Linear Algebra Examples

S ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} a b c \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠ = ⎡ ⎢ ⎣ \begin{matrix} 2 a - 6 b + 6 c a + 2 b + c 2 a + b + 2 c \end{matrix} ⎤ ⎥ ⎦

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

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{matrix} 2 a - 6 b + 6 c a + 2 b + c 2 a + b + 2 c \end{matrix} ⎤ ⎥ ⎦ = 0

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

Step 2

Create a system of equations from the vector equation.

2 a - 6 b + 6 c = 0

$2 a - 6 b + 6 c = 0$

a + 2 b + c = 0

$a + 2 b + c = 0$

2 a + b + 2 c = 0

$2 a + b + 2 c = 0$

Step 3

Write the system as a matrix.

⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 & - 6 & 6 & 0 1 & 2 & 1 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Step 4.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{2}

$\frac{1}{2}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{2}{2} & \frac{- 6}{2} & \frac{6}{2} & \frac{0}{2} 1 & 2 & 1 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

Step 4.1.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 1 & 2 & 1 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 1 & 2 & 1 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.2

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Step 4.2.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 1 - 1 & 2 + 3 & 1 - 3 & 0 - 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.2.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 5 & - 2 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 5 & - 2 & 0 2 & 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.3

Perform the row operation

R_{3} = R_{3} - 2 R_{1}

$R_{3} = R_{3} - 2 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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Step 4.3.1

Perform the row operation

R_{3} = R_{3} - 2 R_{1}

$R_{3} = R_{3} - 2 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 5 & - 2 & 0 2 - 2 \cdot 1 & 1 - 2 \cdot - 3 & 2 - 2 \cdot 3 & 0 - 2 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.3.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 5 & - 2 & 0 0 & 7 & - 4 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ 0 & 5 & - 2 & 0 \\ 0 & 7 & - 4 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 5 & - 2 & 0 0 & 7 & - 4 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ 0 & 5 & - 2 & 0 \\ 0 & 7 & - 4 & 0 \end{array}]$

Step 4.4

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

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Step 4.4.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 \frac{0}{5} & \frac{5}{5} & \frac{- 2}{5} & \frac{0}{5} 0 & 7 & - 4 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ \frac{0}{5} & \frac{5}{5} & \frac{- 2}{5} & \frac{0}{5} \\ 0 & 7 & - 4 & 0 \end{array}]$

Step 4.4.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 7 & - 4 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ 0 & 1 & - \frac{2}{5} & 0 \\ 0 & 7 & - 4 & 0 \end{array}]$

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 7 & - 4 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ 0 & 1 & - \frac{2}{5} & 0 \\ 0 & 7 & - 4 & 0 \end{array}]$

Step 4.5

Perform the row operation

R_{3} = R_{3} - 7 R_{2}

$R_{3} = R_{3} - 7 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

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Step 4.5.1

Perform the row operation

R_{3} = R_{3} - 7 R_{2}

$R_{3} = R_{3} - 7 R_{2}$ to make the entry at

3, 2

$3, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & - 4 - 7 (- \frac{2}{5}) & 0 - 7 \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

$[\begin{array}{c} 1 & - 3 & 3 & 0 \\ 0 & 1 & - \frac{2}{5} & 0 \\ 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & - 4 - 7 (- \frac{2}{5}) & 0 - 7 \cdot 0 \end{array}]$

Step 4.5.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 0 & - \frac{6}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 0 & - \frac{6}{5} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦

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

Step 4.6

Multiply each element of

R_{3}

$R_{3}$ by

- \frac{5}{6}

$- \frac{5}{6}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

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Step 4.6.1

Multiply each element of

R_{3}

$R_{3}$ by

- \frac{5}{6}

$- \frac{5}{6}$ to make the entry at

3, 3

$3, 3$ a

1

$1$ .

⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 - \frac{5}{6} \cdot 0 & - \frac{5}{6} \cdot 0 & - \frac{5}{6} (- \frac{6}{5}) & - \frac{5}{6} \cdot 0 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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

Step 4.6.2

Simplify

R_{3}

$R_{3}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & - \frac{2}{5} & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.7

Perform the row operation

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

$R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

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Step 4.7.1

Perform the row operation

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

$R_{2} = R_{2} + \frac{2}{5} R_{3}$ to make the entry at

2, 3

$2, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 + \frac{2}{5} \cdot 0 & 1 + \frac{2}{5} \cdot 0 & - \frac{2}{5} + \frac{2}{5} \cdot 1 & 0 + \frac{2}{5} \cdot 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.7.2

Simplify

R_{2}

$R_{2}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 3 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.8

Perform the row operation

R_{1} = R_{1} - 3 R_{3}

$R_{1} = R_{1} - 3 R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

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Step 4.8.1

Perform the row operation

R_{1} = R_{1} - 3 R_{3}

$R_{1} = R_{1} - 3 R_{3}$ to make the entry at

1, 3

$1, 3$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 - 3 \cdot 0 & - 3 - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.8.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & - 3 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.9

Perform the row operation

R_{1} = R_{1} + 3 R_{2}

$R_{1} = R_{1} + 3 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

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Step 4.9.1

Perform the row operation

R_{1} = R_{1} + 3 R_{2}

$R_{1} = R_{1} + 3 R_{2}$ to make the entry at

1, 2

$1, 2$ a

0

$0$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 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{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 4.9.2

Simplify

R_{1}

$R_{1}$ .

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 & 0 0 & 1 & 0 & 0 0 & 0 & 1 & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

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

Step 5

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

a = 0

$a = 0$

b = 0

$b = 0$

c = 0

$c = 0$

Step 6

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

⎡ ⎢ ⎣ \begin{matrix} a b c \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦

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

Step 7

Write as a solution set.

⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

Step 8

The kernel of

S

$S$ is the subspace

⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

K (S) = ⎧ ⎪ ⎨ ⎪ ⎩ ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦ ⎫ ⎪ ⎬ ⎪ ⎭

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

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