Linear Algebra Examples

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

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

Step 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 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{array}{c} 1 & 0 \\ 0 & 1 \end{array}]

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

Step 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 3.1

Substitute

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

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

A

$A$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ I_{2})

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ I_{2})$

Step 3.2

Substitute

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

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

I_{2}

$I_{2}$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply

- λ

$- λ$ by each element of the matrix.

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])

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

Step 4.1.2.2

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])

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

Step 4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])

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

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])

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

Step 4.1.2.3

Multiply

- λ \cdot 0

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

0

$0$ by

- 1

$- 1$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])

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

Step 4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])

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

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])

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

Step 4.1.2.4

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

p (λ) = determinant [\begin{array}{c} 3 - λ & 2 + 0 \\ 4 + 0 & 6 - λ \end{array}]

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 + 0 \\ 4 + 0 & 6 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add

2

$2$ and

0

$0$ .

p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 + 0 & 6 - λ \end{array}]

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 + 0 & 6 - λ \end{array}]$

Step 4.3.2

Add

4

$4$ and

0

$0$ .

p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]$

p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]$

p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]$

Step 5

Find the determinant.

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Step 5.1

The determinant of a

2 \times 2

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

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

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

p (λ) = (3 - λ) (6 - λ) - 4 \cdot 2

$p (λ) = (3 - λ) (6 - λ) - 4 \cdot 2$

Step 5.2

Simplify the determinant.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Expand

(3 - λ) (6 - λ)

$(3 - λ) (6 - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

p (λ) = 3 (6 - λ) - λ (6 - λ) - 4 \cdot 2

$p (λ) = 3 (6 - λ) - λ (6 - λ) - 4 \cdot 2$

Step 5.2.1.1.2

Apply the distributive property.

p (λ) = 3 \cdot 6 + 3 (- λ) - λ (6 - λ) - 4 \cdot 2

$p (λ) = 3 \cdot 6 + 3 (- λ) - λ (6 - λ) - 4 \cdot 2$

Step 5.2.1.1.3

Apply the distributive property.

p (λ) = 3 \cdot 6 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2

$p (λ) = 3 \cdot 6 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

p (λ) = 3 \cdot 6 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2

$p (λ) = 3 \cdot 6 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2

Simplify and combine like terms.

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Step 5.2.1.2.1

Simplify each term.

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Step 5.2.1.2.1.1

Multiply

3

$3$ by

6

$6$ .

p (λ) = 18 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2

$p (λ) = 18 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.2

Multiply

- 1

$- 1$ by

3

$3$ .

p (λ) = 18 - 3 λ - λ \cdot 6 - λ (- λ) - 4 \cdot 2

$p (λ) = 18 - 3 λ - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.3

Multiply

6

$6$ by

- 1

$- 1$ .

p (λ) = 18 - 3 λ - 6 λ - λ (- λ) - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ \cdot λ - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ \cdot λ - 4 \cdot 2$

Step 5.2.1.2.1.5

Multiply

λ

$λ$ by

λ

$λ$ by adding the exponents.

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Step 5.2.1.2.1.5.1

Move

λ

$λ$ .

p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 (λ \cdot λ) - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 (λ \cdot λ) - 4 \cdot 2$

Step 5.2.1.2.1.5.2

Multiply

λ

$λ$ by

λ

$λ$ .

p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 2$

p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 2$

Step 5.2.1.2.1.6

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

p (λ) = 18 - 3 λ - 6 λ + 1 λ^{2} - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ + 1 λ^{2} - 4 \cdot 2$

Step 5.2.1.2.1.7

Multiply

λ^{2}

$λ^{2}$ by

1

$1$ .

p (λ) = 18 - 3 λ - 6 λ + λ^{2} - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ + λ^{2} - 4 \cdot 2$

p (λ) = 18 - 3 λ - 6 λ + λ^{2} - 4 \cdot 2

$p (λ) = 18 - 3 λ - 6 λ + λ^{2} - 4 \cdot 2$

Step 5.2.1.2.2

Subtract

6 λ

$6 λ$ from

- 3 λ

$- 3 λ$ .

