All posts by Mr Hartzer

Delta Math

All students are expected to join the appropriate class on Delta Math. There will be assignments; students are also encouraged to use the resources for additional study.

To join your class, go to https://www.deltamath.com/

If you have an existing account, log on and add my teacher code; if you don’t, create one. The teacher code is 559492

There is an assignment that is automatically assigned when you join. To receive full credit, complete a total of 10 total problems (of 16 assigned). Additional exercises earn extra credit. However, only the first four problems you complete in each section will be counted. This assignment is due Monday, October 16.

Khan Academy

All students are expected to join the Khan Academy Algebra II class. There will be assignments; students are also encouraged to use this resource for further study.

To join Mr. Hartzer’s class, login at https://www.khanacademy.org and select “Coaches”. Then enter this under “Join a class”: VR3JMK94

Note that you will not automatically receive assignments that were given before you joined; I’ll have to assign them to you manually.

The first required assignment is on matrices. You must complete a total of four points worth of exercises for full credit; additional exercises earn extra credit! Note that for this assignment, you must complete the exercise, but you don’t have to get 100%.

This is due Friday, October 13.

1 point exercises
Matrix Dimensions
Matrix Elements
Add & Subtract Matrices

2 point exercises
Multiply Matrices by Scalars
Matrix Equations: Addition & Subtraction

4 point exercise
Multiply Matrices

Multiplying Matrices

Multiplying matrices can be confusing, but if you’re organized and disciplined, it’s not difficult. Just make sure to keep things straight.

Let’s multiply two matrices: \[\color{red}{A = \begin{bmatrix} 5 & 1 & 3 \\ 4 & 2 & -1 \end{bmatrix}} \quad \color{blue}{B = \begin{bmatrix} 8 & 11 \\ -6 & 7 \\ 0 & 9 \end{bmatrix}}\]

First, think about \(C = AB\). Can we multiply this? We start by deciding what the size of the product matrix \(C\) will be. Since the dimensions of \(A\) are \(2 \times 3\) and the dimensions of \(B\) are \(3 \times 2\), the dimensions of \(C\) will be \((2 \times 3)(3 \times 2) = 2 \times 2\).

We can only create a product matrix if the number of columns of the first matrix is the same as the number of rows in the second matrix. The size of the product matrix will be the number of rows of the first matrix and the number of columns of the second matrix.

We create a blank 2 x 2 matrix, with enough room to fill in the values: \[\color{green}{C = \begin{bmatrix} \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ \end{bmatrix}}\]

We look at the first matrix (\(A\)) in terms of rows. We will put the values from the first row of \(A\) in the first row of \(C\), and the same for the second row. Since there are three elements in each row of \(A\), we will create three terms in each element of \(C\). That is: \[\color{green}{C = \begin{bmatrix} \color{red}{5}\cdot\_+\color{red}{1}\cdot\_+\color{red}{3}\cdot\_ & \color{red}{5}\cdot\_+\color{red}{1}\cdot\_+\color{red}{3}\cdot\_ \\ \color{red}{4}\cdot\_+\color{red}{2}\cdot\_+\color{red}{-1}\cdot\_ & \color{red}{4}\cdot\_+\color{red}{2}\cdot\_+\color{red}{-1}\cdot\_ \end{bmatrix}}\]

Make sure to leave the blanks! At this point, we’re focusing only on the first matrix, but we want to make sure to leave space for the numbers from the second matrix.

Now we look at the second matrix (\(B\)) in terms of columns. We will put the values from the first column of \(B\) in the first column of \(C\), and so on for the other column. This will give us: \[\color{green}{C = \begin{bmatrix} \color{red}{5}\cdot\color{blue}{8}+\color{red}{1}\cdot\color{blue}{-6}+\color{red}{3}\cdot\color{blue}{0} & \color{red}{5}\cdot\color{blue}{11}+\color{red}{1}\cdot\color{blue}{7}+\color{red}{3}\cdot\color{blue}{9} \\ \color{red}{4}\cdot\color{blue}{8}+\color{red}{2}\cdot\color{blue}{-6}+\color{red}{-1}\cdot\color{blue}{0} & \color{red}{4}\cdot\color{blue}{11}+\color{red}{2}\cdot\color{blue}{7}+\color{red}{-1}\cdot\color{blue}{9} \end{bmatrix}}\]

