9.1+Content+-+Matrix+Theory

Add Matrix Theory content to this page. See specific details on the home page. This unit has been split into parts A & B. Content between the two must be cohesive as it is one continuous, connected material. They are split simply to better organize the page.
 * Quick Instructions**

**Matrix Theory (Unit 9A) Learning Targets**

 * Decompose rational expressions into partial fractions.
 * Solve systems of equations using substitution, linear combination, inverse matrices and reduced row echelon form of matrices.
 * Perform matrix operations and find determinants and the inverse of matrices.

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Partial Fraction Decomposition
Partial fraction decomposition is generally used to separate a fraction into two parts which are called partial fractions

Given the fraction:

math \[\large \frac{3x+11}{x^{2}-2x-3}\] math

In order to begin partial fraction decomposition, one must factor the denominator

math \[\large \frac{3x+11}{(x-3)(x-1)}\] math Note: Partial fraction decomposition can only be used when the denominator is factored into **unique**, **linear** factors

Afterward, the factored fraction is divided into two parts, using variables in place of the numerators.

math \[\frac{A}{x-3}+\frac{B}{x+1}\] math Where A and B are the numerators of the fractions that will be solved for in the next step.

Next, a method called "//Cover Up//" is used to solve for the numerators. First, the denominator of the first fraction is set equal to 0 and solved for. This solution is then substituted for the x in the //**original**// fraction (before it was divided into two parts). However, not all of the fraction is used as the term we previously solved for is "covered up" or ignored. The result is the numerator of the first partial fraction. Therefore:

math \[x-3=0\]\\ math math \[x=3\] math math \[\LARGE \frac{3(3)+11}{3+1}=5\] math Thus, A=5

Afterward, the process is repeated for the second partial fraction.

math \[x+1=0\]\\ math math \[x=-1\] math math \[\LARGE \frac{3(-1)+11}{-1-3}=-2\] math

Thus, B=-2 As a result, we see that:

math \[\LARGE \frac{3x+11}{x^{2}-2x=3}=\frac{5}{x-3}-\frac{2}{x+1}\] math

The following link is a great resource for extra information and practice regarding partial fraction decomposition: http://www.purplemath.com/modules/partfrac.htm

What is a Matrix?
A matrix is defined as "a rectangular array of numbers, symbols, or expressions." Usually it is a set of numbers organized into rows and colums and enclosed in brackets. The individual items in a matrix are called //entries//. Here is an example of a matrix:

math \left[ {\begin{array}{*{20}{c}} 3&4 \\ 9&0 \end{array}} \right] math This matrix has four entries, grouped into two columns and two rows. Because of this, it is known as a 2x2 matrix. When labeling a matrix in this way, take the number of rows and multiply it by the number of columns. To create a matrix on a TI-84, go to second and hit the x inverse button. Then you can go to edit and change set matrices, and then enter them in the column titled name.



On a TI-89, use brackets for the edges of the matrices and commas for each number in a row. To start a new row, use brackets again. Afterward, end the first bracket.



Matrix Addition
Adding matrices together is fairly straightforward. Simply add the entries in corresponding locations:

math \left[ {\begin{array}{*{20}{c}} 3&4 \\ 9&0 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 8&3 \\ 5&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&7 \\ {14}&6 \end{array}} \right] math If both matrices do not have the same dimensions, addition is not possible:

math \left[ {\begin{array}{*{20}{c}} 3&4 \\ 9&0 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} \begin{gathered} 6 \hfill \\ 2 \hfill \\ \end{gathered} &\begin{gathered} 5 \hfill \\ 9 \hfill \\ \end{gathered} \\ 5&4 \end{array}} \right] = {\text{No Solution}} math As you can see, there is no correlation between entries when the matrices have different dimensions.

