Algebra Examples

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

Assign the matrix the name $A$ to simplify the descriptions throughout the problem.

$A = [\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (5) & (\frac{1}{3}) \cdot (0) \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ - 7 \cdot R_{1} + R_{2} & - 7 \cdot R_{1} + R_{2} & - 7 \cdot R_{1} + R_{2} \\ 1 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ (- 7) \cdot (1) + 7 & (- 7) \cdot (\frac{5}{3}) + 5 & (- 7) \cdot (0) + 0 \\ 1 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 0 & - \frac{20}{3} & 0 \\ 1 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 0 & - \frac{20}{3} & 0 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 0 & - \frac{20}{3} & 0 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (\frac{5}{3}) + 1 & (- 1) \cdot (0) + 0 \end{array}]$

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

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 0 & - \frac{20}{3} & 0 \\ 0 & - \frac{2}{3} & 0 \end{array}]$

Perform the row operation $R_{2} = - \frac{3}{20} R_{2}$ 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 $R_{2} = - \frac{3}{20} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ - \frac{3}{20} R_{2} & - \frac{3}{20} R_{2} & - \frac{3}{20} R_{2} \\ 0 & - \frac{2}{3} & 0 \end{array}]$

$R_{2} = - \frac{3}{20} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{3}{20} R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ (- \frac{3}{20}) \cdot (0) & (- \frac{3}{20}) \cdot (- \frac{20}{3}) & (- \frac{3}{20}) \cdot (0) \\ 0 & - \frac{2}{3} & 0 \end{array}]$

$R_{2} = - \frac{3}{20} R_{2}$

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 \\ 0 & 1 & 0 \\ 0 & - \frac{2}{3} & 0 \end{array}]$

Perform the row operation $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ 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 $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{5}{3} R_{2} + R_{1} & - \frac{5}{3} R_{2} + R_{1} & - \frac{5}{3} R_{2} + R_{1} \\ 0 & 1 & 0 \\ 0 & - \frac{2}{3} & 0 \end{array}]$

$R_{1} = - \frac{5}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{5}{3}) \cdot (0) + 1 & (- \frac{5}{3}) \cdot (1) + \frac{5}{3} & (- \frac{5}{3}) \cdot (0) + 0 \\ 0 & 1 & 0 \\ 0 & - \frac{2}{3} & 0 \end{array}]$

$R_{1} = - \frac{5}{3} R_{2} + R_{1}$

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

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

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

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

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

$R_{3} = \frac{2}{3} R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ (\frac{2}{3}) \cdot (0) + 0 & (\frac{2}{3}) \cdot (1) - \frac{2}{3} & (\frac{2}{3}) \cdot (0) + 0 \end{array}]$

$R_{3} = \frac{2}{3} R_{2} + R_{3}$

Simplify $R_{3}$ (row $3$ ).

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

$F$ represents the set of the free variable columns remaining in the row reduced matrix.

$F = {3}$

Nullity is the dimension of the null space of $A$ , which is the same as the number of free variables columns remaining in the row reduced matrix.

$n (A) = d i m (F)$

The nullity of $A$ is $1$ .

$1$

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