{array} \right] = \left[ \begin{array}{c} 1\\ 0\\ 0\\ \end{array} \right] + t \cdot \left[ \begin{array}{c} 2\\ -3\\ 1\\ \end{array} \right] \quad (t \in \mathbb R). \end{split}\]

  1. The system has no solutions.