# Gauss-Jordan Matrix Elimination

**Topics:**Matrix, Invertible matrix, Linear algebra

**Pages:**5 (817 words)

**Published:**June 23, 2013

-This method can be used to solve systems of linear equations involving two or more variables. However, the system must be changed to an augmented matrix. -This method can also be used to find the inverse of a 2x2 matrix or larger matrices, 3x3, 4x4 etc. Note: The matrix must be a square matrix in order to find its inverse. An Augmented Matrix is used to solve a system of linear equations. a1 x + b1 y + c1 z = d1 a 2 x + b2 y + c 2 z = d 2 a3 x + b3 y + c3 z = d 3

System of Equations ⎯ ⎯→

Augmented Matrix ⎯ ⎯→

⎡ a1 ⎢ ⎢a 2 ⎢ a3 ⎣

b1 b2 b3

c1 d1 ⎤ ⎥ c2 d 2 ⎥ c3 d 3 ⎥ ⎦

-When given a system of equations, to write in augmented matrix form, the coefficients of each variable must be taken and put in a matrix. For example, for the following system:

3x + 2 y − z = 3 x − y + 2z = 4 2x + 3 y − z = 3

Augmented Matrix ⎯ ⎯→

⎡ 3 2 − 1 3⎤ ⎢ ⎥ ⎢1 − 1 2 4 ⎥ ⎢ 2 3 − 1 3⎥ ⎣ ⎦

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-There are three different operations known as Elementary Row Operations used when solving or reducing a matrix, using Gauss-Jordan elimination method. 1. Interchanging two rows. 2. Add one row to another row, or multiply one row first and then adding it to another. 3. Multiplying a row by any constant greater than zero. Identity Matrix-is the final result obtained when a matrix is reduced. This matrix consists of ones in the diagonal starting with the first number.

-The numbers in the last column are the answers to the system of equations. ⎡1 0 0 3 ⎤ ⎢ ⎥ ⎯⎯ ⎢0 1 0 2⎥ ← Identity Matrix for a 3x3 ⎢0 0 1 5 ⎥ ⎣ ⎦ ⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎣ 0 0 0 2⎤ ⎥ 1 0 0 6⎥ ← Identity Matrix for a 4x4 ⎯⎯ 0 1 0 1⎥ ⎥ 0 0 1 4⎥ ⎦

-The pattern continues for bigger matrices.

Solving a system using Gauss-Jordan

–The best way to go is to get the ones first in their respective column, and then using that one to get the zeros in that column. -It is very important to understand that there is no exact procedure to follow when using the Gauss-Jordan method to solve for a system. 3x + 2 y − z = 3

x − y + 2z = 4 2x + 3y − z = 3

Write as an augmented matrix.

↓

⎡ 3 2 − 1 3⎤ ⎢ ⎥ ⎢1 − 1 2 4 ⎥ ⎢ 2 3 − 1 3⎥ ⎣ ⎦ ↓ The Math Center ■ Valle Verde ■ Tutorial Support Services ■ EPCC

Switch row 1 with row 2 to get a 1 in the first column

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⎡1 − 1 2 4 ⎤ ⎢ ⎥ ⎢3 2 − 1 3⎥ Multiply row 1 by -3 and add to row 2 to get a zero ⎢ 2 3 − 1 3⎥ ⎣ ⎦ Row 1 multiplied by -3 ⎯ ⎯→ + Row 2 ⎯ ⎯→ New Row 2 ⎯ ⎯→ −3 3 0 3 2 5 −6 -1 -7

− 12 3 -9

-Put the new row 2 in the matrix, note that though row 1 was multiplied by -3, row 1 didn’t change in our matrix. ⎡1 − 1 2 4 ⎤ ⎢ ⎥ ⎢0 5 − 7 − 9⎥ ⎢2 3 − 1 3 ⎥ ⎣ ⎦ Using a similar procedure of multiplying and adding rows, obtain the following matrix 4⎤ ⎡1 − 1 2 ⎢0 5 − 7 − 9⎥ Multiply row 1 by -2 and add to row3 as above. ⎢ ⎥ ⎢2 3 − 1 3 ⎥ ⎣ ⎦ ↓ ⎡1 − 1 2 ⎢ ⎢0 5 − 7 ⎢0 5 − 5 ⎣ ↓ ⎡1 − 1 2 ⎢ ⎢0 5 − 5 ⎢0 5 − 7 ⎣ ↓ ⎡1 − 1 2 ⎢ ⎢0 1 − 1 ⎢0 5 − 7 ⎣ ↓ ⎡1 0 1 3 ⎤ ⎢ ⎥ ⎢0 1 − 1 − 1⎥ Multiply and add like we did earlier, -5 ∗ R2+R3 ⎢0 5 − 7 − 9 ⎥ ⎣ ⎦ ↓ 4⎤ ⎥ − 9⎥ Switch rows 2 and 3 to obtain the following − 5⎥ ⎦ 4⎤ ⎥ − 5⎥ Divide the second row by 5 to obtain a 1 in the second row. − 9⎥ ⎦ 4⎤ ⎥ − 1⎥ Add row 2 to row 1 − 9⎥ ⎦

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⎡1 0 1 3 ⎤ ⎢ ⎥ ⎢0 1 − 1 − 1⎥ Divide row 3 by -2 to obtain a 1 in the third row. ⎢0 0 − 2 − 4 ⎥ ⎣ ⎦ ↓ ⎡1 0 1 3 ⎤ ⎢ ⎥ ⎢0 1 − 1 − 1⎥ ⎢0 0 1 2 ⎥ ⎣ ⎦ -Finally, the matrix can be solved in two different ways: A. Using the 1 in column 3, obtain the other zeros and the solutions.

⎡1 0 0 1 ⎤ ⎢ ⎥ ⎢0 1 0 1 ⎥ ⎢0 0 1 2 ⎥ ⎣ ⎦

x =1 y =1 z = 2

B. Solve by using back substitution.

-The solution to the last row is z = 2 , the answer can be substituted into the equation produced by the second row. y − z = −1 Substituting into this equation, it simplifies to: y − 2 = −1 y =1 -Again, substituting the answer for z into the first...

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