The same solving methods that are used for systems of equations with 2 equations can be used to solve systems with more than 2 equations. For example, a system with 3 equations and 3 unknown variables like this: \begin{cases} x + y + z = 20\\4x - 2y + 3z = 14\\3 \left(x + y\right) = z \end{cases} Can be solved by graphing in a 3D coordinate plane. It can also be solved using matrices.

For a matrix or an augmented matrix, we can perform row operations to create a related matrix.

The operations we can perform are:

• Multiplying a row by a scalar: R_1 \to 2R_1 ; \left[ \begin{matrix} 5 & 4 \\ -1 & 3 \end{matrix} \right] \to \left[ \begin{matrix} 5\cdot 2 & 4 \cdot 2 \\ -1 & 3 \end{matrix} \right] = \left[ \begin{matrix} 10 & 8 \\ -1 & 3 \end{matrix} \right]
• Adding two rows: R_1 \to R_1+R_2 ; \left[ \begin{matrix} 5 & 4 \\ -1 & 3 \end{matrix} \right] \to \left[ \begin{matrix} 5+(-1) & 4+3 \\ -1 & 3 \end{matrix} \right]=\left[ \begin{matrix} 4&7 \\ -1 & 3 \end{matrix} \right]
• Swapping two rows: R_1 \leftrightarrow R_2 ; \left[ \begin{matrix} 5 & 4 \\ -1 & 3 \end{matrix} \right] \to \left[ \begin{matrix} -1 & 3 \\ 5 & 4 \end{matrix} \right]

We can combine multiple operations into one step: R_1 \to R_1-2R_2 ; \left[ \begin{matrix} 5 & 4 \\ -1 & 3 \end{matrix} \right] \to \left[ \begin{matrix} 5-2(-1) & 4-2(3) \\ -1 & 3 \end{matrix} \right]= \left[ \begin{matrix} 7 & -2 \\ -1 & 3 \end{matrix} \right]

We do these steps to help with solving systems of equations represented by matrices.

For equations in standard form, the entries to the left of the vertical line are the coefficients in order and the entries to the right of the vertical line are the constants.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 8\\ 0 & 1 & 2 & -2 \\ 0 & 0 & 1 & 1 \end{array} \right] \iff \begin{alignedat}{4} & x & - & y & + & z & = & 8 \\ & & & y &+2& z & = & -2 \\ & & & & &z & = & 1 \end{alignedat}

We can solve a system in two ways:

1. Getting the augmented matrix to row-echelon form using row operations and then solving using back substitution
2. Getting the augmented matrix to reduced row-echelon form using row operations

The coefficient matrix may also be called an "upper right triangular matrix".

Back substitution can be used because we can identify the value of one of the variables and then substitute that value back into the equation above it and repeat that process until we have solve for all variables.

The coefficient matrix may also be called a "diagonal matrix".

Note that to have a unique solution we need at least the same number of equations as variables and for the reduced row-echelon form to not have any rows that are all zeros for coefficients, but non-zero for the constant.

• Unique solution, for example x=3, y=-2, and z=7: \left[ \begin{array}{ccc|c} 1 &0 & 0&3\\ 0 & 1 &0 &-2\\ 0 & 0 &1&7 \end{array} \right] \text{ or } \left[ \begin{array}{ccc|c} 1 &0 & 0&3\\ 0 & 1 &0 &-2\\ 0 & 0 &1&7 \\ 0 & 0 &0&0 \end{array} \right]
• Infinite number of solution, for example x=3, y=-2, and z=t, where t is any real number: \left[ \begin{array}{ccc|c} 1 &0 & 0&3\\ 0 & 1 &0 &-2 \end{array} \right] \text{ or } \left[ \begin{array}{ccc|c} 1 &0 & 0&3\\ 0 & 1 &0 &-2 \\ 0 & 0 &0 &0 \end{array} \right]
• No solution, where the last equation would be 0x+0y+0z=7, which is not valid: \left[ \begin{array}{ccc|c} 1 &0 & 0&3\\ 0 & 1 &0 &-2\\ 0 & 0 &0&7 \end{array} \right]

Worked examples