Algebra Examples

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

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{3} R_{1}$ 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 $R_{1} = \frac{1}{3} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{1} = \frac{1}{3} R_{1}$

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

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

$R_{1} = \frac{1}{3} R_{1}$

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

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

Perform the row operation $R_{2} = - 6 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 6 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ - 6 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 6 \cdot R_{1} + R_{2}$

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

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

$R_{2} = - 6 \cdot R_{1} + R_{2}$

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

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

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

$x_{1} = 0$

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

${(0)}$

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.

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

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

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

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