Matrix Subtraction
Matrix subtraction is carried out in the same manner as matrix addition, with the subtraction of the values in corresponding locations: math \left[ {\begin{array}{*{20}{c}} 4&7 \\ 6&2 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&6 \\ 7&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {4-1}&7-6 \\ {6-7}&2-2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3}&1 \\ {-1}&0 \end{array}} \right] math

Matrix Multiplication
Multiplying matrices is slightly more difficult than adding them. It requires multiplying the first entry of the first matrix with the first entry of the second matrix and ADDING it to the second entry of the first matrix multiplied by the third entry of the second matrix. This sum is the first entry of the answer. Repeat this pattern until the entire product is found.

math \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} e&f \\ g&h \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {ae + bg}&{af + bh} \\ {ce + dg}&{cf + dh} \end{array}} \right] math

Unfortunately, not all matrices can be multiplied together. Luckily, there is a simple trick to determine of two matrices can be multiplied. If the number of columns in the first matrix matches the number of rows in the second, the matrices can be multiplied. The resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second.

math \begin{gathered} {\text{2x3 * 2x3}} \to {\text{ No Solution}} \hfill \\ {\text{2x3 * 3x2}} \to {\text{ 2x2}} \hfill \\ \end{gathered} math

Below is a source for extra practice and information about matrix multiplication: []

The Identity Matrix
The identity matrix is a matrix in which each entry along the main diagonal (from top left to bottom right) of a square matrix is a 1 and every other entry is a 0.

Examples of identity matrices include: math \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] {\text{ and }} \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0\\ 0&0&1 \end{array}} \right] math

Matrix Inverses
Taking the inverse of a 2x2 matrix can be incredibly helpful when isolating a variable. For example:

math [A][X] = [C] math In order to isolate //X// we must somehow eliminate the A-matrix that is in front of it. Having the inverse of A would be very useful, given the property:

math [A]{[A]^{ - 1}} = [I] math Therefore, we may isolate X by ‍doing the following ‍:

math {[A]^{ - 1}}[A][X] = {[A]^{ - 1}}[C] math math [X] = {[A]^{ - 1}}[C] math

//Note: When trying to isolate for generic matrices (i.e. [A], [C]), or matrices whose dimensions or entries are unknown, one must make sure that it is stated that all matrices are invertible, nxn (square) matrices. If this is not given, the matrix cannot be isolated as the inverse of a non-square matrix does not exist.//

//Note 2: When trying to isolate Matrix X by multiplying both sides by the inverse of Matrix A, the right side of the equation MUST be the inverse of Matrix A multiplying Matrix C and not the other way around. This is because, the commutative property of multiplication does not apply to matrices, and the two products would be different. Thus, if the inverse of Matrix A is to the left of Matrix A, it must also be on the left of Matrix X.//

The process for taking a 2x2 inverse is fairly simple. Take 1/(ad-bc) and multiply it by [d,-b;-c,a]

math {\left[ {\begin{array}{*{20}{c}}

a&b \\

c&d

\end{array}} \right]^{ - 1}} = \frac{1}\left[ {\begin{array}{*{20}{c}}

d&{ - b} \\

{ - c}&a

\end{array}} \right] \[where (ad-bc)\neq 0\]

math

The quantity ad-bc is known as the //determinant//, and is used to determine ‍whether the inverse of a matrix exists ‍ or not. If the determinant equals 0, a //singular matrix error// is given, and it means that every entry in the matrix is 0 or that ad=bc. As a result, the inverse of the matrix does not exist. In the graphic below, the Singular Matrix Error occurs since the determinant equals 0.



