Linear Algebra Examples

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

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).

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

Step 2

Create a system of equations from the vector equation.

$a + 3 b - 6 c = 0$

$2 a + b + c = 0$

$a + 5 b + c = 0$

Step 3

Write the system as a matrix.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.1.1

Perform the row operation $R_{2} = R_{2} - 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 4.1.2

Simplify $R_{2}$ .

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

Step 4.2

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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

Step 4.2.2

Simplify $R_{3}$ .

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

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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

Step 4.3.2

Simplify $R_{2}$ .

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

Step 4.4

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.4.1

Perform the row operation $R_{3} = R_{3} - 2 R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ 0 & 0 & \frac{61}{5} & 0 \end{array}]$

Step 4.5

Multiply each element of $R_{3}$ by $\frac{5}{61}$ to make the entry at $3, 3$ a $1$ .

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Step 4.5.1

Multiply each element of $R_{3}$ by $\frac{5}{61}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 3 & - 6 & 0 \\ 0 & 1 & - \frac{13}{5} & 0 \\ \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot 0 & \frac{5}{61} \cdot \frac{61}{5} & \frac{5}{61} \cdot 0 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

Perform the row operation $R_{2} = R_{2} + \frac{13}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.6.1

Perform the row operation $R_{2} = R_{2} + \frac{13}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 4.6.2

Simplify $R_{2}$ .

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

Step 4.7

Perform the row operation $R_{1} = R_{1} + 6 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{1} = R_{1} + 6 R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Step 4.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}]$

Step 4.8

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.8.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 4.8.2

Simplify $R_{1}$ .

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

Step 5

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

$a = 0$

$b = 0$

$c = 0$

Step 6

Write a solution vector by solving in terms of the free variables in each row.

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

Step 7

Write as a solution set.

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

Step 8

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

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