Linear Algebra Examples

$[\begin{array}{c} 6 & - 2 & - 4 & 4 \\ 3 & - 3 & - 6 & 1 \\ - 12 & 8 & 21 & - 8 \\ - 6 & 0 & - 10 & 7 \end{array}] [\begin{array}{c} x \\ y \\ z \\ w \end{array}] = [\begin{array}{c} 2 \\ - 4 \\ 8 \\ - 43 \end{array}]$

Step 1

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

$[\begin{array}{c} 2 \\ - 4 \\ 8 \\ - 43 \end{array}] = 0$

Step 2

Create a system of equations from the vector equation.

$2 = 0$

$- 4 = 0$

$8 = 0$

$- 43 = 0$

Step 3

Subtract $2$ from both sides of the equation.

$0 = - 2$

$- 4 = 0$

$8 = 0$

$- 43 = 0$

Step 4

Add $4$ to both sides of the equation.

$0 = 4$

$0 = - 2$

$8 = 0$

$- 43 = 0$

Step 5

Subtract $8$ from both sides of the equation.

$0 = - 8$

$0 = - 2$

$0 = 4$

$- 43 = 0$

Step 6

Add $43$ to both sides of the equation.

$0 = 43$

$0 = - 2$

$0 = 4$

$0 = - 8$

Step 7

Write the system of equations in matrix form.

$[\begin{array}{c} - 2 \\ 4 \\ - 8 \\ 43 \end{array}]$

Step 8

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} - \frac{1}{2} R_{1} \\ 4 \\ - 8 \\ 43 \end{array}]$

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

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

$[\begin{array}{c} (- \frac{1}{2}) \cdot (- 2) \\ 4 \\ - 8 \\ 43 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ 4 \\ - 8 \\ 43 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 \\ - 4 \cdot R_{1} + R_{2} \\ - 8 \\ 43 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ (- 4) \cdot (1) + 4 \\ - 8 \\ 43 \end{array}]$

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

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

$[\begin{array}{c} 1 \\ 0 \\ - 8 \\ 43 \end{array}]$

Perform the row operation $R_{3} = 8 \cdot R_{1} + R_{3}$ 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 $R_{3} = 8 \cdot R_{1} + R_{3}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 8 \cdot R_{1} + R_{3} \\ 43 \end{array}]$

$R_{3} = 8 \cdot R_{1} + R_{3}$

Replace $R_{3}$ (row $3$ ) with the actual values of the elements for the row operation $R_{3} = 8 \cdot R_{1} + R_{3}$ .

$[\begin{array}{c} 1 \\ 0 \\ (8) \cdot (1) - 8 \\ 43 \end{array}]$

$R_{3} = 8 \cdot R_{1} + R_{3}$

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

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ 43 \end{array}]$

Perform the row operation $R_{4} = - 43 \cdot R_{1} + R_{4}$ 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 $R_{4} = - 43 \cdot R_{1} + R_{4}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ - 43 \cdot R_{1} + R_{4} \end{array}]$

$R_{4} = - 43 \cdot R_{1} + R_{4}$

Replace $R_{4}$ (row $4$ ) with the actual values of the elements for the row operation $R_{4} = - 43 \cdot R_{1} + R_{4}$ .

$[\begin{array}{c} 1 \\ 0 \\ 0 \\ (- 43) \cdot (1) + 43 \end{array}]$

$R_{4} = - 43 \cdot R_{1} + R_{4}$

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

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

Step 9

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

$0 = 1$

Step 10

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

${}$

Step 11

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.

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

Step 12

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

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

Step 13

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

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