Recall that we worked with systems of linear equations such as the following:

3x + 4y = -6

5x – 2y = 1

5x + 6y – z = 10

7y + 4z = 3
z = -1

We can use a matrix to rewrite these systems in a simpler way. We use what is called an augmented matrix, which has a vertical bar separating the columns of the matrix. The group on the left of the bar are the coefficients of each variable and the group to the right of the bar are the constants (the numbers after the = in each equation in the system). In the systems above, the augmented matrices are as follows:

To solve a system of linear equations in three variables, we wish to produce a matrix with 1's along the diagnoal from the upper left to the lower right of the matrix, with 0's undereath the 1's. Such a matrix will look as follows:

The letters a through f represent real numbers. Recall that the elements of the augemented matrix to the left of the vertical bar represent the coefficients of the variables. Therefore in the above augemented matrix, we can conclude that

x + ay + bz = c

y + dz = e
z = f

Since we know the value of z, we can subsitute that into y + dz = e to solve for y. Then we can substitute the values for y and z into x + ay + bz = c and solve for x.

When trying to remember the ratio for each trigonometric function,  sine, cosine, tangent, cotangent, secant and cosecant, we really only need to know sine, cosine and tangent. Once we know them, cotangent, secant and cosecant are the recipricol of tangent, cosine and sine, respectively.

In a right triangle, the sine of an angle is the length side opposite the angle divided by the length of the hypotenuse (the side opposite the right angle).

Cosine is the length of the side adjacent to the angle divided by the length of the hypotenuse.

Tangent is the length of the side opposite divided by the length of the side adjacent.

Sine = opposite/hypotenuse

Cosine  = adjacent/hypotenuse

Tangent = opposite/adjacent

Cotangent = 1/tangent

Secant = 1/cosine

Cosecant = 1/sine

Notice that tangent is sine/cosine

To answer yesterday's problem, have to know that tan b = sin b/ cos b

Therefore,


sin b/tan b = sin b/ (sin b/cos b)

                  =  sin b/1  * cos b/sin b

                   = cos b

This will be easy for those of you that know trigonometry.

When do we add exponents and when so we multiply exponents?

If we have a variable raised to a power times another variable raised to a power, we add the exponents.

For example,  x^2 * x^5 = x^7

If we have a variable raised to a power and that quantity is raised to another power, then we multiply exponents.

For example,  (x^2)^5 = x^10

It's easy to illustrate this.

x^2 * x^5

Think of x^2 as x * x and x^5 as x * x * x * x * x

Notice there are 7 x's total, therefore when multiplying it's x^7.

(x^2)^5

Think of x^2 * x^2 * x^2 * x^2 * x^2

and x^2 = x * x

Therefore we now have x * x * x * x * x * x * x * x * x * x = x^10

Remember to check out my book at lulu.com. The book is a self help Algebra book covering 1st and some 2nd year high school school algebra concepts. It's also good for a beginner college algebra course. The book is available as a pdf for pc viewing and in paperback.

http://www.lulu.com/spotlight/KKauffman1969

 I'll show you 2 different methods to multiply 2 numbers.

   23
x 15
 115    (multiply 5 by 3 to get 15, carry the one, 5 times 2 is 10 and 1 is 11)
 230   (add a zero under the 5 and multiply 1 by 3 to get 3 and 1 by 2 to get 2)
 345   (add the numbers)

This method is probably the one that the majority of people learned in school.

Here's an alternate method.  We will think of 23 as 20 and 3 and 15 as 10 and 5

Multiply 20 by 10 to get 200
Multiply 20 by 5 to get 100
Multiply 10 by 3 to get 30
Multiply 5 by 3 to get 15

Now we add the numbers to get 345.

Essentially we are using the distributive property twice.

 20(10 + 5) + 3(10+ 5)

200 + 100 + 30 + 15 = 345

What's interesting is we can also distribute as follows and get the same answer

10(20 + 3) + 5(20 + 3)

200 + 30 + 100 + 15 = 345.