Algebra Examples

$[\begin{array}{c} 3 & 7 \\ 7 & 6 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ 7 & 6 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (7) \\ 7 & 6 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{7}{3} \\ 7 & 6 \end{array}]$

Perform the row operation $R_{2} = - 7 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 7 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & \frac{7}{3} \\ - 7 \cdot R_{1} + R_{2} & - 7 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 7 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 7 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} \\ (- 7) \cdot (1) + 7 & (- 7) \cdot (\frac{7}{3}) + 6 \end{array}]$

$R_{2} = - 7 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & \frac{7}{3} \\ 0 & - \frac{31}{3} \end{array}]$

Perform the row operation $R_{2} = - \frac{3}{31} R_{2}$ 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 $R_{2} = - \frac{3}{31} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{7}{3} \\ - \frac{3}{31} R_{2} & - \frac{3}{31} R_{2} \end{array}]$

$R_{2} = - \frac{3}{31} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{3}{31} R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} \\ (- \frac{3}{31}) \cdot (0) & (- \frac{3}{31}) \cdot (- \frac{31}{3}) \end{array}]$

$R_{2} = - \frac{3}{31} R_{2}$

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

$[\begin{array}{c} 1 & \frac{7}{3} \\ 0 & 1 \end{array}]$

Perform the row operation $R_{1} = - \frac{7}{3} R_{2} + R_{1}$ 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 $R_{1} = - \frac{7}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{7}{3} R_{2} + R_{1} & - \frac{7}{3} R_{2} + R_{1} \\ 0 & 1 \end{array}]$

$R_{1} = - \frac{7}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{7}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{7}{3}) \cdot (0) + 1 & (- \frac{7}{3}) \cdot (1) + \frac{7}{3} \\ 0 & 1 \end{array}]$

$R_{1} = - \frac{7}{3} R_{2} + R_{1}$

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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