Linear Algebra Examples

[\begin{matrix} 2 & 1 3 & 2 \end{matrix}]

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

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

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

$A$ .

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

Step 3.2

Substitute

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ for

$I_{2}$ .

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

$- 1$ by

$1$ .

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

Step 4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.2.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.3.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.4

Multiply

$- 1$ by

$1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add

$1$ and

$0$ .

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

Step 4.3.2

Add

$3$ and

$0$ .

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

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

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

$p (λ) = (2 - λ) (2 - λ) - 3 \cdot 1$

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

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

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Step 5.2.1.1.1

Apply the distributive property.

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

Step 5.2.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.3

Apply the distributive property.

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

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

$2$ by

$2$ .

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

Step 5.2.1.2.1.2

Multiply

$- 1$ by

$2$ .

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

Step 5.2.1.2.1.3

Multiply

$2$ by

$- 1$ .

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

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

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

$λ$ .

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

Step 5.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

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

Step 5.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

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

Step 5.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

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

Step 5.2.1.2.2

Subtract

$2 λ$ from

$- 2 λ$ .

$p (λ) = 4 - 4 λ + λ^{2} - 3 \cdot 1$

Step 5.2.1.3

Multiply

$- 3$ by

$1$ .

$p (λ) = 4 - 4 λ + λ^{2} - 3$

Step 5.2.2

Subtract

$3$ from

$4$ .

$p (λ) = - 4 λ + λ^{2} + 1$

Step 5.2.3

Reorder

$- 4 λ$ and

$λ^{2}$ .

$p (λ) = λ^{2} - 4 λ + 1$

Step 6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$λ^{2} - 4 λ + 1 = 0$

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

$a = 1$ ,

$b = - 4$ , and

$c = 1$ into the quadratic formula and solve for

$λ$ .

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

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

$- 4$ to the power of

$2$ .

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

Step 7.3.1.2

Multiply

$- 4 \cdot 1 \cdot 1$ .

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Step 7.3.1.2.1

Multiply

$- 4$ by

$1$ .

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

Step 7.3.1.2.2

Multiply

$- 4$ by

$1$ .

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

Step 7.3.1.3

Subtract

$4$ from

$16$ .

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

Step 7.3.1.4

Rewrite

$12$ as

$2^{2} \cdot 3$ .

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Step 7.3.1.4.1

Factor

$4$ out of

$12$ .

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

Step 7.3.1.4.2

Rewrite

$4$ as

$2^{2}$ .

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

Step 7.3.1.5

Pull terms out from under the radical.

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

Step 7.3.2

Multiply

$2$ by

$1$ .

$λ = \frac{4 \pm 2 \sqrt{3}}{2}$

Step 7.3.3

Simplify

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

$λ = 2 \pm \sqrt{3}$

Step 7.4

The final answer is the combination of both solutions.

$λ = 2 + \sqrt{3}, 2 - \sqrt{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = 2 + \sqrt{3}, 2 - \sqrt{3}$

Decimal Form:

$λ = 3.73205080 \dots, 0.26794919 \dots$