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Vectors and Matrices

Session Preparation:

Brooks: Chapter 8

Some of the exercises may require you to use Python. You may also need to install the numpy and sympy libraries if you haven't already.

Resources Danish Class:

Session notes

Session Resources

Exercises

Exercise 1: Recap

You are given the augmented matrix

\(A =\begin{bmatrix} 1 & -2 & 0 & -1 & 1\\ -4 & 8 & 2 & 4 & -8\\ -2 & 4 & 0 & 1 & -2 \end{bmatrix}\)​

  1. What is the size of the matrix? (1)

    1. \(3\times5\)

  2. Reduce the matrix to echelon form (either by hand or by doing seperate row operations in Python)

     

    Note. There are many ways to reduce the matrix to echelon form. Here is one way to do it.

    \(\begin{bmatrix} 1 & -2 & 0 & -1 & 1 \\ 0 & 0 & 2 & 0 & -4 \\ 0 & 0 & 0 & -1 & 0 \end{bmatrix}\)

  3. Use the echelon form to find the basic variables and free variables

     

    The basic variables are \(x_1\), \(x_3\) and \(x_4\) and the free variable is \(x_2\)

  4. Find the general solution to the system and write the basic variables in terms of the free variables

     

    Express \(x_1, x_3\), and \(x_4\) in terms of the free variable \(x_2\) : 1. From Row 1: \(x_1-2 x_2=1 \Rightarrow x_1=2 x_2+1\) 2. From Row 2: \(x_3=-2\) 3. From Row 3: \(x_4=0\)

  5. Mark each of the following statements as TRUE or FALSE

    (i) The matrix A describes a system of five equations and three variables (1)

    1. FALSE

    (ii) The matrix A describes a consistent system of equations (1)

    1. TRUE

    (iii) A 3x5 augmented matrix can not have a unique solution (1)

    1. TRUE

    (iv) A 3x5 augmented matrix will always be consistent (1)

    1. FALSE

Exercise 2: Matrix and vector equations

For each of the following systems of linear equations write it as (i) an augmented matrix (ii) a vector equation and (iii) a matrix equation.

  1. \(\begin{cases} 2x_1 +3x_2 - 4x_3 =8\\ -x_2 +x_3 + 2x_4 = -2\\ -x_1 + 2x_4 = 3 \end{cases}\)

     

    (i) Augmented matrix: \(\begin{bmatrix} 2 & 3 & -4 & 0 & 8\\ 0 & -1& 1 & 2 & -2\\ -1& 0 & 0 & 2 & 3 \end{bmatrix}\)

    (ii) Vector equation: \(x_1\begin{bmatrix}2\\\ 0\\\ -1\end{bmatrix} + x_2\begin{bmatrix}3\\\ -1\\\ 0\end{bmatrix}+ x_3\begin{bmatrix}-4\\\ 1\\\ 0\end{bmatrix}+ x_4\begin{bmatrix}0\\\ 2\\\ 2\end{bmatrix}=\begin{bmatrix}8\\\ -2\\\ 3\end{bmatrix}\)

    (iii) Matrix equation: \(\begin{bmatrix} 2 & 3 & -4 & 0 \\ 0 & -1& 1 & 2 \\ -1& 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix} 8\\-2\\3 \end{bmatrix}\)

  2. \(\begin{cases} x_1 + x_2 = 2\\ x_1 -x_2 = 4 \end{cases}\)

     

    (i) Augmented matrix: \(\begin{bmatrix} 1 & 1 & 2\\ 1 & -1& 4 \end{bmatrix}\)

    (ii) Vector equation: \(x_1\begin{bmatrix}1\\1\end{bmatrix}+ x_2\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}2\\4\end{bmatrix}\)

    (iii) Matrix equation: \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \begin{bmatrix} 2\\4 \end{bmatrix}\)

  3. \(\begin{cases} x_1 = 4\\ x_2 = -1\\ x_3 = 10 \end{cases}\)

     

    (i) Augmented matrix: \(\begin{bmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 10 \end{bmatrix}\)

    (ii) Vector equation: \(x_1\begin{bmatrix}1\\0\\0\end{bmatrix}+ x_2\begin{bmatrix}0\\1\\0\end{bmatrix}+x_3\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}4\\-1\\10\end{bmatrix}\)

    (iii) Matrix equation: \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 4\\-1\\10 \end{bmatrix}\)

Exercise 3: Linear combinations

Let

\(\mathbf{v}_1=\begin{bmatrix}1\\ -2\\ 4\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}3\\ 0\\ 2\end{bmatrix}\) and \(\mathbf{v}_3=\begin{bmatrix}-1\\ 5\\ 1\end{bmatrix}\).

Calculate the linear combinations below.

