The new second row translates into −5 y = −5, which means y = 1. Adding −3 times the first row of the augmented matrix to the second row yields Likewise, the counterpart of adding a multiple of one equation to another is adding a multiple of one row to another row. Now, the counterpart of eliminating a variable from an equation in the system is changing one of the entries in the coefficient matrix to zero. You may choose to include a vertical line-as shown above-to separate the coefficients of the unknowns from the extra column representing the constants. The first row, r 1 = (1, 1, 3), corresponds to the first equation, 1 x + 1 y = 3, and the second row, r 2 = (3, −2, 4), corresponds to the second equation, 3 x − 2 y = 4. This is called the augmented matrix, and each row corresponds to an equation in the given system. Next, the coefficient matrix is augmented by writing the constants that appear on the right‐hand sides of the equations as an additional column: This is called the coefficient matrix of the system. The first step is to write the coefficients of the unknowns in a matrix: The previous example will be redone using matrices. This method reduces the effort in finding the solutions by eliminating the need to explicitly write the variables at each step. Gaussian elimination is usually carried out using matrices.

(Back‐substitution of y = 1 into the original second equation, 3 x − 2 y = 4, would also yeild x = 2.) The solution of this system is therefore ( x, y) = (2, 1), as noted in Example 1.

Back‐substitution of y = 1 into the original first equation, x + y = 3, yields x = 2. This final equation, −5 y = −5, immediately implies y = 1. Multiplying the first equation by −3 and adding the result to the second equation eliminates the variable x:

Gauss jordan elimination example