Algebra Examples

$B = [\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}]$

Step 1

Find the eigenvalues.

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Step 1.1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{2})$

Step 1.2

The identity matrix or unit matrix of size $2$ is the $2 \times 2$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{2})$ .

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Step 1.3.1

Substitute $[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] - λ I_{2})$

Step 1.3.2

Substitute $[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ for $I_{2}$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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 $- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 + 0 \\ 3 + 0 & 4 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 \\ 3 + 0 & 4 - λ \end{array}]$

Step 1.4.3.2

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 \\ 3 & 4 - λ \end{array}]$

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

$p (λ) = (1 - λ) (4 - λ) - 3 \cdot 2$

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 $(1 - λ) (4 - λ)$ using the FOIL Method.

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Step 1.5.2.1.1.1

Apply the distributive property.

$p (λ) = 1 (4 - λ) - λ (4 - λ) - 3 \cdot 2$

Step 1.5.2.1.1.2

Apply the distributive property.

$p (λ) = 1 \cdot 4 + 1 (- λ) - λ (4 - λ) - 3 \cdot 2$

Step 1.5.2.1.1.3

Apply the distributive property.

$p (λ) = 1 \cdot 4 + 1 (- λ) - λ \cdot 4 - λ (- λ) - 3 \cdot 2$

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 $4$ by $1$ .

$p (λ) = 4 + 1 (- λ) - λ \cdot 4 - λ (- λ) - 3 \cdot 2$

Step 1.5.2.1.2.1.2

Multiply $- λ$ by $1$ .

$p (λ) = 4 - λ - λ \cdot 4 - λ (- λ) - 3 \cdot 2$

Step 1.5.2.1.2.1.3

Multiply $4$ by $- 1$ .

$p (λ) = 4 - λ - 4 λ - λ (- λ) - 3 \cdot 2$

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 4 - λ - 4 λ - 1 \cdot - 1 λ \cdot λ - 3 \cdot 2$

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 $λ$ .

$p (λ) = 4 - λ - 4 λ - 1 \cdot - 1 (λ \cdot λ) - 3 \cdot 2$

Step 1.5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 4 - λ - 4 λ - 1 \cdot - 1 λ^{2} - 3 \cdot 2$

Step 1.5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 4 - λ - 4 λ + 1 λ^{2} - 3 \cdot 2$

Step 1.5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 4 - λ - 4 λ + λ^{2} - 3 \cdot 2$

Step 1.5.2.1.2.2

Subtract $4 λ$ from $- λ$ .

$p (λ) = 4 - 5 λ + λ^{2} - 3 \cdot 2$

Step 1.5.2.1.3

Multiply $- 3$ by $2$ .

$p (λ) = 4 - 5 λ + λ^{2} - 6$

Step 1.5.2.2

Subtract $6$ from $4$ .

$p (λ) = - 5 λ + λ^{2} - 2$

Step 1.5.2.3

Reorder $- 5 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 5 λ - 2$

Step 1.6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{2} - 5 λ - 2 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.7.2

Substitute the values $a = 1$ , $b = - 5$ , and $c = - 2$ into the quadratic formula and solve for $λ$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 1.7.3

Simplify.

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Step 1.7.3.1

Simplify the numerator.

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Step 1.7.3.1.1

Raise $- 5$ to the power of $2$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 1.7.3.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 1.7.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Step 1.7.3.1.2.2

Multiply $- 4$ by $- 2$ .

$λ = \frac{5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Step 1.7.3.1.3

Add $25$ and $8$ .

$λ = \frac{5 \pm \sqrt{33}}{2 \cdot 1}$

Step 1.7.3.2

Multiply $2$ by $1$ .

$λ = \frac{5 \pm \sqrt{33}}{2}$

Step 1.7.4

The final answer is the combination of both solutions.

$λ = \frac{5 + \sqrt{33}}{2}, \frac{5 - \sqrt{33}}{2}$

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.

$ε_{B} = N (B - λ I_{2})$

Step 3

Find the eigenvector using the eigenvalue $λ = \frac{5 + \sqrt{33}}{2}$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] - \frac{5 + \sqrt{33}}{2} [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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 $- \frac{5 + \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} \cdot 1 & - \frac{5 + \sqrt{33}}{2} \cdot 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & - \frac{5 + \sqrt{33}}{2} \cdot 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- \frac{5 + \sqrt{33}}{2} \cdot 0$ .

