Algebra Examples

$S ([\begin{array}{c} a \\ b \\ c \end{array}]) = [\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}]$

The kernel of a transformation is a vector that makes the transformation equal to the zero vector (the pre-image of the transformation).

$[\begin{array}{c} a + 3 b - 6 c \\ 2 a + b + c \\ a + 5 b + c \end{array}] = 0$

Create a system of equations from the vector equation.

$a + 3 b - 6 c = 0$

$2 a + b + c = 0$

$a + 5 b + c = 0$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 2 & 1 & 1 & 0 \\ 1 & 5 & 1 & 0 \end{array}]$

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \\ 1 & 5 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (3) + 1 & (- 2) \cdot (- 6) + 1 & (- 2) \cdot (0) + 0 \\ 1 & 5 & 1 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 1 & 5 & 1 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} & - 1 \cdot R_{1} + R_{3} \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (3) + 5 & (- 1) \cdot (- 6) + 1 & (- 1) \cdot (0) + 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & - 5 & 13 & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} & - \frac{1}{5} R_{2} \\ 0 & 2 & 7 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ (- \frac{1}{5}) \cdot (0) & (- \frac{1}{5}) \cdot (- 5) & (- \frac{1}{5}) \cdot (13) & (- \frac{1}{5}) \cdot (0) \\ 0 & 2 & 7 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

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

$[\begin{array}{c} - 3 \cdot R_{2} + R_{1} & - 3 \cdot R_{2} + R_{1} & - 3 \cdot R_{2} + R_{1} & - 3 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

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

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

$[\begin{array}{c} (- 3) \cdot (0) + 1 & (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{13}{5}) - 6 & (- 3) \cdot (0) + 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 2 & 7 & 0 \end{array}]$

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

Tap for more steps...

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

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ - 2 \cdot R_{2} + R_{3} & - 2 \cdot R_{2} + R_{3} & - 2 \cdot R_{2} + R_{3} & - 2 \cdot R_{2} + R_{3} \end{array}]$

$R_{3} = - 2 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ (- 2) \cdot (0) + 0 & (- 2) \cdot (1) + 2 & (- 2) \cdot (- \frac{13}{5}) + 7 & (- 2) \cdot (0) + 0 \end{array}]$

$R_{3} = - 2 \cdot R_{2} + R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]$

Perform the row operation $R_{3} = \frac{5}{61} R_{3}$ on $R_{3}$ (row $3$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{3}$ (row $3$ ) with the row operation $R_{3} = \frac{5}{61} R_{3}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ \frac{5}{61} R_{3} & \frac{5}{61} R_{3} & \frac{5}{61} R_{3} & \frac{5}{61} R_{3} \end{array}]$

$R_{3} = \frac{5}{61} R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = \frac{5}{61} R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ (\frac{5}{61}) \cdot (0) & (\frac{5}{61}) \cdot (0) & (\frac{5}{61}) \cdot (\frac{61}{5}) & (\frac{5}{61}) \cdot (0) \end{array}]$

$R_{3} = \frac{5}{61} R_{3}$

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

$[\begin{array}{c} 1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Perform the row operation $R_{1} = - \frac{9}{5} R_{3} + R_{1}$ 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 $R_{1} = - \frac{9}{5} R_{3} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{9}{5} R_{3} + R_{1} & - \frac{9}{5} R_{3} + R_{1} & - \frac{9}{5} R_{3} + R_{1} & - \frac{9}{5} R_{3} + R_{1} \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

$R_{1} = - \frac{9}{5} R_{3} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{9}{5} R_{3} + R_{1}$ .

$[\begin{array}{c} (- \frac{9}{5}) \cdot (0) + 1 & (- \frac{9}{5}) \cdot (0) + 0 & (- \frac{9}{5}) \cdot (1) + \frac{9}{5} & (- \frac{9}{5}) \cdot (0) + 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

$R_{1} = - \frac{9}{5} R_{3} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Perform the row operation $R_{2} = \frac{13}{5} R_{3} + R_{2}$ 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 $R_{2} = \frac{13}{5} R_{3} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ \frac{13}{5} R_{3} + R_{2} & \frac{13}{5} R_{3} + R_{2} & \frac{13}{5} R_{3} + R_{2} & \frac{13}{5} R_{3} + R_{2} \\ 0 & 0 & 1 & 0 \end{array}]$

$R_{2} = \frac{13}{5} R_{3} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{13}{5} R_{3} + R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ (\frac{13}{5}) \cdot (0) + 0 & (\frac{13}{5}) \cdot (0) + 1 & (\frac{13}{5}) \cdot (1) - \frac{13}{5} & (\frac{13}{5}) \cdot (0) + 0 \\ 0 & 0 & 1 & 0 \end{array}]$

$R_{2} = \frac{13}{5} R_{3} + R_{2}$

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

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

Use the result matrix to declare the final solutions to the system of equations.

$x_{1} = 0$

$x_{2} = 0$

$x_{3} = 0$

This expression is the solution set for the system of equations.

${(0, 0, 0)}$

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.

$X = [\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}] = [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]$

The null space of the set is the set of vectors created from the free variables of the system.

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

The kernel of $S$ is the subspace ${[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}$ .

$K (S) = {[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]}$

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Algebra Examples

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