Initial set of equations in matrix form

[ 3 4 5 ] [X1] [ 27 ] [ 6 12 13 ] [X2] = [ 68 ] [ 9 20 22 ] [X3] [ 111 ]

Usually, we abbreviate the system by leaving out the X vector

[ 3 4 5 ] [ 27 ] [ 6 12 13 ] [ 68 ] [ 9 20 22 ] [ 111 ]

Step 1: subtract twice the first row from the second, and three times
the first row from the third. This produces 0s in the first column
under the 3 entry, which is referred to as the *pivot*.
Note that the column of constants is manipulated in the same way
as the matrix columns.

[ 3 4 5 ] [ 27 ] [ 0 4 3 ] [ 14 ] [ 0 8 7 ] [ 30 ]

Step 2: Subtract twice the second row from the third to produce
a zero in the second column under the 4. The matrix now has all 0s
below the main diagonal. This is referred to as *upper triangular*
form, sometimes denoted simply as *U*.

[ 3 4 5 ] [ 27 ] [ 0 4 3 ] [ 14 ] [ 0 0 1 ] [ 2 ]

Step 3: The upper triangular form allows us to find the values of the
variables by progressive *back substitution*.

The last row represents the equation
1 * X3 = 2, from which
we obtain *X3 = 2*
directly.

The second row represents the equation

4 * X2 + 3 * X3 = 14.

Using X3 = 2, we obtain

4 * X2 + 6 = 14 ⇒ 4 * X2 = 8
⇒ *X2 = 2*.

The first row represents the equation

3 * X1 + 4 * X2 + 5 * X3 = 27.

Substituting in the previously obtained values for X2 and X3 gives

3 * X1 + 4 * 2 + 5 * 2 = 27 ⇒ 3 * X1 = 9
⇒ *X1 = 3*.

This completes the solution of the system, which can be checked by substituting the values obtained back into the original system.