To reiterate: We fill in the values from each ROW of \(A\) in every element of the matching ROW of \(C\), and the values from each COLUMN of \(B\) in every element of the matching COLUMN of \(C\). At this point, let’s get rid of the color and see what we have: \[C = \begin{bmatrix} 5\cdot8+1\cdot-6+3\cdot0 & 5\cdot11+1\cdot7+3\cdot9 \\ 4\cdot8+2\cdot-6-1\cdot0 & 4\cdot11+2\cdot7-1\cdot9 \end{bmatrix}\]

Finally, we evaluate each element of the matrix for our solution: \[C = \begin{bmatrix} 34 & 89 \\ 20 & 49 \end{bmatrix}\]

Let’s try it the other way: \(D = BA\). What will be the size of \(D\)? Since the dimensions of \(B\) are \(3 \times 2\) and the dimensions of \(A\) are \(2 \times 3\), the dimensions of \(D\) will be \((3 \times 2)(2 \times 3) = 3 \times 3\).

We create a blank 3 x 3 matrix: \[\color{purple}{D = \begin{bmatrix} \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \\ \_\_\_\_ & \_\_\_\_ & \_\_\_\_ \end{bmatrix}}\]

This time, we look at \(B\) in terms of rows. We put the values from the first row of \(B\) in the first row of \(D\), and so on for the other two rows, just as before. Since there are two elements in each row of \(B\), we will create two terms in eacn element of \(D\). That is: \[\color{purple}{D = \begin{bmatrix} \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ & \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ & \color{blue}{8}\cdot\_ + \color{blue}{11}\cdot\_ \\ \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ & \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ & \color{blue}{-6}\cdot\_ + \color{blue}{7}\cdot\_ \\ \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ & \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ & \color{blue}{0}\cdot\_ + \color{blue}{9}\cdot\_ \end{bmatrix}}\]

Now, we look at \(A\) in terms of columns, placing the values of each column in the blanks in \(D\): \[\color{purple}{D = \begin{bmatrix} \color{blue}{8}\cdot\color{red}{5} + \color{blue}{11}\cdot\color{red}{4} & \color{blue}{8}\cdot\color{red}{1} + \color{blue}{11}\cdot\color{red}{2} & \color{blue}{8}\cdot\color{red}{3} + \color{blue}{11}\cdot\color{red}{-1} \\ \color{blue}{-6}\cdot\color{red}{5} + \color{blue}{7}\cdot\color{red}{4} & \color{blue}{-6}\cdot\color{red}{1} + \color{blue}{7}\cdot\color{red}{2} & \color{blue}{-6}\cdot\color{red}{3} + \color{blue}{7}\cdot\color{red}{-1} \\ \color{blue}{0}\cdot\color{red}{5} + \color{blue}{9}\cdot\color{red}{4} & \color{blue}{0}\cdot\color{red}{1} + \color{blue}{9}\cdot\color{red}{2} & \color{blue}{0}\cdot\color{red}{3} + \color{blue}{9}\cdot\color{red}{-1} \end{bmatrix}}\]

Without the color, this is: \[D = \begin{bmatrix} 8\cdot 5 + 11\cdot 4 & 8\cdot 1 + 11 \cdot 2 & 8 \cdot 3 + 11 \cdot -1 \\ -6 \cdot 5 + 7 \cdot 4 & -6 \cdot 1 + 7 \cdot 2 &  -6 \cdot 3 +  7 \cdot -1 \\ 0 \cdot 5 + 9 \cdot 4 & 0 \cdot 1 + 9 \cdot 2 & 0 \cdot 3 + 9 \cdot -1 \end{bmatrix}\]

Evaluating each element gives us: \[D = \begin{bmatrix} 84 & 30 & 13 \\ -2 & 8 & -25 \\ 36 & 18 & -9 \end{bmatrix}\]

If you would like to check your work, here is an online matrix multiplication calculator. Make up some small matrices and practice multiplying them, then check your answer! (Remember: These tools are provided so you can check your work, not to help you cheat.)