Linear Algebra Examples

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

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

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

Step 3.2

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $3$ and $0$ .

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

Step 4.3.2

Add $2$ and $0$ .

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

Step 5

Find the determinant.

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Step 5.1

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

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

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

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Step 5.2.1.1.1

Apply the distributive property.

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

Step 5.2.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.3

Apply the distributive property.

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

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 $- 1$ by $1$ .

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

Step 5.2.1.2.1.2

Multiply $- λ$ by $1$ .

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

Step 5.2.1.2.1.3

Multiply $- λ \cdot - 1$ .

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Step 5.2.1.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.1.2.1.3.2

Multiply $λ$ by $1$ .

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

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.2.1.2.1.5

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

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Step 5.2.1.2.1.5.1

Move $λ$ .

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

Step 5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 5.2.1.2.2

Add $- λ$ and $λ$ .

$p (λ) = - 1 + 0 + λ^{2} - 2 \cdot 3$

Step 5.2.1.2.3

Add $- 1$ and $0$ .

$p (λ) = - 1 + λ^{2} - 2 \cdot 3$

Step 5.2.1.3

Multiply $- 2$ by $3$ .

$p (λ) = - 1 + λ^{2} - 6$

Step 5.2.2

Subtract $6$ from $- 1$ .

$p (λ) = λ^{2} - 7$

Step 6

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

$λ^{2} - 7 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Add $7$ to both sides of the equation.

$λ^{2} = 7$

Step 7.2

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

$λ = \pm \sqrt{7}$

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 $\pm$ to find the first solution.

$λ = \sqrt{7}$

Step 7.3.2

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

$λ = - \sqrt{7}$

Step 7.3.3

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

$λ = \sqrt{7}, - \sqrt{7}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = \sqrt{7}, - \sqrt{7}$

Decimal Form:

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