Linear Algebra Examples

[\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}]

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

Step 1

Find the eigenvalues.

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Step 1.1

Set up the formula to find the characteristic equation

p (λ)

$p (λ)$ .

p (λ) = determinant (A - λ I_{2})

$p (λ) = determinant (A - λ I_{2})$

Step 1.2

The identity matrix or unit matrix of size

2

$2$ is the

2 \times 2

$2 \times 2$ square matrix with ones on the main diagonal and zeros elsewhere.

[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

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

Step 1.3

Substitute the known values into

p (λ) = determinant (A - λ I_{2})

$p (λ) = determinant (A - λ I_{2})$ .

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Step 1.3.1

Substitute

[\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}]

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

A

$A$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] - λ I_{2})

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] - λ I_{2})$

Step 1.3.2

Substitute

[\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

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

I_{2}

$I_{2}$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply

- λ

$- λ$ by each element of the matrix.

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 λ - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 λ & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 & - λ \cdot 1 \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 & - λ \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 & - λ \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

p (λ) = determinant ([\begin{matrix} - 1 & 2 - 6 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 & - λ \end{matrix}])

$p (λ) = determinant ([\begin{array}{c} - 1 & 2 \\ - 6 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 1 - λ & 2 + 0 \\ - 6 + 0 & 6 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add

$2$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 1 - λ & 2 \\ - 6 + 0 & 6 - λ \end{array}]$

Step 1.4.3.2

Add

$- 6$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 1 - λ & 2 \\ - 6 & 6 - λ \end{array}]$

Step 1.5

Find the determinant.

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Step 1.5.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 1 - λ) (6 - λ) - (- 6 \cdot 2)$

Step 1.5.2

Simplify the determinant.

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Step 1.5.2.1

Simplify each term.

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Step 1.5.2.1.1

Expand

$(- 1 - λ) (6 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = - 1 (6 - λ) - λ (6 - λ) - (- 6 \cdot 2)$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = - 1 \cdot 6 - 1 (- λ) - λ (6 - λ) - (- 6 \cdot 2)$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = - 1 \cdot 6 - 1 (- λ) - λ \cdot 6 - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2

Simplify and combine like terms.

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Step 1.5.2.1.2.1

Simplify each term.

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Step 1.5.2.1.2.1.1

Multiply

$- 1$ by

$6$ .

$p (λ) = - 6 - 1 (- λ) - λ \cdot 6 - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.2

Multiply

$- 1 (- λ)$ .

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Step 1.5.2.1.2.1.2.1

Multiply

$- 1$ by

$- 1$ .

$p (λ) = - 6 + 1 λ - λ \cdot 6 - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.2.2

Multiply

$λ$ by

$1$ .

$p (λ) = - 6 + λ - λ \cdot 6 - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.3

Multiply

$6$ by

$- 1$ .

$p (λ) = - 6 + λ - 6 λ - λ (- λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 6 + λ - 6 λ - 1 \cdot - 1 λ \cdot λ - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.5

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 1.5.2.1.2.1.5.1

Move

$λ$ .

$p (λ) = - 6 + λ - 6 λ - 1 \cdot - 1 (λ \cdot λ) - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = - 6 + λ - 6 λ - 1 \cdot - 1 λ^{2} - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

$p (λ) = - 6 + λ - 6 λ + 1 λ^{2} - (- 6 \cdot 2)$

Step 1.5.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = - 6 + λ - 6 λ + λ^{2} - (- 6 \cdot 2)$

Step 1.5.2.1.2.2

Subtract

$6 λ$ from

$λ$ .

$p (λ) = - 6 - 5 λ + λ^{2} - (- 6 \cdot 2)$

Step 1.5.2.1.3

Multiply

$- (- 6 \cdot 2)$ .

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Step 1.5.2.1.3.1

Multiply

$- 6$ by

$2$ .

$p (λ) = - 6 - 5 λ + λ^{2} - - 12$

Step 1.5.2.1.3.2

Multiply

$- 1$ by

$- 12$ .

$p (λ) = - 6 - 5 λ + λ^{2} + 12$

Step 1.5.2.2

Add

$- 6$ and

$12$ .

$p (λ) = - 5 λ + λ^{2} + 6$

Step 1.5.2.3

Reorder

$- 5 λ$ and

$λ^{2}$ .

$p (λ) = λ^{2} - 5 λ + 6$

Step 1.6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$λ^{2} - 5 λ + 6 = 0$

Step 1.7

Solve for

$λ$ .

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Step 1.7.1

Factor

$λ^{2} - 5 λ + 6$ using the AC method.

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Step 1.7.1.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$6$ and whose sum is

$- 5$ .

$- 3, - 2$

Step 1.7.1.2

Write the factored form using these integers.

$(λ - 3) (λ - 2) = 0$

Step 1.7.2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$λ - 3 = 0$

$λ - 2 = 0$

Step 1.7.3

Set

$λ - 3$ equal to

$0$ and solve for

$λ$ .

