Algebra Examples

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

Step 1

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

$p (λ) = determinant (A - λ I_{2})$

Step 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 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $- λ$ by each element of the 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 $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Subtract $λ$ from $0$ .

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

Step 4.3.2

Add $1$ and $0$ .

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

Step 4.3.3

Add $- 1$ and $0$ .

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

Step 5

Find the determinant.

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Step 5.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = - λ (6 - λ) - (- 1 \cdot 1)$

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.

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

Step 5.2.1.2

Multiply $6$ by $- 1$ .

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

Step 5.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.2.1.4

Simplify each term.

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Step 5.2.1.4.1

Multiply $λ$ by $λ$ by adding the exponents.

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Step 5.2.1.4.1.1

Move $λ$ .

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

Step 5.2.1.4.1.2

Multiply $λ$ by $λ$ .

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

Step 5.2.1.4.2

Multiply $- 1$ by $- 1$ .

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

Step 5.2.1.4.3

Multiply $λ^{2}$ by $1$ .

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

Step 5.2.1.5

Multiply $- (- 1 \cdot 1)$ .

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Step 5.2.1.5.1

Multiply $- 1$ by $1$ .

$p (λ) = - 6 λ + λ^{2} - - 1$

Step 5.2.1.5.2

Multiply $- 1$ by $- 1$ .

$p (λ) = - 6 λ + λ^{2} + 1$

Step 5.2.2

Reorder $- 6 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 6 λ + 1$

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