Linear Algebra Examples

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

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

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} 1 & 3 \\ 2 & - 1 \end{array}]

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

A

$A$ .

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

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

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

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

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

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

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

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

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

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

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

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

Step 4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

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

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

Step 4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

Step 4.1.2.4

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

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

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

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

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

Step 4.2

Add the corresponding elements.

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

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add

3

$3$ and

0

$0$ .

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

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

Step 4.3.2

Add

2

$2$ and

0

$0$ .

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

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

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

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

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

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

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) = (1 - λ) (- 1 - λ) - 2 \cdot 3

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

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

(1 - λ) (- 1 - λ)

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

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Step 5.2.1.1.1

Apply the distributive property.

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

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

Step 5.2.1.1.2

Apply the distributive property.

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

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

Step 5.2.1.1.3

Apply the distributive property.

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

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

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

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

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

- 1

$- 1$ by

1

$1$ .

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

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

Step 5.2.1.2.1.2

Multiply

- λ

$- λ$ by

1

$1$ .

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

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

Step 5.2.1.2.1.3

Multiply

- λ \cdot - 1

$- λ \cdot - 1$ .

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Step 5.2.1.2.1.3.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

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

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

Step 5.2.1.2.1.3.2

Multiply

λ

$λ$ by

1

$1$ .

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

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

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

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

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

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

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) = - 1 - λ + λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 3

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

Step 5.2.1.2.1.5.2

Multiply

λ

$λ$ by

λ

$λ$ .

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

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

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

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

Step 5.2.1.2.1.6

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

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

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

Step 5.2.1.2.1.7

Multiply

λ^{2}

$λ^{2}$ by

1

$1$ .

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

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

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

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

Step 5.2.1.2.2

Add

- λ

$- λ$ and

λ

$λ$ .

p (λ) = - 1 + 0 + λ^{2} - 2 \cdot 3

$p (λ) = - 1 + 0 + λ^{2} - 2 \cdot 3$

Step 5.2.1.2.3

Add

- 1

$- 1$ and

0

$0$ .

p (λ) = - 1 + λ^{2} - 2 \cdot 3

$p (λ) = - 1 + λ^{2} - 2 \cdot 3$

p (λ) = - 1 + λ^{2} - 2 \cdot 3

$p (λ) = - 1 + λ^{2} - 2 \cdot 3$

Step 5.2.1.3

Multiply

- 2

$- 2$ by

3

$3$ .

p (λ) = - 1 + λ^{2} - 6

$p (λ) = - 1 + λ^{2} - 6$

p (λ) = - 1 + λ^{2} - 6

$p (λ) = - 1 + λ^{2} - 6$

Step 5.2.2

Subtract

6

$6$ from

- 1

$- 1$ .

p (λ) = λ^{2} - 7

$p (λ) = λ^{2} - 7$

p (λ) = λ^{2} - 7

$p (λ) = λ^{2} - 7$

p (λ) = λ^{2} - 7

$p (λ) = λ^{2} - 7$

Step 6

Set the characteristic polynomial equal to

0

$0$ to find the eigenvalues

λ

$λ$ .

λ^{2} - 7 = 0

$λ^{2} - 7 = 0$

Step 7

Solve for

λ

$λ$ .

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Step 7.1

Add

7

$7$ to both sides of the equation.

λ^{2} = 7

$λ^{2} = 7$

Step 7.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

λ = \pm \sqrt{7}

$λ = \pm \sqrt{7}$

Step 7.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 7.3.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

λ = \sqrt{7}

$λ = \sqrt{7}$

Step 7.3.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

λ = - \sqrt{7}

$λ = - \sqrt{7}$

Step 7.3.3

The complete solution is the result of both the positive and negative portions of the solution.

λ = \sqrt{7}, - \sqrt{7}

$λ = \sqrt{7}, - \sqrt{7}$

λ = \sqrt{7}, - \sqrt{7}

$λ = \sqrt{7}, - \sqrt{7}$

λ = \sqrt{7}, - \sqrt{7}

$λ = \sqrt{7}, - \sqrt{7}$

Step 8

The result can be shown in multiple forms.

Exact Form:

λ = \sqrt{7}, - \sqrt{7}

$λ = \sqrt{7}, - \sqrt{7}$

Decimal Form:

λ = 2.64575131 \dots, - 2.64575131 \dots

$λ = 2.64575131 \dots, - 2.64575131 \dots$