Matrix Algebra and Determinants

Session Preparation

Brooks: Chapter 9

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

Python solutions will be available after the session. [Python] Means that you are advised to use Python to solve the exercise. You can find the Python solutions here or download them here

Exercise 1

1. Determine by inspection whether the following sets of vectors are linearly independent or dependent. Justify each answer.

   i. \(\left[\begin{array}{l}5 \\ 1\end{array}\right],\left[\begin{array}{l}2 \\ 8\end{array}\right],\left[\begin{array}{l}1 \\ 3\end{array}\right],\left[\begin{array}{r}-1 \\ 7\end{array}\right]\) (1)

   1. Dependent

   ii. \(\left[\begin{array}{r}2 \\ -4 \\ 8\end{array}\right],\left[\begin{array}{r}-3 \\ 6 \\ -12\end{array}\right]\) (1)

   1. Dependent

   iii. \(\left[\begin{array}{r}5 \\ -3 \\ -1\end{array}\right],\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right],\left[\begin{array}{r}-7 \\ 2 \\ 4\end{array}\right]\) (1)

   1. Dependent

   iv. \(\left[\begin{array}{l}3 \\ 4\end{array}\right],\left[\begin{array}{r}-1 \\ 5\end{array}\right],\left[\begin{array}{l}3 \\ 5\end{array}\right],\left[\begin{array}{l}7 \\ 1\end{array}\right]\) (1)

   1. Dependent

   v. \(\left[\begin{array}{r}-8 \\ 12 \\ -4\end{array}\right],\left[\begin{array}{r}2 \\ -3 \\ -1\end{array}\right]\) (1)

   1. Independent

   vi. \(\left[\begin{array}{r}1 \\ 4 \\ -7\end{array}\right],\left[\begin{array}{r}-2 \\ 5 \\ 3\end{array}\right],\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]\) (1)

   1. Dependent

2. Let \(A = \left[\begin{array}{cccccc}1 & -4 & -2 & 0 & 3 & -5 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\)

   Describe all solutions to \(A \mathbf{x}=\mathbf{0}\) in parametric form.

    

   The solution in parametric form is:

   \[ \mathbf{x} = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{array}\right] = x_2 \left[\begin{array}{c} 4 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right] + x_4 \left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right] + x_6 \left[\begin{array}{c} -5 \\ 0 \\ 1 \\ 0 \\ 4 \\ 1 \end{array}\right] \]

   where \(x_2\), \(x_4\), and \(x_6\) are free variables.

3. [Python] Use as many columns of \(A\) as possible to construct a matrix \(B\) with the property that the equation \(B \mathbf{x}=\mathbf{0}\) has only the trivial solution. Solve \(B \mathbf{x}=\mathbf{0}\) to verify your work.

\(A=\left[\begin{array}{rrrrr}3 & -4 & 10 & 7 & -4 \\ -5 & -3 & -7 & -11 & 15 \\ 4 & 3 & 5 & 2 & 1 \\ 8 & -7 & 23 & 4 & 15\end{array}\right]\)

 

\(\left[\begin{array}{ccc}3 & -4 & 7 \\ -5 & -3 & -11 \\ 4 & 3 & 2 \\ 8 & -7 & 4\end{array}\right]\)

Exercise 2

In (a) and (b) compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\[ \begin{aligned} & A=\left[\begin{array}{rrr} 2 & 0 & -1 \\ 4 & -5 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 7 & -5 & 1 \\ 1 & -4 & -3 \end{array}\right], \\ & C=\left[\begin{array}{rr} 1 & 2 \\ -2 & 1 \end{array}\right], \quad D=\left[\begin{array}{rr} 3 & 5 \\ -1 & 4 \end{array}\right], \quad E=\left[\begin{array}{r} -5 \\ 3 \end{array}\right] \end{aligned} \]

1. \(-2 A, \; B-2 A, \; A C, \; C D\)

    

   \[-2 A=\left[\begin{array}{rrr}-4 & 0 & 2 \\ -8 & 10 & -4\end{array}\right]\]

   \[ B-2 A=\left[\begin{array}{rrr} 3 & -5 & 3 \\ -7 & 6 & -7 \end{array}\right] \]

   \(A C\) is not defined.

   \[C D=\left[\begin{array}{rr}1 & 13 \\ -7 & -6\end{array}\right]\]

2. [Python] \(A+3 B, \; 2 C-3 E, \; D B, \; E C\)

    

   \[ A+3 B=\left[\begin{array}{rrr} 23 & -15 & 2 \\ 7 & -17 & -7 \end{array}\right] \]

   \(2 C-3 E\) is not defined.

   \[ D B=\left[\begin{array}{rrr} 26 & -35 & -12 \\ -3 & -11 & -13 \end{array}\right] \]

   \(E C\) is not defined.

