Exercises – Short Day

Exercise 1: Pivots in 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 6.3.2 in the textbook are satisfied.

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

The matrix has three pivots. These are in the second, fourth and seventh columns of \(\mathbf B\). 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 \(4 \times 4\) matrix with complex entries 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 6.3.3 defines the rank of a matrix and explains how it can be read directly from the reduced row-echelon form of the matrix. So, first, compute the reduced row-echelon form.

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, count how many pivots the reduced row-echelon form of matrix \(\mathbf C\) has and then use Definition 6.3.3.

Answer

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

Exercise 3: Parametric representation with one 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 system of linear equations also appeared in Exercises 4 and 7 on Long Day.

Question a

What is the system of linear equations 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 more compactly:

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

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 system of linear equations that has the augmented matrix \([{\mathbf A} |{\mathbf b}]\).

Question c

In Exercise 4 on Long Day, we found that the matrix \([{\mathbf A} |{\mathbf b}]\) has rank \(3\) 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 this reduced row-echelon form to describe the general solution to the homogeneous system of linear equations with coefficient matrix \(\mathbf A\).

Hint

The reduced row-echelon form of the coefficient matrix \(\mathbf A\) is obtained 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 system of linear equations? Which columns contain a pivot? Now proceed as in Example 6.4.3 in the textbook.

Answer

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

It is also fine to use a symbol other than \(t\) for the free parameter, such as \(t_1\) or \(s\).

Question d

What is the general solution to the system of linear equations over \(\mathbb R\) that corresponds to the given augmented matrix \([{\mathbf A}|{\mathbf b}]\)? To check your answer, compare it with the answer to Exercise 7b from Long Day.

Hint

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

Answer

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

This should match your answer to Exercise 7b from Long Day, apart from possibly being written in a different form, such as \((-t,1-5t,1+9t,t)\) or the like.

Exercise 4: Parametric representation with several 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

The matrix \({\mathbf B}\) can be viewed as the augmented matrix of a system of linear equations over \(\mathbb C\). Is the system homogeneous or inhomogeneous? How many unknowns/variables does the system have?

Answer

The system is inhomogeneous. It has seven unknowns.

Question b

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

Answer

If the seven unknowns are denoted by \(x_1,x_2,x_3,x_4,x_5,x_6,x_7\) as usual, 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 in 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 we set \(x_1\), \(x_3\), \(x_5\) and \(x_6\) equal to zero, we can then compute the values of \(x_2\), \(x_4\) and \(x_7\) from the system of equations.

Answer

If you follow the above hint, 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 system of linear equations 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. It suggests that you should first find the reduced row-echelon form of the system’s augmented matrix.

Answer

The system has no solutions.

Exercise 6: Three examples of systems of linear equations

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

Question a

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

Answer

The general solution is

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

Question b

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

Answer

The general solution is

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

Question c

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

Answer

The system has no solutions.

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