Linear Algebra Examples

$[\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}]$

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 $[\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] - λ I_{2})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 8 & - 3 \\ 3 & 2 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 8 - λ & - 3 + 0 \\ 3 + 0 & 2 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 8 - λ & - 3 \\ 3 + 0 & 2 - λ \end{array}]$

Step 4.3.2

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 8 - λ & - 3 \\ 3 & 2 - λ \end{array}]$

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

$p (λ) = (8 - λ) (2 - λ) - 3 \cdot - 3$

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 $(8 - λ) (2 - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

$p (λ) = 8 (2 - λ) - λ (2 - λ) - 3 \cdot - 3$

Step 5.2.1.1.2

Apply the distributive property.

$p (λ) = 8 \cdot 2 + 8 (- λ) - λ (2 - λ) - 3 \cdot - 3$

Step 5.2.1.1.3

Apply the distributive property.

$p (λ) = 8 \cdot 2 + 8 (- λ) - λ \cdot 2 - λ (- λ) - 3 \cdot - 3$

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 $8$ by $2$ .

$p (λ) = 16 + 8 (- λ) - λ \cdot 2 - λ (- λ) - 3 \cdot - 3$

Step 5.2.1.2.1.2

Multiply $- 1$ by $8$ .

$p (λ) = 16 - 8 λ - λ \cdot 2 - λ (- λ) - 3 \cdot - 3$

Step 5.2.1.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = 16 - 8 λ - 2 λ - λ (- λ) - 3 \cdot - 3$

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 λ \cdot λ - 3 \cdot - 3$

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 $λ$ .

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 (λ \cdot λ) - 3 \cdot - 3$

Step 5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 16 - 8 λ - 2 λ - 1 \cdot - 1 λ^{2} - 3 \cdot - 3$

Step 5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 16 - 8 λ - 2 λ + 1 λ^{2} - 3 \cdot - 3$

Step 5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 16 - 8 λ - 2 λ + λ^{2} - 3 \cdot - 3$

Step 5.2.1.2.2

Subtract $2 λ$ from $- 8 λ$ .

$p (λ) = 16 - 10 λ + λ^{2} - 3 \cdot - 3$

Step 5.2.1.3

Multiply $- 3$ by $- 3$ .

$p (λ) = 16 - 10 λ + λ^{2} + 9$

Step 5.2.2

Add $16$ and $9$ .

$p (λ) = - 10 λ + λ^{2} + 25$

Step 5.2.3

Reorder $- 10 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 10 λ + 25$

Step 6

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

$λ^{2} - 10 λ + 25 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor using the perfect square rule.

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Step 7.1.1

Rewrite $25$ as $5^{2}$ .

$λ^{2} - 10 λ + 5^{2} = 0$

Step 7.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 λ = 2 \cdot λ \cdot 5$

Step 7.1.3

Rewrite the polynomial.

$λ^{2} - 2 \cdot λ \cdot 5 + 5^{2} = 0$

Step 7.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = λ$ and $b = 5$ .

${(λ - 5)}^{2} = 0$

Step 7.2

Set the $λ - 5$ equal to $0$ .

$λ - 5 = 0$

Step 7.3

Add $5$ to both sides of the equation.

$λ = 5$