Linear Algebra Examples

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

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.

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

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

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

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

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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Simplify element $0, 0$ by multiplying $- λ \cdot 1$ .

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

Simplify element $0, 1$ by multiplying $- λ \cdot 0$ .

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

Simplify element $1, 0$ by multiplying $- λ \cdot 0$ .

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

Simplify element $1, 1$ by multiplying $- λ \cdot 1$ .

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

Combine the similar matrices with each others.

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

Simplify each element of the matrix $[\begin{array}{c} - λ + 3 & 0 + 2 \\ 0 + 4 & - λ + 6 \end{array}]$ .

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Combine the same size matrices $[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$ and $[\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$ by adding the corresponding elements of each.

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

Simplify element $0, 1$ of the matrix.

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

Simplify element $1, 0$ of the matrix.

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

The determinant of $[\begin{array}{c} - λ + 3 & 2 \\ 4 & - λ + 6 \end{array}]$ is $λ^{2} - 9 λ + 10$ .

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

$determinant [\begin{array}{c} - L + 3 & 2 \\ 4 & - L + 6 \end{array}] = | \begin{array}{c} - L + 3 & 2 \\ 4 & - L + 6 \end{array} |$

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

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

Simplify the determinant.

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Simplify each term.

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

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Remove parentheses.

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

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

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

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

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Combine $λ$ and $λ$

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Raise $λ$ to the power of $1$ .

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

Raise $λ$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Multiply $- 1$ by $- 1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Multiply $6$ by $- 1$ .

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

Multiply $- 1$ by $3$ .

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

Multiply $3$ by $6$ .

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

Subtract $3 λ$ from $- 6 λ$ .

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

Multiply $- 4$ by $2$ .

$p (λ) = λ^{2} - 9 λ + 18 - 8$

Subtract $8$ from $18$ .

$p (λ) = λ^{2} - 9 λ + 10$

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

$λ^{2} - 9 λ + 10 = 0$

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

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

Simplify.

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Simplify the numerator.

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

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

Multiply $10$ by $1$ .

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

Multiply $- 4$ by $10$ .

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

Subtract $40$ from $81$ .

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

Multiply $2$ by $1$ .

$λ = \frac{9 \pm \sqrt{41}}{2}$

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

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Simplify the numerator.

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

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

Multiply $10$ by $1$ .

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

Multiply $- 4$ by $10$ .

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

Subtract $40$ from $81$ .

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

Multiply $2$ by $1$ .

$λ = \frac{9 \pm \sqrt{41}}{2}$

Change the $\pm$ to $+$ .

$λ = \frac{9 + \sqrt{41}}{2}$

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

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Simplify the numerator.

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

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

Multiply $10$ by $1$ .

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

Multiply $- 4$ by $10$ .

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

Subtract $40$ from $81$ .

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

Multiply $2$ by $1$ .

$λ = \frac{9 \pm \sqrt{41}}{2}$

Change the $\pm$ to $-$ .

$λ = \frac{9 - \sqrt{41}}{2}$

The final answer is the combination of both solutions.

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

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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