Algebra Examples

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

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

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

Create a system of equations from the vector equation.

$a - b - c = 0$

$a - b + c = 0$

Write the system of equations in matrix form.

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

Find the reduced row echelon form of the matrix.

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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 $0$ .

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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 $0$ .

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

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

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

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

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

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

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

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 $0$ .

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

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

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

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

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

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

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

Exchange row $3$ and row $2$ to organize the zeros into position.

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

$R_{3} \leftrightarrow R$

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

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

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

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

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

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

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

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

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

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

$R_{1} = R_{2} + R_{1}$

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

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

$R_{1} = R_{2} + R_{1}$

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

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

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

$x_{1} - x_{2} = 0$

$x_{3} = 0$

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

${(x_{2}, x_{2}, 0) | x_{2} \in ℝ}$

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.

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

Express the vector as a linear combination of column vector using the properties of vector column addition.

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

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

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

The kernel of $S$ is the subspace ${[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$ .

$K (S) = {[\begin{array}{c} 1 \\ 1 \\ 0 \end{array}]}$

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