Linear Algebra Examples

$[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$

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

$A = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$

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 $1$ .

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

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

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

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (2) \\ 4 & 6 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{2}{3} \\ 4 & 6 \end{array}]$

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

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

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

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

$[\begin{array}{c} 1 & \frac{2}{3} \\ (- 4) \cdot (1) + 4 & (- 4) \cdot (\frac{2}{3}) + 6 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{2}{3} \\ 0 & \frac{10}{3} \end{array}]$

Perform the row operation $R_{2} = \frac{3}{10} R_{2}$ 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 $R_{2} = \frac{3}{10} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} \\ \frac{3}{10} R_{2} & \frac{3}{10} R_{2} \end{array}]$

$R_{2} = \frac{3}{10} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{3}{10} R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} \\ (\frac{3}{10}) \cdot (0) & (\frac{3}{10}) \cdot (\frac{10}{3}) \end{array}]$

$R_{2} = \frac{3}{10} R_{2}$

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

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

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

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

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

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

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

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

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

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

$D$ represents the set of the pivot columns in the row-reduced form of the matrix.

$D = {1, 2}$

Rank is the dimension of the column space of $A$ .

$r (A) = d i m (D)$

The rank of $A$ is $2$ .

$2$

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