We know that an array of numbers arranged in rows and columns and placed in brac

Question

We know that an array of numbers arranged in rows and columns and placed in brackets, is an example of a matrix (plural: matrices). The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and to solve systems of linear equations. How would you explain solving a linear system using either Gaussian Elimination or Gauss-Jordan Elimination? We know that an array of numbers arranged in rows and columns and placed in brackets, is an example of a matrix (plural: matrices). The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and to solve systems of linear equations. How would you explain solving a linear system using either Gaussian Elimination or Gauss-Jordan Elimination? We know that an array of numbers arranged in rows and columns and placed in brackets, is an example of a matrix (plural: matrices). The numbers inside the brackets are called elements of the matrix. Matrices are used to display information and to solve systems of linear equations. How would you explain solving a linear system using either Gaussian Elimination or Gauss-Jordan Elimination?

Explanation / Answer

Answer:

Solving three-variable, three-equation linear systems is more difficult, at least initially, than solving the two-variable systems, because the computations involved are more messy. You will need to be very neat in your working, and you should plan to use lots of scratch paper. The method for solving these systems is an extension of the two-variable solving-by-addition method, so make sure you know this method well and can use it consistently correctly.

Though the method of solution is based on addition/elimination, trying to do actual addition tends to get very messy, so there is a systematized method for solving the three-or-more-variables systems. This method is called "Gaussian elimination" (with the equations ending up in what is called "row-echelon form").

It's fairly easy to see how to proceed in this case. We'll just back-substitute the z-value from the third equation into the second equation, solve the result for y, and then plug z and y into the first equation and solve the result for x. Below is an example;

Solve the following system of equations.

5x + 4y –    z = 0
      10y – 3z = 11
                  z = 3

10y – 3(3) = 11
10y – 9 = 11
10y = 20
y = 2

5x + 4(2) – (3) = 0
5x + 8 – 3 = 0
5x + 5 = 0
5x = –5
x = –1

Then the solution is (x, y, z) = (–1, 2, 3).

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