Matrices and Vectors - Pre-Calculus

By definition, an inverse matrix is the matrix B that you would need to multiply matrix A by to get the identity. Since the identity matrix yields whatever matrix it is being multiplied by, the answer is the identity itself.

First, draw the vector = (3, 4); this is represented in red below. Then, plot the point P1 (1, 4), and draw a line (represented in blue) through it that is parallel to the vector .

We must find the equation of line . For any point P2 (x, y) on , . Since is on line and is parallel to , for some value of t. By substitution, we have . Therefore, the equation is a vector equation describing all of the points (x, y) on line parallel to through P1 (1, 4).

This is true. The independent variable in this equation is called a parameter.

A line through a point (x1,y1) that is parallel to the vector = (a1, a2) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 3t

y = 5 + 2t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

A line through a point (x1,y1) that is parallel to the vector = (a1, a2) has the following parametric equations, where t is any real number.

Using the given vector and point, we get the following:

x = 4 - 7t

y = -3 + 3.5t

Each value of t creates a distinct (x, y) ordered pair. You can think of these points as representing positions of an object, and of t as representing time in seconds. Evaluating the parametric equations for a value of t gives us the coordinates of the position of the object after t seconds have passed.

In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = 5t - 3

In the equation y = -3x +1.5, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = -3t +1.5

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 6:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

Start by solving each parametric equation for t:

Next, write an equation involving the expressions for t; since both are equal to t, we can set them equal to one another:

Multiply both sides by the LCD, 4:

Get y by itself to represent this as an equation in slope-intercept (y = mx + b) form:

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

As you set up the equation for y, use the value g = -32.

Finally, find x and y when t = .05:

Use a calculator to solve, making sure you are in degree mode:

This means that after 0.5 seconds, the ball has travelled 17.5 feet horizontally and 5.7 feet vertically.

To solve this problem, we need to know that the path of a projectile can be described with the following equations:

In these equations, t is time and g is the acceleration due to gravity.

First, you need to write the position of the ball as a pair of parametric equations that define the path of the ball at anytime, t, in seconds:

As you set up the equation for y, use the value g = -32.

Finally, find x and y when t = .05:

To find the height that the ball will be at when it has travelled horizontally, plug 400 feet (horizontal distance) in for x.

This tells us that the ball has travelled 400 feet horizontally when 2.84 seconds have passed. Let's use the value of t = 2.84 and plug it into our equation for y to see how high the ball is at this time:

This means that when the ball has travelled 400 feet horizontally, 2.84 seconds have passed, and the ball is 16.65 feet above the ground.

Note the first and the last columns are equal.

Therefore, when we try to find the determinant using the following formula we get the determinant equaling 0:

This means simply, that the matrix does not have an inverse.

For a 2x2 matrix

the inverse can be found by

Because the determinant is equal to zero in this problem, or

,

the inverse does not exist.

There are a couple of ways to do this. I will use the determinant method.

First we need to find the determinant of this matrix, which is

for a matrix in the form:

.

Substituting in our values we find the determinant to be:

Now one formula for finding the inverse of the matrix is

.

We use the inverse of a 2x2 matrix formula to determine the answer. Given a matrix

it's inverse is given by the formula:

First we define the determinant of our matrix:

Then,

This matrix has no inverse because the columns are not linearly independent. This means if you row reduce to try to compute the inverse, one of the rows will have only zeros, which means there is no inverse.

By writing the augmented matrix , and reducing the left side to the identity matrix, we can implement the same operations onto the right side, and we arrive at , with the right side representing the inverse of the original matrix.

It will be helpful to first draw the vector so we can see what quadrant the angle is in:

Since the vector is pointing up and to the right, it is in the first quadrant. To determine the angle, set up a trig equation with tangent, since the component 5 is opposite and the component 4 is adjacent to the angle we are looking for:

to solve for theta, take the inverse tangent of both sides:

Now we have the direction, and we can solve for the magnitude using Pythagorean Theorem:

take the square root of both sides

The vector in polar form is

To convert to polar form, we need to find the magnitude of the vector, , and the angle it forms with the positive -axis going counterclockwise, or . This is shown in the figure below.

We find the angle using trigonometric identities:

Using a calculator,

To find the magnitude of a vector, we add up the squares of each component and take the square root:

.

So, our vector written in polar form is

To convert a point or a vector to its polar form, use the following equations to determine the magnitude and the direction.

Substitute the vector to the equations to find the magnitude and the direction.

The polar form is: