Find the Eigenvalues [[two , one],[three , two]]

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

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

$p (λ) = determinant (A + (- λ I_{2}))$

Substitute the known values in the formula.

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

Subtract the eigenvalue times the identity matrix from the original matrix.

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First, multiply $- λ$ by each element of the matrix.

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

Clarify each element of the matrix $[\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$ .

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Rearrange $- λ \cdot 1$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 1$ .

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

Add the corresponding elements of $[\begin{array}{c} 2 & 1 \\ 3 & 2 \end{array}]$ to each element of $[\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$ .

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

Simplify each element of the matrix $[\begin{array}{c} - λ + 2 & 0 + 1 \\ 0 + 3 & - λ + 2 \end{array}]$ .

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Simplify $0 + 1$ .

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

Simplify $0 + 3$ .

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

The determinant of $[\begin{array}{c} - λ + 2 & 1 \\ 3 & - λ + 2 \end{array}]$ is $λ^{2} - 4 λ + 1$ .

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These are both valid notations for the determinant of a matrix.

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

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

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

Clarify the determinant.

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Clarify each term.

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Expand $(- λ + 2) (- λ + 2)$ using the FOIL Method.

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Use the distributive law.

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

Apply the distributive property.

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

Apply the distributive property.

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

Clarify and Add like terms.

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Simplify each term.

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Multiply $λ$ by $λ$ .

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

Multiply $- 1$ by $- 1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Initially, multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Multiply $2$ by $2$ .

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

Subtract $2 λ$ from $- 2 λ$ .

$p (λ) = λ^{2} - 4 λ + 4 - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$p (λ) = λ^{2} - 4 λ + 4 - 3$

Subtract $3$ from $4$ .

$p (λ) = λ^{2} - 4 λ + 1$

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

$λ^{2} - 4 λ + 1 = 0$

Find solution to the equation for $L$ .

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Apply the quadratic formula to solve the equation.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $λ$ .

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

Simplify.

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Clarify the numerator.

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Raise $- 4$ to the power of $2$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

First, multiply $1$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Subtract $4$ from $16$ .

$λ = \frac{4 \pm \sqrt{12}}{2 \cdot 1}$

Should be rewritten $12$ as $2^{2} \cdot 3$ .

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Factor $4$ out of $12$ .

$λ = \frac{4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$λ = \frac{4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Withdraw terms from under the radical.

$λ = \frac{4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{4 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{3}}{2}$ .

$λ = 2 \pm \sqrt{3}$

Clarify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Multiply $1$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Subtract $4$ from $16$ .

$λ = \frac{4 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

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Factor $4$ out of $12$ .

$λ = \frac{4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$λ = \frac{4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$λ = \frac{4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{4 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{3}}{2}$ .

$λ = 2 \pm \sqrt{3}$

Change the $\pm$ to $+$ .

$λ = 2 + \sqrt{3}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Multiply $1$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Subtract $4$ from $16$ .

$λ = \frac{4 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

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Factor $4$ out of $12$ .

$λ = \frac{4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$λ = \frac{4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$λ = \frac{4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{4 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{3}}{2}$ .

$λ = 2 \pm \sqrt{3}$

Change the $\pm$ to $-$ .

$λ = 2 - \sqrt{3}$

The combination of 2 solutions is the answer.

$λ = 2 + \sqrt{3}, 2 - \sqrt{3}$

The result can be displayed in a variety of ways.

Exact Form:

$λ = 2 + \sqrt{3}, 2 - \sqrt{3}$

Decimal Form:

$λ = 3.73205080 \dots, 0.26794919 \dots$

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