Linear Algebra Examples

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

Step 1

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

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

Step 2

Find the reduced row echelon form of the matrix.

Tap for more steps...

Perform the row operation $R_{2} = - 1 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Tap for more steps...

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

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

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

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

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

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

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

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

Perform the row operation $R_{2} = - 1 \cdot R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Tap for more steps...

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 1 \cdot R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

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

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

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

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

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

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

Perform the row operation $R_{1} = - 1 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Tap for more steps...

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

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

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

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

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

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

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

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

Step 3

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

$D = (S E T, 1, 3)$

Step 4

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

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

Step 5

The rank of $A$ is $2$ .

$2$

Enter YOUR Problem