3. [Python] Let \(A=\left[\begin{array}{rr}3 & -6 \\ -1 & 2\end{array}\right], B=\left[\begin{array}{rr}-1 & 1 \\ 3 & 4\end{array}\right]\), and \(C= \left[\begin{array}{rr}-3 & -5 \\ 2 & 1\end{array}\right]\). Verify that \(A B=A C\) and yet \(B \neq C\).

    

   Since \( AB = \left[\begin{array}{rr} -21 & -21 \\ 7 & 7 \end{array}\right] = AC \), we conclude \( AB = AC \).

   Thus, \( AB = AC \) but \( B \neq C \).

4. [Python] If \(A=\left[\begin{array}{rr}1 & -3 \\ -3 & 5\end{array}\right]\) and \(A B=\left[\begin{array}{rr}-3 & -11 \\ 1 & 17\end{array}\right]\), determine the matrix \(B\).

    

   \(\left[\begin{array}{ll}3 & 1 \\ 2 & 4\end{array}\right]\)

Exercise 3

1. Find the inverses of \(\begin{array}{cc}{\left[\begin{array}{l}8 \\ 5\end{array}\right.} & \left.\begin{array}{c}6 \\ 4\end{array}\right]\end{array}\) and \(\begin{array}{cc}{\left[\begin{array}{l}3 \\ 8\end{array}\right.} & \left.\begin{array}{c}2 \\ 5\end{array}\right]\end{array}\).

    

   \(\left[\begin{array}{ll}8 & 6 \\ 5 & 4\end{array}\right]^{-1}=\frac{1}{32-30}\left[\begin{array}{rr}4 & -6 \\ -5 & 8\end{array}\right]=\left[\begin{array}{cr}2 & -3 \\ -5 / 2 & 4\end{array}\right]\)

   \(\left[\begin{array}{ll}3 & 2 \\ 8 & 5\end{array}\right]^{-1}=\frac{1}{15-16}\left[\begin{array}{rr}5 & -2 \\ -8 & 3\end{array}\right]=\left[\begin{array}{cc}-5 & 2 \\ 8 & -3\end{array}\right]\)

2. [Python] Use the inverse found in Exercise 3.a to solve the system

   \(\begin{aligned} & 8 x_1+6 x_2=2 \\ & 5 x_1+4 x_2=-1 \end{aligned}\)

    

   Solution: \(x_1=7\) and \(x_2=-9\).

3. [Python] Find the inverse of the matrix. Use the algorithm for finding the inverse of a \(n \times n\) matrix.

   \(\left[\begin{array}{rrr} 1 & 0 & -2 \\ -3 & 1 & 4 \\ 2 & -3 & 4 \end{array}\right]\)

    

   \[\left[\begin{array}{ccc} 8 & 3 & 1 \\ 10 & 4 & 1 \\ 7 / 2 & 3 / 2 & 1 / 2 \end{array}\right] \]

Exercise 4

In this exercise the matrices are all \(n \times n\). Each part of the exercise is an implication of the form "If (statement l), then (statement 2)." Mark an implication as True if the truth of (statement 2) always follows whenever ( statement 1) happens to be true. An implication is False if there is an instance in which ( statement 2) is false but (statement l) is true. Justify each answer.

1. If the equation \(A \mathbf{x}=\mathbf{0}\) has only the trivial solution, then \(A\) is row equivalent to the \(n \times n\) identity matrix. (1)

   1. True

2. If the columns of \(A\) span \(\mathbb{R}^n\), then the columns are linearly independent. (1)

   1. True

3. If \(A\) is an \(n \times n\) matrix, then the equation \(A \mathbf{x}=\mathbf{b}\) has at least one solution for each \(\mathbf{b}\) in \(\mathbb{R}^n\). (1)

   1. False

4. If the equation \(A \mathbf{x}=\mathbf{0}\) has a nontrivial solution, then \(A\) has fewer than \(n\) pivot positions. (1)

   1. True

5. If \(A^T\) is not invertible, then \(A\) is not invertibl5. (1)

   1. True

Exercise 5

[Python] Given

\(A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -3\end{array}\right]\) and \(B=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\).

Solve the matrix equation \(X+A=2(X-B)\).

 

\(X=\left[\begin{array}{ccc}3 & 2 & 2 \\ 2 & 5 & 2 \\ 2 & 2 & -1\end{array}\right]\)

Exercise 6

[Python] Let the matrix \(A\) be given by

\(A=\left[\begin{array}{cc} 3-2 q & 1 \\ 4 & 3+2 q \end{array}\right]\)

where \(q\) is a scalar. Calculate \(q\) so that

\(A^2=\left[\begin{array}{cc} 29 & 6 \\ 24 & 125 \end{array}\right]\)

 

\(q=4\)