  1. \(2\mathbf{v}_1+3\mathbf{v}_2+\mathbf{v}_3\)

     

    \(\begin{bmatrix}10\\1\\15\end{bmatrix}\)

  2. \(\mathbf{v}_1-\mathbf{v}_3\)

     

    \(\begin{bmatrix}2\\-7\\3\end{bmatrix}\)

  3. \(\frac{1}{2}\mathbf{v}_1+\frac{3}{2}\mathbf{v}_2+\mathbf{v}_3\)

     

    \(\begin{bmatrix}4\\4\\6\end{bmatrix}\)

  4. Find the solution to the linear system described by the augmented matrix

    \[ \begin{bmatrix} 1 & 3 & -1 & 10 \\ -2 & 0 & 5 & 1 \\ 4 & 2 & 1 & 15 \end{bmatrix} \]
     

    \(\begin{cases} x_1 = 2\\ x_2 = 3\\ x_3 = 1 \end{cases}\)

Exercise 4: Parametric vector form

In the following exercises, the solution to a set of linear equations has already been found. Your job is to write the solution on parametric vector form.

  1. A system has been found to have the solution

    \[ \begin{aligned} x_1 &= 4 - x_2 \\ x_2 &= x_2 \\ x_3 &= -1 + 3x_2 \end{aligned} \]

    where \(x_2\) is a free variable. Write the solution in parametric vector form.

     

    \(\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{c} 4 \\ 0 \\ -1 \end{array}\right]+x_2\left[\begin{array}{c} -1 \\ 1 \\ 3 \end{array}\right]\)

  2. A system of equations has been found to have the solution

    \[ \begin{aligned} x_1 &= 5 + 4x_4 \\ x_2 &= 2 \\ x_3 &= x_3 \\ x_4 &= x_4 \\ x_5 &= -8 + x_3 - 7x_4 \end{aligned} \]

    where \(x_3\) and \(x_4\) are free variables. Write the solution in parametric vector form.

     

    \(\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{array}\right]=\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 0 \\ -8 \end{array}\right]+x_3\left[\begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \\ 1 \end{array}\right]+x_4\left[\begin{array}{c} 4 \\ 0 \\ 0 \\ 1 \\ -7 \end{array}\right]\)

Exercise 5: Homogenous equation

Solve the homogenous equation \(A\mathbf{x}=\mathbf{0}\) for the following matrices. Write the solution in parametric vector form.

  1. \(\begin{bmatrix} 4 & -8\\ -3 & 6\end{bmatrix}\)

     

    \(\mathbf{x}=x_2\left[\begin{array}{l}2 \\ 1\end{array}\right]\)

  2. \(\begin{bmatrix} 4 & -9 & 1 \\ 2 & -5 & 1 \\ -3 & 1 & 5 \end{bmatrix}\)

     

    \(\mathbf{x}=x_3\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\)

  3. \(\begin{bmatrix} 2 & -10 & 6 \\ 1 & -5 & 3 \end{bmatrix}\)

     

    \(\mathbf{x}=x_2\left[\begin{array}{l}5 \\ 1 \\ 0\end{array}\right]+x_3\left[\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right]\)

    \(\mathbf{x}=x_2\left[\begin{array}{l}5 \\\ 1 \\\ 0\end{array}\right]+x_3\left[\begin{array}{c}-3 \\\ 0 \\\ 1\end{array}\right]\)

  4. For the exercises a)-c) give a geometric interpretation of the solution set.

     

    a) The solution set is a line in \(\mathbb{R}^2\), as it has one free variable, representing a one-dimensional subspace.
    b) The solution set is a line in \(\mathbb{R}^3\), as it has one free variable, representing a one-dimensional subspace.
    c) The solution set is a plane in \(\mathbb{R}^3\), as it has two free variables, representing a two-dimensional subspace.

Exercise 6: Linear Independence

  1. For \(\mathbf{a}_1 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}\), \(\mathbf{a}_2 = \begin{bmatrix} 1\\ 2\\ -5 \end{bmatrix}\), \(\mathbf{a}_3 = \begin{bmatrix} 4\\ 1\\ 2 \end{bmatrix}\), and \(\mathbf{b} = \begin{bmatrix} 8\\ 6\\ 12 \end{bmatrix}\) determine if \(\mathbf{b}\) is a linear combination of \(\mathbf{a}_1,\mathbf{a}_2\) and \(\mathbf{a}_3\).

     

    Yes, since \(\mathbf{b}=3\mathbf{a}_1-2\mathbf{a}_2+\mathbf{a}_3\)

  2. Let \(A = \begin{bmatrix}1 & -2 & 4 \\ 0 & 4 & -5 \\ -3 & 6 & -12 \end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}1 \\ -4 \\ -1\end{bmatrix}\) be given. Determine whether \(\mathbf{b}\) is a linear combination of the columns in \(A\).

     

    No, \(\mathbf{b}\) is not a linear combination of the columns of \(A\)

  3. For \(A = \begin{bmatrix}2 & -6 & 5 \\ 1 & -5 & 1 \\ -2 & 6 & p \end{bmatrix}\) and \(\mathbf{b}= \begin{bmatrix}3 \\ 0 \\ q\end{bmatrix}\) determine the values of \(p\) and \(q\) such that \(\mathbf{b}\) is not a linear combination of the columns of \(A\).

     

    \(p=-5\) and \(q\neq-3\)