Linear Algebra Examples

$5 [\begin{array}{c} x & y \\ y & z \end{array}] + 2 [\begin{array}{c} 2 x & - y \\ 3 y & - 4 z \end{array}] = [\begin{array}{c} 45 & 12 - y \\ 33 & 15 \end{array}]$

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).

$[\begin{array}{c} 45 & 12 - y \\ 33 & 15 \end{array}] = 0$

Step 2

Create a system of equations from the vector equation.

$45 = 0$

$33 = 0$

Step 3

Subtract $45$ from both sides of the equation.

$0 = - 45$

$33 = 0$

Step 4

Subtract $33$ from both sides of the equation.

$0 = - 33$

$0 = - 45$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} - 45 \\ - 33 \end{array}]$

Step 6

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{1} = - \frac{1}{45} R_{1}$ 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 $R_{1} = - \frac{1}{45} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{1}{45} R_{1} \\ - 33 \end{array}]$

$R_{1} = - \frac{1}{45} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{45} R_{1}$ .

$[\begin{array}{c} (- \frac{1}{45}) \cdot (- 45) \\ - 33 \end{array}]$

$R_{1} = - \frac{1}{45} R_{1}$

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

$[\begin{array}{c} 1 \\ - 33 \end{array}]$

Perform the row operation $R_{2} = 33 \cdot R_{1} + R_{2}$ 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 $R_{2} = 33 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 \\ 33 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = 33 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 33 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 \\ (33) \cdot (1) - 33 \end{array}]$

$R_{2} = 33 \cdot R_{1} + R_{2}$

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

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

Step 7

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

$0 = 1$

Step 8

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

${}$

Step 9

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 10

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 11

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

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