Linear Algebra Examples

$[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$

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 $[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ I_{2})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 + 0 \\ 4 + 0 & 6 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 + 0 & 6 - λ \end{array}]$

Step 4.3.2

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 2 \\ 4 & 6 - λ \end{array}]$

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

$p (λ) = (3 - λ) (6 - λ) - 4 \cdot 2$

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 $(3 - λ) (6 - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

$p (λ) = 3 (6 - λ) - λ (6 - λ) - 4 \cdot 2$

Step 5.2.1.1.2

Apply the distributive property.

$p (λ) = 3 \cdot 6 + 3 (- λ) - λ (6 - λ) - 4 \cdot 2$

Step 5.2.1.1.3

Apply the distributive property.

$p (λ) = 3 \cdot 6 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

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 $3$ by $6$ .

$p (λ) = 18 + 3 (- λ) - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = 18 - 3 λ - λ \cdot 6 - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.3

Multiply $6$ by $- 1$ .

$p (λ) = 18 - 3 λ - 6 λ - λ (- λ) - 4 \cdot 2$

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ \cdot λ - 4 \cdot 2$

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 $λ$ .

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 (λ \cdot λ) - 4 \cdot 2$

Step 5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 18 - 3 λ - 6 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 2$

Step 5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 18 - 3 λ - 6 λ + 1 λ^{2} - 4 \cdot 2$

Step 5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 18 - 3 λ - 6 λ + λ^{2} - 4 \cdot 2$

Step 5.2.1.2.2

Subtract $6 λ$ from $- 3 λ$ .

$p (λ) = 18 - 9 λ + λ^{2} - 4 \cdot 2$

Step 5.2.1.3

Multiply $- 4$ by $2$ .

$p (λ) = 18 - 9 λ + λ^{2} - 8$

Step 5.2.2

Subtract $8$ from $18$ .

$p (λ) = - 9 λ + λ^{2} + 10$

Step 5.2.3

Reorder $- 9 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 9 λ + 10$

Step 6

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

$λ^{2} - 9 λ + 10 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.2

Substitute the values $a = 1$ , $b = - 9$ , and $c = 10$ into the quadratic formula and solve for $λ$ .

$\frac{9 \pm \sqrt{{(- 9)}^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

Raise $- 9$ to the power of $2$ .

$λ = \frac{9 \pm \sqrt{81 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Step 7.3.1.2

Multiply $- 4 \cdot 1 \cdot 10$ .

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Step 7.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{9 \pm \sqrt{81 - 4 \cdot 10}}{2 \cdot 1}$

Step 7.3.1.2.2

Multiply $- 4$ by $10$ .

$λ = \frac{9 \pm \sqrt{81 - 40}}{2 \cdot 1}$

Step 7.3.1.3

Subtract $40$ from $81$ .

$λ = \frac{9 \pm \sqrt{41}}{2 \cdot 1}$

Step 7.3.2

Multiply $2$ by $1$ .

$λ = \frac{9 \pm \sqrt{41}}{2}$

Step 7.4

The final answer is the combination of both solutions.

$λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = \frac{9 + \sqrt{41}}{2}, \frac{9 - \sqrt{41}}{2}$

Decimal Form:

$λ = 7.70156211 \dots, 1.29843788 \dots$

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