p (λ) = 18 - 9 λ + λ^{2} - 4 \cdot 2

$p (λ) = 18 - 9 λ + λ^{2} - 4 \cdot 2$

p (λ) = 18 - 9 λ + λ^{2} - 4 \cdot 2

$p (λ) = 18 - 9 λ + λ^{2} - 4 \cdot 2$

Step 5.2.1.3

Multiply

- 4

$- 4$ by

2

$2$ .

p (λ) = 18 - 9 λ + λ^{2} - 8

$p (λ) = 18 - 9 λ + λ^{2} - 8$

p (λ) = 18 - 9 λ + λ^{2} - 8

$p (λ) = 18 - 9 λ + λ^{2} - 8$

Step 5.2.2

Subtract

8

$8$ from

18

$18$ .

p (λ) = - 9 λ + λ^{2} + 10

$p (λ) = - 9 λ + λ^{2} + 10$

Step 5.2.3

Reorder

- 9 λ

$- 9 λ$ and

λ^{2}

$λ^{2}$ .

p (λ) = λ^{2} - 9 λ + 10

$p (λ) = λ^{2} - 9 λ + 10$

p (λ) = λ^{2} - 9 λ + 10

$p (λ) = λ^{2} - 9 λ + 10$

p (λ) = λ^{2} - 9 λ + 10

$p (λ) = λ^{2} - 9 λ + 10$

Step 6

Set the characteristic polynomial equal to

0

$0$ to find the eigenvalues

λ

$λ$ .

λ^{2} - 9 λ + 10 = 0

$λ^{2} - 9 λ + 10 = 0$

Step 7

Solve for

λ

$λ$ .

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Step 7.1

Use the quadratic formula to find the solutions.

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

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

Step 7.2

Substitute the values

a = 1

$a = 1$ ,

b = - 9

$b = - 9$ , and

c = 10

$c = 10$ into the quadratic formula and solve for

λ

$λ$ .

\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

Raise

- 9

$- 9$ to the power of

2

$2$ .

λ = \frac{9 \pm \sqrt{81 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{81 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Step 7.3.1.2

Multiply

- 4 \cdot 1 \cdot 10

$- 4 \cdot 1 \cdot 10$ .

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Step 7.3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

λ = \frac{9 \pm \sqrt{81 - 4 \cdot 10}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{81 - 4 \cdot 10}}{2 \cdot 1}$

Step 7.3.1.2.2

Multiply

- 4

$- 4$ by

10

$10$ .

λ = \frac{9 \pm \sqrt{81 - 40}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{81 - 40}}{2 \cdot 1}$

λ = \frac{9 \pm \sqrt{81 - 40}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{81 - 40}}{2 \cdot 1}$

Step 7.3.1.3

Subtract

40

$40$ from

81

$81$ .

λ = \frac{9 \pm \sqrt{41}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{41}}{2 \cdot 1}$

λ = \frac{9 \pm \sqrt{41}}{2 \cdot 1}

$λ = \frac{9 \pm \sqrt{41}}{2 \cdot 1}$

Step 7.3.2

Multiply

2

$2$ by

1

$1$ .

λ = \frac{9 \pm \sqrt{41}}{2}

$λ = \frac{9 \pm \sqrt{41}}{2}$

λ = \frac{9 \pm \sqrt{41}}{2}

$λ = \frac{9 \pm \sqrt{41}}{2}$

Step 7.4

The final answer is the combination of both solutions.

λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}

$λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}$

λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}

$λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}

$λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}$

Decimal Form:

λ = 7.70156211 \dots, 1.29843788 \dots

$λ = 7.70156211 \dots, 1.29843788 \dots$

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