Linear Algebra Examples

$[\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}]$

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 $[\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] - λ I_{2})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.4 & 1 - c \\ 0.6 & c \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0.4 - λ & 1 - c + 0 \\ 0.6 + 0 & c - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $1 - c$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.4 - λ & 1 - c \\ 0.6 + 0 & c - λ \end{array}]$

Step 4.3.2

Add $0.6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.4 - λ & 1 - c \\ 0.6 & c - λ \end{array}]$

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

$p (λ) = (0.4 - λ) (c - λ) - 0.6 (1 - c)$

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 $(0.4 - λ) (c - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

$p (λ) = 0.4 (c - λ) - λ (c - λ) - 0.6 (1 - c)$

Step 5.2.1.1.2

Apply the distributive property.

$p (λ) = 0.4 c + 0.4 (- λ) - λ (c - λ) - 0.6 (1 - c)$

Step 5.2.1.1.3

Apply the distributive property.

$p (λ) = 0.4 c + 0.4 (- λ) - λ c - λ (- λ) - 0.6 (1 - c)$

Step 5.2.1.2

Simplify each term.

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Step 5.2.1.2.1

Multiply $- 1$ by $0.4$ .

$p (λ) = 0.4 c - 0.4 λ - λ c - λ (- λ) - 0.6 (1 - c)$

Step 5.2.1.2.2

Rewrite using the commutative property of multiplication.

$p (λ) = 0.4 c - 0.4 λ - λ c - 1 \cdot - 1 λ \cdot λ - 0.6 (1 - c)$

Step 5.2.1.2.3

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

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Step 5.2.1.2.3.1

Move $λ$ .

$p (λ) = 0.4 c - 0.4 λ - λ c - 1 \cdot - 1 (λ \cdot λ) - 0.6 (1 - c)$

Step 5.2.1.2.3.2

Multiply $λ$ by $λ$ .

$p (λ) = 0.4 c - 0.4 λ - λ c - 1 \cdot - 1 λ^{2} - 0.6 (1 - c)$

Step 5.2.1.2.4

Multiply $- 1$ by $- 1$ .

$p (λ) = 0.4 c - 0.4 λ - λ c + 1 λ^{2} - 0.6 (1 - c)$

Step 5.2.1.2.5

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0.4 c - 0.4 λ - λ c + λ^{2} - 0.6 (1 - c)$

Step 5.2.1.3

Apply the distributive property.

$p (λ) = 0.4 c - 0.4 λ - λ c + λ^{2} - 0.6 \cdot 1 - 0.6 (- c)$

Step 5.2.1.4

Multiply $- 0.6$ by $1$ .

$p (λ) = 0.4 c - 0.4 λ - λ c + λ^{2} - 0.6 - 0.6 (- c)$

Step 5.2.1.5

Multiply $- 1$ by $- 0.6$ .

$p (λ) = 0.4 c - 0.4 λ - λ c + λ^{2} - 0.6 + 0.6 c$

Step 5.2.2

Add $0.4 c$ and $0.6 c$ .

$p (λ) = c - 0.4 λ - λ c + λ^{2} - 0.6$

Step 5.2.3

Move $λ$ .

$p (λ) = c - 0.4 λ - 1 c λ + λ^{2} - 0.6$

Step 5.2.4

Move $- 0.4 λ$ .

$p (λ) = c - 1 c λ + λ^{2} - 0.4 λ - 0.6$

Step 5.2.5

Move $c$ .

$p (λ) = - 1 c λ + λ^{2} + c - 0.4 λ - 0.6$

Step 6

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

$- 1 c λ + λ^{2} + c - 0.4 λ - 0.6 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Rewrite $- 1 c$ as $- c$ .

$- c λ + λ^{2} + c - 0.4 λ - 0.6 = 0$

Step 7.2

Use the quadratic formula to find the solutions.

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

Step 7.3

Substitute the values $a = 1$ , $b = - c - 0.4$ , and $c = c - 0.6$ into the quadratic formula and solve for $λ$ .

$\frac{- (- c - 0.4) \pm \sqrt{{(- c - 0.4)}^{2} - 4 \cdot (1 \cdot (c - 0.6))}}{2 \cdot 1}$

Step 7.4

Simplify.

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Step 7.4.1

Simplify the numerator.

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Step 7.4.1.1

Apply the distributive property.

$λ = \frac{c + 0.4 \pm \sqrt{{(- c - 0.4)}^{2} - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.2

Multiply $- - c$ .

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Step 7.4.1.2.1

Multiply $- 1$ by $- 1$ .

