1.2 Identity Matrix and Gauss-Jordan Elimination

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A square matrix, a matrix that has the same number of rows as columns, that has 1s along it's main diagonal and 0s elswhere is called an Identity Matrix

Example
[1 0 0]
[0 1 0]
[0 0 1]

It is denoted by I_n for the n x n matrix.

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We can use matrices to describe a system of linear equations. Using the coefficients of the variables, we can create a coefficient matrix

2x + 2y - 4z = 14
3x + y + z = 8
2x - y + 2z = -1

This turns into:

[2  2 -4]
[3  1  1]
[2 -1  2]

The augmented matrix uses both the coefficients and the constant terms.

[2  2 -4 : 14]
[3  1  1 : 8]
[2 -1  2 : -1]

Note: The coefficient matrix is a submatrix of the augmented matrix.

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Transformations of the matrix, called elementary transformations, can be used to change a sustem of linear equations into another sustem that has the same solution. It is simpler to work in terms of matrices using equivalent transfomations called elementary row operations. The transformations are as follows:

1. Interchange two rows of a matrix.
2. Multiply the elements of a row by a nonzero constant.
3. Add a multiple of one row to the elements of another row.

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Gauss-Jordon Elimination

Using these operations, we can now solve a system of equations by turning it into an augmented matrix and transforming that matrix into an augmented identity matrix. We will use the earlier equation and it's augmented matrix.

2x + 2y - 4z = 14
3x + y + z = 8
2x - y + 2z = -1

[2  2 -4 : 14]
[3  1  1 : 8]
[2 -1  2 : -1]

Our first step is to use operation 2 to get a one in position (1,1). (That is row 1, column 1.)

[2  2 -4 : 14] x 1/2
[3  1  1 : 8]
[2 -1  2 : -1]

This turns the matrix into a new, equivalent matrix.

[1  1 -2 : 7]
[3  1  1 : 8]
[2 -1  2 : -1]

Now by using operation 3, we can eliminate the other numbers in column one and turn them into zeros.

[1  1 -2 : 7]
[3  1  1 : 8] - (3)R₁
[2 -1  2 : -1] - (2)R₁

Or we take element (1,1) and multiply it by 3 then subtract it from element (2,1). We also take (1,2), multiply by 3 and subtract it from (2,2). We repeat this to all corresponding elements in row 2. Then we repeat with row three using the different multiple. This changes the matrix into the following:

[1  1 -2 : 7]
[0 -2  7 :-13]
[0 -3  6 :-15]

Now, to make the solution simpler, we use operation 1 to swap row 2 and row 3 in order to avoid fractions.

[1  1 -2 : 7]
[0 -3  6 :-15]
[0 -2  7 :-13]

And then multiply row 2 by -1/3 to get a 1 in the (2,2) position.

[1  1 -2 : 7]
[0 1 -2 :  5]
[0 -2  7 :-13]

Using operation 3 again, we can get zeros in the other two positions in column two by subtracting row 2 from row 1 and adding 2 x row 2 to row 3.

[1  0  0:  2]
[0 1 -2 :  5]
[0  0  3: -3]

Now to get a 1 in position (3,3), we multiply row 3 by 1/3.

[1  0  0:  2]
[0 1 -2 :  5]
[0  0  1: -1]

And the final step is to get the zeros in the rest of column three. We'll add 2 x row 3 to row 2.

[1  0  0:  2]
[0 1  0 :  3]
[0  0  1: -1]

This matrix is almost done. Each 1 in the coefficient matrix corresponds to a variable in the original equation. The numbers in the augmented column are the solutions. The solution to this system of equations is:

x = 2
y = 3
z = -1

Note: It looks more difficult than it actually is. The more you practice with it, the easier it becomes. Also, this method is most useful to solve systems with a unique solution, though it can also be used to find a general solution, which will be covered in the next section.

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If there are any questions, please post them in the comment area.

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