Algebra Examples

$[\begin{array}{c} 2 & 8 \\ 3 & 6 \end{array}]$

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{2} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{2} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ 3 & 6 \end{array}]$

$R_{1} = \frac{1}{2} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{2} R_{1}$ .

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (8) \\ 3 & 6 \end{array}]$

$R_{1} = \frac{1}{2} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 4 \\ 3 & 6 \end{array}]$

Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 4 \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 4 \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (4) + 6 \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 4 \\ 0 & - 6 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{6} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - \frac{1}{6} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 4 \\ - \frac{1}{6} R_{2} & - \frac{1}{6} R_{2} \end{array}]$

$R_{2} = - \frac{1}{6} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{6} R_{2}$ .

$[\begin{array}{c} 1 & 4 \\ (- \frac{1}{6}) \cdot (0) & (- \frac{1}{6}) \cdot (- 6) \end{array}]$

$R_{2} = - \frac{1}{6} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & 4 \\ 0 & 1 \end{array}]$

Perform the row operation $R_{1} = - 4 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - 4 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - 4 \cdot R_{2} + R_{1} & - 4 \cdot R_{2} + R_{1} \\ 0 & 1 \end{array}]$

$R_{1} = - 4 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - 4 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (- 4) \cdot (0) + 1 & (- 4) \cdot (1) + 4 \\ 0 & 1 \end{array}]$

$R_{1} = - 4 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x_{1} = 0$

$x_{2} = 0$

This expression is the solution set for the system of equations.

${(0, 0)}$

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x_{1} \\ x_{2} \end{array}] = [\begin{array}{c} 0 \\ 0 \end{array}]$

The null space of the set is the set of vectors created from the free variables of the system.

${[\begin{array}{c} 0 \\ 0 \end{array}]}$

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