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Step 3.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \frac{5 + \sqrt{33}}{2} \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2.2.2

Multiply $0$ by $\frac{5 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ - \frac{5 + \sqrt{33}}{2} \cdot 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- \frac{5 + \sqrt{33}}{2} \cdot 0$ .

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Step 3.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ 0 \frac{5 + \sqrt{33}}{2} & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2.3.2

Multiply $0$ by $\frac{5 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 + \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 1 - \frac{5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Write $1$ as a fraction with a common denominator.

$[\begin{array}{c} \frac{2}{2} - \frac{5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.2

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{2 - (5 + \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.3

Simplify the numerator.

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Step 3.2.3.3.1

Apply the distributive property.

$[\begin{array}{c} \frac{2 - 1 \cdot 5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.3.2

Multiply $- 1$ by $5$ .

$[\begin{array}{c} \frac{2 - 5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.3.3

Subtract $5$ from $2$ .

$[\begin{array}{c} \frac{- 3 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.4

Rewrite $- 3$ as $- 1 (3)$ .

$[\begin{array}{c} \frac{- 1 (3) - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.5

Factor $- 1$ out of $- \sqrt{33}$ .

$[\begin{array}{c} \frac{- 1 (3) - (\sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.6

Factor $- 1$ out of $- 1 (3) - (\sqrt{33})$ .

$[\begin{array}{c} \frac{- 1 (3 + \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.7

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.8

Add $2$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 + 0 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.9

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & 4 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.10

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & 4 \cdot \frac{2}{2} - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.11

Combine $4$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{4 \cdot 2}{2} - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.12

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{4 \cdot 2 - (5 + \sqrt{33})}{2} \end{array}]$

Step 3.2.3.13

Simplify the numerator.

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Step 3.2.3.13.1

Multiply $4$ by $2$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - (5 + \sqrt{33})}{2} \end{array}]$

Step 3.2.3.13.2

Apply the distributive property.

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 1 \cdot 5 - \sqrt{33}}{2} \end{array}]$

Step 3.2.3.13.3

Multiply $- 1$ by $5$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 5 - \sqrt{33}}{2} \end{array}]$

Step 3.2.3.13.4

Subtract $5$ from $8$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 \\ 3 & \frac{3 - \sqrt{33}}{2} \end{array}]$

Step 3.3

Find the null space when $λ = \frac{5 + \sqrt{33}}{2}$ .

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Step 3.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{33}}{2} & 2 & 0 \\ 3 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

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 $- \frac{2}{3 + \sqrt{33}}$ 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 $- \frac{2}{3 + \sqrt{33}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{3 + \sqrt{33}} (- \frac{3 + \sqrt{33}}{2}) & - \frac{2}{3 + \sqrt{33}} \cdot 2 & - \frac{2}{3 + \sqrt{33}} \cdot 0 \\ 3 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3 - \sqrt{33}}{6} & 0 \\ 3 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

$[\begin{array}{c} 1 & \frac{3 - \sqrt{33}}{6} & 0 \\ 3 - 3 \cdot 1 & \frac{3 - \sqrt{33}}{2} - 3 \frac{3 - \sqrt{33}}{6} & 0 - 3 \cdot 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 - \sqrt{33}}{6} & 0 \\ 0 & 0 & 0 \end{array}]$

Step 3.3.3

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

$x + \frac{3 - \sqrt{33}}{6} y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{y}{2} + \frac{\sqrt{33} y}{6} \\ y \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - \frac{1}{2} + \frac{\sqrt{33}}{6} \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} - \frac{1}{2} + \frac{\sqrt{33}}{6} \\ 1 \end{array}] | y \in R}$

Step 3.3.7

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

${[\begin{array}{c} - \frac{1}{2} + \frac{\sqrt{33}}{6} \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue $λ = \frac{5 - \sqrt{33}}{2}$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] - \frac{5 - \sqrt{33}}{2} [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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 $- \frac{5 - \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} \cdot 1 & - \frac{5 - \sqrt{33}}{2} \cdot 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & - \frac{5 - \sqrt{33}}{2} \cdot 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- \frac{5 - \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \frac{5 - \sqrt{33}}{2} \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2.2

Multiply $0$ by $\frac{5 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ - \frac{5 - \sqrt{33}}{2} \cdot 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- \frac{5 - \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ 0 \frac{5 - \sqrt{33}}{2} & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3.2

Multiply $0$ by $\frac{5 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 1 & 2 \\ 3 & 4 \end{array}] + [\begin{array}{c} - \frac{5 - \sqrt{33}}{2} & 0 \\ 0 & - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 1 - \frac{5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Write $1$ as a fraction with a common denominator.

