Linear Algebra Examples

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

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

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

Step 3.2

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

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Subtract $λ$ from $0$ .

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

Step 4.3.2

Add $1$ and $0$ .

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

Step 4.3.3

Add $1$ and $0$ .

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

Step 4.3.4

Subtract $λ$ from $0$ .

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

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

$p (λ) = - λ (- λ) - 1 \cdot 1$

Step 5.2

Simplify each term.

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Step 5.2.1

Rewrite using the commutative property of multiplication.

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

Step 5.2.2

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

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Step 5.2.2.1

Move $λ$ .

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

Step 5.2.2.2

Multiply $λ$ by $λ$ .

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

Step 5.2.3

Multiply $- 1$ by $- 1$ .

$p (λ) = 1 λ^{2} - 1 \cdot 1$

Step 5.2.4

Multiply $λ^{2}$ by $1$ .

$p (λ) = λ^{2} - 1 \cdot 1$

Step 5.2.5

Multiply $- 1$ by $1$ .

$p (λ) = λ^{2} - 1$

Step 6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{2} - 1 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Add $1$ to both sides of the equation.

$λ^{2} = 1$

Step 7.2

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

$λ = \pm \sqrt{1}$

Step 7.3

Any root of $1$ is $1$ .

$λ = \pm 1$

Step 7.4

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

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Step 7.4.1

First, use the positive value of the $\pm$ to find the first solution.

$λ = 1$

Step 7.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$λ = - 1$

Step 7.4.3

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

$λ = 1, - 1$