Linear Algebra Examples

$[\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}]$

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.

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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

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

$p (λ) = [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Simplify 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 $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}]$

Rearrange $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$

Add the corresponding elements of $[\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}]$ to each element of $[\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$ .

$p (λ) = [\begin{array}{c} - λ + 1 & 0 + 2 \\ 0 + 3 & - λ + 5 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} - λ + 1 & 0 + 2 \\ 0 + 3 & - λ + 5 \end{array}]$ .

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Simplify $0 + 2$ .

$p (λ) = [\begin{array}{c} - λ + 1 & 2 \\ 0 + 3 & - λ + 5 \end{array}]$

Simplify $0 + 3$ .

$p (λ) = [\begin{array}{c} - λ + 1 & 2 \\ 3 & - λ + 5 \end{array}]$

The determinant of $[\begin{array}{c} - λ + 1 & 2 \\ 3 & - λ + 5 \end{array}]$ is $λ^{2} - 6 λ - 1$ .

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

$determinant [\begin{array}{c} - λ + 1 & 2 \\ 3 & - λ + 5 \end{array}] = [\begin{array}{c} - λ + 1 & 2 \\ 3 & - λ + 5 \end{array}]$

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

$p (λ) = (- λ + 1) (- λ + 5) - 3 \cdot 2$

Simplify the determinant.

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Simplify each term.

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Expand $(- λ + 1) (- λ + 5)$ using the FOIL Method.

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Apply the distributive property.

$p (λ) = - λ (- λ + 5) + 1 (- λ + 5) - 3 \cdot 2$

Apply the distributive property.

$p (λ) = - λ (- λ) - λ \cdot 5 + 1 (- λ + 5) - 3 \cdot 2$

Apply the distributive property.

$p (λ) = - λ (- λ) - λ \cdot 5 + 1 (- λ) + 1 \cdot 5 - 3 \cdot 2$

Simplify and combine like terms.

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Simplify each term.

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Multiply $λ$ by $λ$ .

$p (λ) = - 1 \cdot (- 1 λ^{2}) - λ \cdot 5 + 1 (- λ) + 1 \cdot 5 - 3 \cdot 2$

Multiply $- 1$ by $- 1$ .

$p (λ) = 1 λ^{2} - λ \cdot 5 + 1 (- λ) + 1 \cdot 5 - 3 \cdot 2$

Multiply $λ^{2}$ by $1$ .

$p (λ) = λ^{2} - λ \cdot 5 + 1 (- λ) + 1 \cdot 5 - 3 \cdot 2$

Multiply $5$ by $- 1$ .

$p (λ) = λ^{2} - 5 λ + 1 (- λ) + 1 \cdot 5 - 3 \cdot 2$

Multiply $- λ$ by $1$ .

$p (λ) = λ^{2} - 5 λ - λ + 1 \cdot 5 - 3 \cdot 2$

Multiply $5$ by $1$ .

$p (λ) = λ^{2} - 5 λ - λ + 5 - 3 \cdot 2$

Subtract $λ$ from $- 5 λ$ .

$p (λ) = λ^{2} - 6 λ + 5 - 3 \cdot 2$

Multiply $- 3$ by $2$ .

$p (λ) = λ^{2} - 6 λ + 5 - 6$

Subtract $6$ from $5$ .

$p (λ) = λ^{2} - 6 λ - 1$

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

$λ^{2} - 6 λ - 1 = 0$

Solve the equation for $L$ .

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Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $λ$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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Raise $- 6$ to the power of $2$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Multiply $- 1$ by $1$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$λ = \frac{6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Add $36$ and $4$ .

$λ = \frac{6 \pm \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

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Factor $4$ out of $40$ .

$λ = \frac{6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

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

$λ = \frac{6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$λ = \frac{6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{6 \pm 2 \sqrt{10}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{10}}{2}$ .

$λ = 3 \pm \sqrt{10}$

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

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Simplify the numerator.

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Raise $- 6$ to the power of $2$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Multiply $- 1$ by $1$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$λ = \frac{6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Add $36$ and $4$ .

$λ = \frac{6 \pm \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

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Factor $4$ out of $40$ .

$λ = \frac{6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

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

$λ = \frac{6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$λ = \frac{6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{6 \pm 2 \sqrt{10}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{10}}{2}$ .

$λ = 3 \pm \sqrt{10}$

Change the $\pm$ to $+$ .

$λ = 3 + \sqrt{10}$

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

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Simplify the numerator.

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Raise $- 6$ to the power of $2$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Multiply $- 1$ by $1$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$λ = \frac{6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Add $36$ and $4$ .

$λ = \frac{6 \pm \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

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Factor $4$ out of $40$ .

$λ = \frac{6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

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

$λ = \frac{6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$λ = \frac{6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$λ = \frac{6 \pm 2 \sqrt{10}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{10}}{2}$ .

$λ = 3 \pm \sqrt{10}$

Change the $\pm$ to $-$ .

$λ = 3 - \sqrt{10}$

The final answer is the combination of both solutions.

$λ = 3 + \sqrt{10}, 3 - \sqrt{10}$

The result can be shown in multiple forms.

Exact Form:

$λ = 3 + \sqrt{10}, 3 - \sqrt{10}$

Decimal Form:

$λ = 6.16227766 \dots, - 0.16227766 \dots$

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