Exercises – Long Day

Exercise 1: Properties of systems of linear equations

We are given the following system of linear equations 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 system of linear equations is inhomogeneous since not all the constants on the right-hand sides are 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 satisfies the first equation but not the second).

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

Question c

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

Hint

The homogeneous system corresponding to a given system of linear equations is described in Theorem 6.1.2.

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 another solution to the given inhomogeneous system of linear equations.

Hint

If needed, first read Theorem 6.1.2 in the textbook.

Answer

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

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 exactly a row operation like \(R_3 \leftarrow R_3+2R_1\) means, have a look at the text just before Example 6.2.4 in the textbook and at the example itself, where a similar row operation is 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 need to start from scratch. You can continue from 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 required reduced row-echelon form.

Question e

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

Answer

The system of equations that corresponds to this 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 substitution, we see that the triple is a solution to the original system. This is as expected since Theorem 6.2.1 in the textbook states that elementary row operations do not change the solution set to a system of linear equations.

Exercise 3: Row-echelon forms

We define the following matrices:

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

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, the textbook describes the conditions that a matrix in row-echelon form satisfies.

Answer

The matrices \(\mathbf A\), \(\mathbf B\) and \(\mathbf D\) are in row-echelon form. \(\mathbf C\) is not. Among these, 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 a \(4\times 5\) 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. Incidentally, this number is called the rank of the matrix (see Definition 6.3.3).

Exercise 5: Row-echelon form vs reduced row-echelon form

We define the following matrix, which is already 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, show that \(\mathbf B\) has a reduced row-echelon form of the form:

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

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

Question b

Explain why, more generally, 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 system of linear equations over \(\mathbb R\) in the three unknowns \(x_1\), \(x_2\) and \(x_3\) that has no solutions.

Hint

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

Question b

Can a homogeneous system of linear equations have no solutions?

Question c

Provide an example of a homogeneous system of linear equations 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 system of linear equations over \(\mathbb R\) in the unknowns \(x_1\), \(x_2\), \(x_3\) and \(x_4\).

Question a

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

Question b

Let \(t\) be a real number. Find a solution to the system that satisfies \(x_4=t\).

Exercise 8: Square matrices and corresponding systems of linear equations

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

Let \({\mathbf A} \in \mathbb{R}^{4 \times 4}\) be a square matrix and \(\mathbf b\in\mathbb R^4\) a vector. 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

Use the given fact 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 see that the reduced row-echelon form of the augmented matrix \({[{\mathbf A}|{\mathbf b}]}\) has the form:

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

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

Exercise 9: Reduced row-echelon form

Question a

Give an example of a \(2 \times 3\) matrix \(\mathbf B\) with entries in \(\mathbb C\) that is in reduced row-echelon form, but such that the \(2 \times 2\) matrix we would obtain by deleting the first column of \({\mathbf B}\) is not in reduced row-echelon form.

Answer

There are many possible answers. One of them is

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

Question b

Let \(m\) and \(n\) be natural numbers and \({\mathbf B}\) a given \(m \times (n+1)\) matrix. We are informed that \(\mathbf B\) is in reduced row-echelon form. Show that the \(m \times n\) matrix \(\mathbf C\) you obtain by deleting the last column in \({\mathbf B}\) is also in reduced row-echelon form. Conclude that if \({\mathbf B}\) is the reduced row-echelon form of an augmented matrix \([\mathbf A | \mathbf b]\) of a system of linear equations, then the reduced row-echelon form of the coefficient matrix \(\mathbf A\) is obtained by deleting the last column in \(\mathbf B\).

Hint

A matrix in reduced row-echelon form satisfies the four requirements in Definition 6.3.2 in the textbook. Convince yourself that if \({\mathbf B}\) satisfies these requirements, then \({\mathbf C}\) does so too.

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