Row Echelon Form
Row Echelon Form is a format used to describe systems of equations with matrices. It is generally useful for systems with three or more variables. The coefficients of x, y, and z are rewritten in a matrix, and the solutions are separated from them by a solid line. For example:

math \left\{ \begin{gathered} x + 4z = 1 \hfill \\ 2x - y - 6c = 4 \hfill \\ 2x + 3y - 2z = 8 \hfill \\ \end{gathered} \right\} math

Can be rewritten as

math \left[ \begin{array}{ccc|c} 1 & 0 & 4 & 1 \\ 2 & -1 & -6 & 4 \\ 2 & 3 & -2 & 8 \end{array} \right] math

This is not yet considered row echelon form. First we must reduce it by multiplying the first row by two and subtracting the second row. We place the result in the second row. This process of multiplying, adding, and subtracting rows will be used repeatedly to manipulate (reduce) the matrix and achieve both Row Echelon Form and Reduced Row Echelon Form:

math - {R_2} + 2{R_1} \to {R_2} math

This yields:

math \left[ \begin{array}{ccc|c} 1 & 0 & -4 & 1 \\ 0 & 1 & -2 & -2 \\ 2 & 3 & -2 & 8 \end{array} \right] math

Continue reducing until you reach this:

math \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & -3 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{array} \right] math

This is row echelon form; the bottom number is solved for completely, the middle requires basic substitution, and the top requires two substitutions. In polynomial form, we have achieved this:

math \begin{gathered} x - \frac{1}{2}y - 3z = 2 \hfill \\ y + z = 1 \hfill \\ z = 1 \hfill \\ \end{gathered} math

This can be easily solved to find

math \begin{gathered} x = 5 \hfill \\ y = 0 \hfill \\ z = 1 \hfill \\ \end{gathered} math

Once the matrix has been manipulated such that:
 * There is only one 1 in each row and column
 * Every entry that is not a 1 must be a 0
 * The 1 in each row must be farther to the right in the matrix than the one in the previous row

//Note: The above rules do not take the last column of an augmented matrix into account, as that column contains the solutions of the system.//

We may rewrite the matrix in "reduced" row echelon form:

math \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array} \right] math

Some tips to achieving the Echelon Forms:


 * When manipulating the matrix by adding or subtracting rows, be sure not to change entries that are already 0's to other numbers
 * If all of the entries in a matrix share a common factor, one can quickly reduce by dividing the entries by that number
 * Reducing to the Echelon Forms follows a similar process as solving a systems of equations. Though, with matrices, the equations are in the form of rows, one can follow the same steps to manipulate the matrix as they would to manipulate a system

On a TI-89, in order to access the ref and rref functions, one may go to MATH (2nd, 5) and then Matrix.



On a TI-84, in order to access the ref and rref functions, one may go to MATRIX (2nd, X inverse) and then arrow over to MATH.



Below are several informative links about Row Echelon Form, Reduced Row Echelon Form, and how to solve systems of equations with matrices by hand. They have information, practice problems with the answers worked out, and a quiz: [] [] []

Using Parameters
In some cases, you will get an irregular matrix that cannot be satisfied by individual values for x, y, and z

math \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 1 \end{array} \right] math

Our third equation states 0z=7. No value of z can satisfy this equation, so we say there is NO SOLUTION. This is known as an "inconsistent" matrix because no answer can be found.

math \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & 1 \end{array} \right] \to {\text{No Solution}} math

Here's another tricky case math \left[ \begin{array}{ccc|c} 1 & 2 & 9 & 5 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right] math

Our third equation states 0z=0. ANY value of z would satisfy this condition. A value of z can be found, so this matrix is known as "consistant," and because we have //infinitely many// possible answers, we say the matrix is and "dependant". To simplify we use a parameter "p"

math z = p math

Therefore;

math \left\{ \begin{gathered} x = 5 - 2(1 - 3p) - 9p \hfill \\ y = 1 - 3p \hfill \\ z = p \hfill \\ \end{gathered} \right\} math Of course the top equation may be simplified. math x = 3 - 3p math

We have solved for x and y in terms of p. P is sometimes used to measure time, and there is an entire unit devoted to working with parameters; please see link for further information:

10.1 Content - Parametrics


 * Primary authors of this page (as of 06/02/12):**


 * **Anvesh Jalasutram**
 * **Aidan Murphy**