Algebra Examples

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

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

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

$A$ .

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

Step 1.3.2

Substitute

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

$I_{2}$ .

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply

$- 1$ by

$1$ .

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

Step 1.4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply

$0$ by

$- 1$ .

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

Step 1.4.1.2.2.2

Multiply

$0$ by

$λ$ .

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

Step 1.4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply

$0$ by

$- 1$ .

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

Step 1.4.1.2.3.2

Multiply

$0$ by

$λ$ .

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

Step 1.4.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add

$2$ and

$0$ .

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

Step 1.4.3.2

Add

$3$ and

$0$ .

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

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

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

$p (λ) = (4 - λ) (1 - λ) - 3 \cdot 2$

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

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

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Step 1.5.2.1.1.1

Apply the distributive property.

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

Step 1.5.2.1.1.2

Apply the distributive property.

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

Step 1.5.2.1.1.3

Apply the distributive property.

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

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

$1$ .

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

Step 1.5.2.1.2.1.2

Multiply

$- 1$ by

$4$ .

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

Step 1.5.2.1.2.1.3

Multiply

$- 1$ by

$1$ .

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

Step 1.5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

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

$λ$ .

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

Step 1.5.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

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

Step 1.5.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

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

Step 1.5.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

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

Step 1.5.2.1.2.2

Subtract

$λ$ from

$- 4 λ$ .

$p (λ) = 4 - 5 λ + λ^{2} - 3 \cdot 2$

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.

$ε_{A} = N (A - λ I_{2})$

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.

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

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.

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

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply

$- 1$ by

$1$ .

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

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

$- 1$ .

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

Step 3.2.1.2.2.2

Multiply

$0$ by

$\frac{5 + \sqrt{33}}{2}$ .

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

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

$- 1$ .

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

Step 3.2.1.2.3.2

Multiply

$0$ by

$\frac{5 + \sqrt{33}}{2}$ .

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

Step 3.2.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 3.2.2

Add the corresponding elements.

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

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

To write

$4$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

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

Step 3.2.3.2

Combine

$4$ and

$\frac{2}{2}$ .

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

Step 3.2.3.3

Combine the numerators over the common denominator.

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

Step 3.2.3.4

Simplify the numerator.

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Step 3.2.3.4.1

Multiply

$4$ by

$2$ .

$[\begin{array}{c} \frac{8 - (5 + \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.4.2

Apply the distributive property.

$[\begin{array}{c} \frac{8 - 1 \cdot 5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.4.3

Multiply

$- 1$ by

$5$ .

$[\begin{array}{c} \frac{8 - 5 - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 + \sqrt{33}}{2} \end{array}]$

Step 3.2.3.4.4

Subtract

$5$ from

$8$ .

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

Step 3.2.3.5

Add

$2$ and

$0$ .

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

Step 3.2.3.6

Add

$3$ and

$0$ .

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

Step 3.2.3.7

Write

$1$ as a fraction with a common denominator.

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

Step 3.2.3.8

Combine the numerators over the common denominator.

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

Step 3.2.3.9

Simplify the numerator.

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Step 3.2.3.9.1

Apply the distributive property.

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

Step 3.2.3.9.2

Multiply

$- 1$ by

$5$ .

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

Step 3.2.3.9.3

Subtract

$5$ from

$2$ .

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

Step 3.2.3.10

Rewrite

$- 3$ as

$- 1 (3)$ .

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

Step 3.2.3.11

Factor

$- 1$ out of

$- \sqrt{33}$ .

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

Step 3.2.3.12

Factor

$- 1$ out of

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

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

Step 3.2.3.13

Move the negative in front of the fraction.

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

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

$1$ .

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

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.

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

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.

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

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.

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

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply

$- 1$ by

$1$ .

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

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

$- 1$ .

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

Step 4.2.1.2.2.2

Multiply

$0$ by

$\frac{5 - \sqrt{33}}{2}$ .

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

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

$- 1$ .

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

Step 4.2.1.2.3.2

Multiply

$0$ by

$\frac{5 - \sqrt{33}}{2}$ .

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

Step 4.2.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 4.2.2

Add the corresponding elements.

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

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

To write

$4$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

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

Step 4.2.3.2

Combine

$4$ and

$\frac{2}{2}$ .

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

Step 4.2.3.3

Combine the numerators over the common denominator.

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

Step 4.2.3.4

Simplify the numerator.

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Step 4.2.3.4.1

Multiply

$4$ by

$2$ .

$[\begin{array}{c} \frac{8 - (5 - \sqrt{33})}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.2

Apply the distributive property.

$[\begin{array}{c} \frac{8 - 1 \cdot 5 - - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.3

Multiply

$- 1$ by

$5$ .

$[\begin{array}{c} \frac{8 - 5 - - \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.4

Multiply

$- - \sqrt{33}$ .

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Step 4.2.3.4.4.1

Multiply

$- 1$ by

$- 1$ .

$[\begin{array}{c} \frac{8 - 5 + 1 \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.4.2

Multiply

$\sqrt{33}$ by

$1$ .

$[\begin{array}{c} \frac{8 - 5 + \sqrt{33}}{2} & 2 + 0 \\ 3 + 0 & 1 - \frac{5 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.5

Subtract

$5$ from

$8$ .

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

Step 4.2.3.5

Add

$2$ and

$0$ .

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

Step 4.2.3.6

Add

$3$ and

$0$ .

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

Step 4.2.3.7

Write

$1$ as a fraction with a common denominator.

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

Step 4.2.3.8

Combine the numerators over the common denominator.

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

Step 4.2.3.9

Simplify the numerator.

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Step 4.2.3.9.1

Apply the distributive property.

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

Step 4.2.3.9.2

Multiply

$- 1$ by

$5$ .

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

Step 4.2.3.9.3

Multiply

$- - \sqrt{33}$ .

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Step 4.2.3.9.3.1

Multiply

$- 1$ by

$- 1$ .

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

Step 4.2.3.9.3.2

Multiply

$\sqrt{33}$ by

$1$ .

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

Step 4.2.3.9.4

Subtract

$5$ from

$2$ .

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

Step 4.2.3.10

Rewrite

$- 3$ as

$- 1 (3)$ .

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

Step 4.2.3.11

Factor

$- 1$ out of

$\sqrt{33}$ .

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

Step 4.2.3.12

Factor

$- 1$ out of

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

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

Step 4.2.3.13

Move the negative in front of the fraction.

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

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

$1$ .

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

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.

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

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

$A$ 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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