$[\begin{array}{c} \frac{2}{2} - \frac{5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.2

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{2 - (5 - \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3

Simplify the numerator.

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Step 4.2.3.3.1

Apply the distributive property.

$[\begin{array}{c} \frac{2 - 1 \cdot 5 - - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3.2

Multiply $- 1$ by $5$ .

$[\begin{array}{c} \frac{2 - 5 - - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3.3

Multiply $- - \sqrt{33}$ .

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Step 4.2.3.3.3.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} \frac{2 - 5 + 1 \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3.3.2

Multiply $\sqrt{33}$ by $1$ .

$[\begin{array}{c} \frac{2 - 5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3.4

Subtract $5$ from $2$ .

$[\begin{array}{c} \frac{- 3 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4

Rewrite $- 3$ as $- 1 (3)$ .

$[\begin{array}{c} \frac{- 1 (3) + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.5

Factor $- 1$ out of $\sqrt{33}$ .

$[\begin{array}{c} \frac{- 1 (3) - 1 (- \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.6

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{33})$ .

$[\begin{array}{c} \frac{- 1 (3 - \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.7

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.8

Add $2$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 + 0 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.9

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & 4 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.10

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & 4 \cdot \frac{2}{2} - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.11

Combine $4$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{4 \cdot 2}{2} - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.12

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{4 \cdot 2 - (5 - \sqrt{33})}{2} \end{array}]$

Step 4.2.3.13

Simplify the numerator.

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Step 4.2.3.13.1

Multiply $4$ by $2$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - (5 - \sqrt{33})}{2} \end{array}]$

Step 4.2.3.13.2

Apply the distributive property.

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 1 \cdot 5 - - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.13.3

Multiply $- 1$ by $5$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 5 - - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.13.4

Multiply $- - \sqrt{33}$ .

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Step 4.2.3.13.4.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 5 + 1 \sqrt{33}}{2} \end{array}]$

Step 4.2.3.13.4.2

Multiply $\sqrt{33}$ by $1$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{8 - 5 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.13.5

Subtract $5$ from $8$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 \\ 3 & \frac{3 + \sqrt{33}}{2} \end{array}]$

Step 4.3

Find the null space when $λ = \frac{5 - \sqrt{33}}{2}$ .

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Step 4.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{33}}{2} & 2 & 0 \\ 3 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

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 $- \frac{2}{3 - \sqrt{33}}$ 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 $- \frac{2}{3 - \sqrt{33}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{3 - \sqrt{33}} (- \frac{3 - \sqrt{33}}{2}) & - \frac{2}{3 - \sqrt{33}} \cdot 2 & - \frac{2}{3 - \sqrt{33}} \cdot 0 \\ 3 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3 + \sqrt{33}}{6} & 0 \\ 3 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

$[\begin{array}{c} 1 & \frac{3 + \sqrt{33}}{6} & 0 \\ 3 - 3 \cdot 1 & \frac{3 + \sqrt{33}}{2} - 3 \frac{3 + \sqrt{33}}{6} & 0 - 3 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 + \sqrt{33}}{6} & 0 \\ 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x + \frac{3 + \sqrt{33}}{6} y = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{y}{2} - \frac{\sqrt{33} y}{6} \\ y \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - \frac{1}{2} - \frac{\sqrt{33}}{6} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${y [\begin{array}{c} - \frac{1}{2} - \frac{\sqrt{33}}{6} \\ 1 \end{array}] | y \in R}$

Step 4.3.7

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

${[\begin{array}{c} - \frac{1}{2} - \frac{\sqrt{33}}{6} \\ 1 \end{array}]}$

Step 5

The eigenspace of $B$ is the list of the vector space for each eigenvalue.

${[\begin{array}{c} - \frac{1}{2} + \frac{\sqrt{33}}{6} \\ 1 \end{array}], [\begin{array}{c} - \frac{1}{2} - \frac{\sqrt{33}}{6} \\ 1 \end{array}]}$

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