$λ = \frac{1 c + 0.4 \pm \sqrt{{(- c - 0.4)}^{2} - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.2.2

Multiply $c$ by $1$ .

$λ = \frac{c + 0.4 \pm \sqrt{{(- c - 0.4)}^{2} - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.3

Multiply $- 1$ by $- 0.4$ .

$λ = \frac{c + 0.4 \pm \sqrt{{(- c - 0.4)}^{2} - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.4

Rewrite ${(- c - 0.4)}^{2}$ as $(- c - 0.4) (- c - 0.4)$ .

$λ = \frac{c + 0.4 \pm \sqrt{(- c - 0.4) (- c - 0.4) - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.5

Expand $(- c - 0.4) (- c - 0.4)$ using the FOIL Method.

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Step 7.4.1.5.1

Apply the distributive property.

$λ = \frac{c + 0.4 \pm \sqrt{- c (- c - 0.4) - 0.4 (- c - 0.4) - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.5.2

Apply the distributive property.

$λ = \frac{c + 0.4 \pm \sqrt{- c (- c) - c \cdot - 0.4 - 0.4 (- c - 0.4) - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.5.3

Apply the distributive property.

$λ = \frac{c + 0.4 \pm \sqrt{- c (- c) - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6

Simplify and combine like terms.

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Step 7.4.1.6.1

Simplify each term.

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Step 7.4.1.6.1.1

Rewrite using the commutative property of multiplication.

$λ = \frac{c + 0.4 \pm \sqrt{- 1 \cdot (- 1 c \cdot c) - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.2

Multiply $c$ by $c$ by adding the exponents.

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Step 7.4.1.6.1.2.1

Move $c$ .

$λ = \frac{c + 0.4 \pm \sqrt{- 1 \cdot (- 1 (c \cdot c)) - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.2.2

Multiply $c$ by $c$ .

$λ = \frac{c + 0.4 \pm \sqrt{- 1 \cdot (- 1 c^{2}) - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.3

Multiply $- 1$ by $- 1$ .

$λ = \frac{c + 0.4 \pm \sqrt{1 c^{2} - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.4

Multiply $c^{2}$ by $1$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} - c \cdot - 0.4 - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.5

Multiply $- 0.4$ by $- 1$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.4 c - 0.4 (- c) - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.6

Multiply $- 1$ by $- 0.4$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.4 c + 0.4 c - 0.4 \cdot - 0.4 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.1.7

Multiply $- 0.4$ by $- 0.4$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.4 c + 0.4 c + 0.16 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.6.2

Add $0.4 c$ and $0.4 c$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.8 c + 0.16 - 4 \cdot 1 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.7

Multiply $- 4$ by $1$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.8 c + 0.16 - 4 \cdot (c - 0.6)}}{2 \cdot 1}$

Step 7.4.1.8

Apply the distributive property.

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.8 c + 0.16 - 4 c - 4 \cdot - 0.6}}{2 \cdot 1}$

Step 7.4.1.9

Multiply $- 4$ by $- 0.6$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} + 0.8 c + 0.16 - 4 c + 2.4}}{2 \cdot 1}$

Step 7.4.1.10

Subtract $4 c$ from $0.8 c$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} - 3.2 c + 0.16 + 2.4}}{2 \cdot 1}$

Step 7.4.1.11

Add $0.16$ and $2.4$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} - 3.2 c + 2.56}}{2 \cdot 1}$

Step 7.4.1.12

Factor using the perfect square rule.

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Step 7.4.1.12.1

Rewrite $2.56$ as ${1.6}^{2}$ .

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} - 3.2 c + {1.6}^{2}}}{2 \cdot 1}$

Step 7.4.1.12.2

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

$3.2 c = 2 \cdot c \cdot 1.6$

Step 7.4.1.12.3

Rewrite the polynomial.

$λ = \frac{c + 0.4 \pm \sqrt{c^{2} - 2 \cdot c \cdot 1.6 + {1.6}^{2}}}{2 \cdot 1}$

Step 7.4.1.12.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = c$ and $b = 1.6$ .

$λ = \frac{c + 0.4 \pm \sqrt{{(c - 1.6)}^{2}}}{2 \cdot 1}$

Step 7.4.1.13

Pull terms out from under the radical, assuming positive real numbers.

$λ = \frac{c + 0.4 \pm (c - 1.6)}{2 \cdot 1}$

Step 7.4.2

Multiply $2$ by $1$ .

$λ = \frac{c + 0.4 \pm (c - 1.6)}{2}$

Step 7.5

The final answer is the combination of both solutions.

$λ = \frac{5 c - 3}{5}$

$λ = 1$

$λ = \frac{5 c - 3}{5}$

$λ = 1$