Algebra Examples

$[\begin{array}{c} 4 \\ 0 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

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

$A = [\begin{array}{c} 4 \\ 0 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = \frac{1}{4} R_{1}$ 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 $R_{1} = \frac{1}{4} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{4} R_{1} \\ 0 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{1} = \frac{1}{4} R_{1}$

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

$[\begin{array}{c} (\frac{1}{4}) \cdot (4) \\ 0 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{1} = \frac{1}{4} R_{1}$

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

$[\begin{array}{c} 1 \\ 0 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

Perform the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ 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 $R_{3} = - 4 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ - 4 \cdot R_{1} + R_{3} \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{3} = - 4 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = - 4 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 \\ 0 \\ (- 4) \cdot (1) + 4 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{3} = - 4 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 5 \\ 6 \\ 7 \\ 8 \end{array}]$

Perform the row operation $R_{4} = - 5 \cdot R_{1} + R_{4}$ on $R_{4}$ (row $4$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{4}$ (row $4$ ) with the row operation $R_{4} = - 5 \cdot R_{1} + R_{4}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ - 5 \cdot R_{1} + R_{4} \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{4} = - 5 \cdot R_{1} + R_{4}$

Replace $R_{4}$ (row $4$ ) with the actual values of the elements for the row operation $R_{4} = - 5 \cdot R_{1} + R_{4}$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ (- 5) \cdot (1) + 5 \\ 6 \\ 7 \\ 8 \end{array}]$

$R_{4} = - 5 \cdot R_{1} + R_{4}$

Simplify $R_{4}$ (row $4$ ).

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 6 \\ 7 \\ 8 \end{array}]$

Perform the row operation $R_{5} = - 6 \cdot R_{1} + R_{5}$ on $R_{5}$ (row $5$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{5}$ (row $5$ ) with the row operation $R_{5} = - 6 \cdot R_{1} + R_{5}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ - 6 \cdot R_{1} + R_{5} \\ 7 \\ 8 \end{array}]$

$R_{5} = - 6 \cdot R_{1} + R_{5}$

Replace $R_{5}$ (row $5$ ) with the actual values of the elements for the row operation $R_{5} = - 6 \cdot R_{1} + R_{5}$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ (- 6) \cdot (1) + 6 \\ 7 \\ 8 \end{array}]$

$R_{5} = - 6 \cdot R_{1} + R_{5}$

Simplify $R_{5}$ (row $5$ ).

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 7 \\ 8 \end{array}]$

Perform the row operation $R_{6} = - 7 \cdot R_{1} + R_{6}$ on $R_{6}$ (row $6$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{6}$ (row $6$ ) with the row operation $R_{6} = - 7 \cdot R_{1} + R_{6}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ - 7 \cdot R_{1} + R_{6} \\ 8 \end{array}]$

$R_{6} = - 7 \cdot R_{1} + R_{6}$

Replace $R_{6}$ (row $6$ ) with the actual values of the elements for the row operation $R_{6} = - 7 \cdot R_{1} + R_{6}$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ (- 7) \cdot (1) + 7 \\ 8 \end{array}]$

$R_{6} = - 7 \cdot R_{1} + R_{6}$

Simplify $R_{6}$ (row $6$ ).

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 8 \end{array}]$

Perform the row operation $R_{7} = - 8 \cdot R_{1} + R_{7}$ on $R_{7}$ (row $7$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{7}$ (row $7$ ) with the row operation $R_{7} = - 8 \cdot R_{1} + R_{7}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ - 8 \cdot R_{1} + R_{7} \end{array}]$

$R_{7} = - 8 \cdot R_{1} + R_{7}$

Replace $R_{7}$ (row $7$ ) with the actual values of the elements for the row operation $R_{7} = - 8 \cdot R_{1} + R_{7}$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ (- 8) \cdot (1) + 8 \end{array}]$

$R_{7} = - 8 \cdot R_{1} + R_{7}$

Simplify $R_{7}$ (row $7$ ).

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

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

$F = {}$

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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