*John Hamilton*Show bio

John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.

Instructor:
*John Hamilton*
Show bio

John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.

The following homeschool assignment deals with several precalculus concepts, including vectors on the plane, solving linear systems, inverse matrices, and Cramer's rule. The assignment is appropriate for the 12th grade student.
Updated: 05/12/2021

Are your students determined to learn about determinants? Am I determined to annoy you with more of my lame math jokes? Of course! Anyway, **vectors** are often used to represent position, while **matrices and determinants** can be useful for solving linear equations.

In addition to what is mentioned above, this assignment will include augmented matrices.

By the end of this assignment, students will have completed three steps, solved three problems, and turned in an appropriate presentation.

Note: The answers with explanations are situated at the bottom of the page.

- For even further assistance, be sure to go online and visit our outstanding High School Precalculus: Homeschool Curriculum.

**Augmented matrix:**used to solve linear equations**Higher-order determinants:**involve 3 x 3 and larger matrices**Inverse matrix:**multiplies by a given matrix to obtain the identity matrix

- Graph paper
- Internet capability
- Paper
- Pencil
- Ruler

Time / Length

- Five days to complete this mathematics assignment
- Ten days to turn in an appropriate presentation

Let's get started! Are you ready to rumble? As you may know, I never get tired of that cool wrestling expression. Anyway, let's **perform operations on vectors in the plane**. The good news is adding two vectors is pretty much the same as adding two numbers.

**Problem # 1:**

Add vector *A* (2, 3) and vector *B* (5, 6).

Okay, that was pretty easy, so now let's **find the dot product of vectors**.

**Problem # 2:**

Find the dot product given both a vector's beginning points and end points.

Vector *A* is in Quadrant I and moves from (0, 0) to (4, 7).

Vector *B* is in Quadrant II and moves from (0, 0) to (-8, 9).

We can't take the determinant of a rectangular matrix, but we can take the determinant of a square matrix.

**Problem # 3:**

Find the determinant of:

Next, let's write an augmented matrix for a linear system:

**Example # 1:**

*x* + *y* = 12

3*x* - 2*y* = 16

That wasn't too difficult, was it? Notice each vertical line represents an equal sign, and take special care that your number 2 has a negative sign.

**Example # 2:**

Now, let's perform math operations with matrices. If two matrices are the same size, it's pretty easy to add or subtract them, such as:

**Example # 3:**

However, multiplying or dividing matrices is a bit more involved:

We just learned how to solve the determinant for a 2 x 2 matrix, but how do we evaluate higher-order determinants? Take a deep breath! This is a bit challenging.

Your formula for the determinant of a given 3 x 3 matrix is:

Which equals:

*a*(*e**i*-*f**h*) - *b*(*d**i*-*f**g*) + *c*(*d**h*-*e**g*)

Whew! I told you to take a deep breath. However, just plug any set of numbers into that formula, and follow the procedure exactly.

Let's solve a linear system using Gaussian elimination:

**Example # 4:**

3*x* + 2*y* = 7

*x* - *y* = -1

Write your augmented matrix:

3 2 | 7

1 -1 | -1

Switch your rows:

1 -1 | -1

3 2 | 7

Multiply row 1 by -3 and add the result to row 2:

1 -1 | -1

0 5 | 10

Multiply row 2 by 1/5:

1 -1 | -1

0 1 | 2

We're almost there! What was the whole point of those computations? Did you notice we now have all ones on the diagonal, and a zero in the bottom left corner? Good! These two equations really represent in linear form:

(1)*x* (-1 )*y* = (-1)

0*x* (+1)*y* = (2)

Or:

*x* - *y* = -1

*y* = 2

Lastly, substitute your 2nd equation into your 1st equation:

*x* - (2) = -1

*x* = 1

Your answer is:

(1, 2)

Check your answer:

Plug (1, 2) back into your original equations:

3*x* + 2*y* = 7

*x* - *y* = -1

3(1) + 2(2) = 7

(1) - (2) = -1

7 = 7

-1 = -1

Hey, it worked again! We are so good at this precalculus. Let's move on to calculus. Okay, not right now, but maybe next semester.

How do we solve systems of linear equations using determinants? Cramer's rule can be applied.

Let's say you are given this system of linear equations:

**Example # 5:**

2*x* + *y* = 80

*x* - 3*y* = -30

Now we utilize Cramer's rule.

Your coefficient determinant *D* would be:

= 2 * -3 - 1 * 1

= -6 - 1

= -7

Next, your *D**x* determinant is found by substituting your constant numbers into your first column:

See what we did there? We replaced the 2 in your matrix with an 80, and the 1 in your matrix with a -30.

So:

80 * -3 - 1 * -30 =

-240 - -30 =

-210

Now we need to find our *D**y* determinant. You guessed it! We substitute the constant numbers into the second column.

So:

2 * -30 - 80 * 1 =

-60 - 80 =

-140

Let's recap:

*D* = -7

*D**x* = -210

*D**y* = -140

We're almost done!

*x* = *D**x* / *D*

*y* = *D**y* / *D*

Therefore:

*x* = -210 / -7

*y* = -140 / -7

And:

*x* = 30

*y* = 20

Check your answer by plugging back into your original equations:

2*x* + *y* = 80

*x* - 3*y* = -30

2(30) + (20) = 80

(30) - 3(20) = -30

80 = 80

-30 = -30

We're so good at this precalculus thing it's scary!

Hey! You know what would be fun? Binge-watching Netflix for eight straight hours would be fun. Instead, we're going to apply Cramer's rule as it relates to inconsistent and dependent systems. Isn't that great? Hello? Wake up, student. Here we go!

Well, here's the catch. The term **inconsistent systems** refers to those which contain no solutions. The term **dependent systems** refers to those with an infinite number of solutions.

Finally, as a matter of fact, you actually can't use Cramer's rule for these two systems, because your coefficient determinant would be zero, and of course you can't divide by zero. That's all you really need to know about this particular subject.

Now it's time to demonstrate your command of this precalculus material via a quality presentation:

Create a ''how-to'' guide explaining these math processes and skills, which could be in the format of a booklet, poster, wall mural, manual, a YoutTube video or a TED Talk podcast.

**Solution # 1:**

Add vector *A* (2, 3) and vector *B* (5, 6).

Well, vector *A* + *B* =

(2 + 5, 3 + 6) =

(7, 9)

**Solution # 2:**

Find the dot product given both a vector's beginning points and end points.

Vector *A* is in Quadrant I and moves from (0, 0) to (4, 7).

Vector *B* is in Quadrant II and moves from (0, 0) to (-8, 9).

4 * -8 + 7 * 9 =

Do you see how you multiplied your *x*-values, multiplied your *y*-values, and then added them together? Good! That's pretty simple, isn't it?

-32 + 63 =

31

**Solution # 3:**

Find the determinant of:

Our formula for finding the determinant of:

Is:

*a* * *d* - *b* * *c*

So:

9 * 2 - 1 * 8 =

18 - 8 =

10

Requirements | 0 - 5 points |
---|---|

Problem # 1 is solved correctly | |

Problem # 2 is solved correctly | |

Problem # 3 is solved correctly | |

Deliverable is clear and concise | |

Total: |
/ 20 points |

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