Algebra Examples

[\begin{matrix} 0 & 1 - 1 & 6 \end{matrix}]

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

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

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

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{matrix} 0 & 1 - 1 & 6 \end{matrix}]

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

A

$A$ .

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

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

Step 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} 0 & 1 - 1 & 6 \end{matrix}] - λ [\begin{matrix} 1 & 0 0 & 1 \end{matrix}])

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

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

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

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{matrix} 0 & 1 - 1 & 6 \end{matrix}] + [\begin{matrix} - λ \cdot 1 & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

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{matrix} 0 & 1 - 1 & 6 \end{matrix}] + [\begin{matrix} - λ & - λ \cdot 0 - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

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{matrix} 0 & 1 - 1 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 λ - λ \cdot 0 & - λ \cdot 1 \end{matrix}])

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

Step 4.1.2.2.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

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{matrix} 0 & 1 - 1 & 6 \end{matrix}] + [\begin{matrix} - λ & 0 0 λ & - λ \cdot 1 \end{matrix}])

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

Step 4.1.2.3.2

Multiply

0

$0$ by

λ

$λ$ .

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

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

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

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

Step 4.1.2.4

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

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

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

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

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

Step 4.2

Add the corresponding elements.

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

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

Step 4.3

Simplify each element.

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Step 4.3.1

Subtract

λ

$λ$ from

0

$0$ .

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

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

Step 4.3.2

Add

1

$1$ and

0

$0$ .

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

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

Step 4.3.3

Add

- 1

$- 1$ and

0

$0$ .

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

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

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

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

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

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

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{matrix} a & b c & d \end{matrix} ∣ ∣ ∣ = a d - c b

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

p (λ) = - λ (6 - λ) - (- 1 \cdot 1)

$p (λ) = - λ (6 - λ) - (- 1 \cdot 1)$

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

Apply the distributive property.

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

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

Step 5.2.1.2

Multiply

6

$6$ by

- 1

$- 1$ .

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

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

Step 5.2.1.3

Rewrite using the commutative property of multiplication.

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

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

Step 5.2.1.4

Simplify each term.

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Step 5.2.1.4.1

Multiply

λ

$λ$ by

λ

$λ$ by adding the exponents.

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Step 5.2.1.4.1.1

Move

λ

$λ$ .

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

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

Step 5.2.1.4.1.2

Multiply

λ

$λ$ by

λ

$λ$ .

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

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

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

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

Step 5.2.1.4.2

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

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

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

Step 5.2.1.4.3

Multiply

λ^{2}

$λ^{2}$ by

1

$1$ .

p (λ) = - 6 λ + λ^{2} - (- 1 \cdot 1)

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

p (λ) = - 6 λ + λ^{2} - (- 1 \cdot 1)

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

Step 5.2.1.5

Multiply

- (- 1 \cdot 1)

$- (- 1 \cdot 1)$ .

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Step 5.2.1.5.1

Multiply

- 1

$- 1$ by

1

$1$ .

p (λ) = - 6 λ + λ^{2} - - 1

$p (λ) = - 6 λ + λ^{2} - - 1$

Step 5.2.1.5.2

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

p (λ) = - 6 λ + λ^{2} + 1

$p (λ) = - 6 λ + λ^{2} + 1$

p (λ) = - 6 λ + λ^{2} + 1

$p (λ) = - 6 λ + λ^{2} + 1$

p (λ) = - 6 λ + λ^{2} + 1

$p (λ) = - 6 λ + λ^{2} + 1$

Step 5.2.2

Reorder

- 6 λ

$- 6 λ$ and

λ^{2}

$λ^{2}$ .

p (λ) = λ^{2} - 6 λ + 1

$p (λ) = λ^{2} - 6 λ + 1$

p (λ) = λ^{2} - 6 λ + 1

$p (λ) = λ^{2} - 6 λ + 1$

p (λ) = λ^{2} - 6 λ + 1

$p (λ) = λ^{2} - 6 λ + 1$

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