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Step 1.7.3.1

Set

$λ - 3$ equal to

$0$ .

$λ - 3 = 0$

Step 1.7.3.2

Add

$3$ to both sides of the equation.

$λ = 3$

Step 1.7.4

Set

$λ - 2$ equal to

$0$ and solve for

$λ$ .

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Step 1.7.4.1

Set

$λ - 2$ equal to

$0$ .

$λ - 2 = 0$

Step 1.7.4.2

Add

$2$ to both sides of the equation.

$λ = 2$

Step 1.7.5

The final solution is all the values that make

$(λ - 3) (λ - 2) = 0$ true.

$λ = 3, 2$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where

$N$ is the null space and

$I$ is the identity matrix.

$ε_{A} = N (A - λ I_{2})$

Step 3

Find the eigenvector using the eigenvalue

$λ = 3$ .

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Step 3.1

Substitute the known values into the formula.

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

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply

$- 3$ by each element of the matrix.

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

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply

$- 3$ by

$1$ .

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

Step 3.2.1.2.2

Multiply

$- 3$ by

$0$ .

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

Step 3.2.1.2.3

Multiply

$- 3$ by

$0$ .

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

Step 3.2.1.2.4

Multiply

$- 3$ by

$1$ .

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

Step 3.2.2

Add the corresponding elements.

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

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract

$3$ from

$- 1$ .

$[\begin{array}{c} - 4 & 2 + 0 \\ - 6 + 0 & 6 - 3 \end{array}]$

Step 3.2.3.2

Add

$2$ and

$0$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 + 0 & 6 - 3 \end{array}]$

Step 3.2.3.3

Add

$- 6$ and

$0$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 & 6 - 3 \end{array}]$

Step 3.2.3.4

Subtract

$3$ from

$6$ .

$[\begin{array}{c} - 4 & 2 \\ - 6 & 3 \end{array}]$

Step 3.3

Find the null space when

$λ = 3$ .

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Step 3.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} - 4 & 2 & 0 \\ - 6 & 3 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{4}$ to make the entry at

$1, 1$ a

$1$ .

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Step 3.3.2.1.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{4}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 2 & - \frac{1}{4} \cdot 0 \\ - 6 & 3 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify

$R_{1}$ .

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

Step 3.3.2.2

Perform the row operation

$R_{2} = R_{2} + 6 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 3.3.2.2.1

Perform the row operation

$R_{2} = R_{2} + 6 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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

Step 3.3.2.2.2

Simplify

$R_{2}$ .

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

Step 3.3.3

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

$x - \frac{1}{2} y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{y}{2} \\ y \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} \frac{1}{2} \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} \frac{1}{2} \\ 1 \end{array}] | y \in R}$

Step 3.3.7

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

${[\begin{array}{c} \frac{1}{2} \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue

$λ = 2$ .

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Step 4.1

Substitute the known values into the formula.

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

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply

$- 2$ by each element of the matrix.

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

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply

$- 2$ by

$1$ .

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

Step 4.2.1.2.2

Multiply

$- 2$ by

$0$ .

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

Step 4.2.1.2.3

Multiply

$- 2$ by

$0$ .

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

Step 4.2.1.2.4

Multiply

$- 2$ by

$1$ .

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

Step 4.2.2

Add the corresponding elements.

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

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract

$2$ from

$- 1$ .

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

Step 4.2.3.2

Add

$2$ and

$0$ .

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

Step 4.2.3.3

Add

$- 6$ and

$0$ .

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

Step 4.2.3.4

Subtract

$2$ from

$6$ .

$[\begin{array}{c} - 3 & 2 \\ - 6 & 4 \end{array}]$

Step 4.3

Find the null space when

$λ = 2$ .

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Step 4.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} - 3 & 2 & 0 \\ - 6 & 4 & 0 \end{array}]$

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{3}$ to make the entry at

$1, 1$ a

$1$ .

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Step 4.3.2.1.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{3}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 2 & - \frac{1}{3} \cdot 0 \\ - 6 & 4 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 \\ - 6 & 4 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation

$R_{2} = R_{2} + 6 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 4.3.2.2.1

Perform the row operation

$R_{2} = R_{2} + 6 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & 0 \\ - 6 + 6 \cdot 1 & 4 + 6 (- \frac{2}{3}) & 0 + 6 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify

$R_{2}$ .

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

Step 4.3.3

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

$x - \frac{2}{3} y = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{2 y}{3} \\ y \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${y [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}] | y \in R}$

Step 4.3.7

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

${[\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]}$

Step 5

The eigenspace of

$A$ is the list of the vector space for each eigenvalue.

${[\begin{array}{c} \frac{1}{2} \\ 1 \end{array}], [\begin{array}{c} \frac{2}{3} \\ 1 \end{array}]}$

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