Linear Algebra Examples

$A = [\begin{array}{c} 5 & - 2 \\ - 2 & 2 \end{array}]$

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 $[\begin{array}{c} 5 & - 2 \\ - 2 & 2 \end{array}]$ for $A$ .

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

Step 3.2

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

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $- 2$ and $0$ .

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

Step 4.3.2

Add $- 2$ and $0$ .

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

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

$p (λ) = (5 - λ) (2 - λ) - (- 2 \cdot - 2)$

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 $(5 - λ) (2 - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

$p (λ) = 5 (2 - λ) - λ (2 - λ) - (- 2 \cdot - 2)$

Step 5.2.1.1.2

Apply the distributive property.

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

Step 5.2.1.1.3

Apply the distributive property.

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

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 $5$ by $2$ .

$p (λ) = 10 + 5 (- λ) - λ \cdot 2 - λ (- λ) - (- 2 \cdot - 2)$

Step 5.2.1.2.1.2

Multiply $- 1$ by $5$ .

$p (λ) = 10 - 5 λ - λ \cdot 2 - λ (- λ) - (- 2 \cdot - 2)$

Step 5.2.1.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = 10 - 5 λ - 2 λ - λ (- λ) - (- 2 \cdot - 2)$

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 10 - 5 λ - 2 λ - 1 \cdot - 1 λ \cdot λ - (- 2 \cdot - 2)$

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 $λ$ .

$p (λ) = 10 - 5 λ - 2 λ - 1 \cdot - 1 (λ \cdot λ) - (- 2 \cdot - 2)$

Step 5.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 10 - 5 λ - 2 λ - 1 \cdot - 1 λ^{2} - (- 2 \cdot - 2)$

Step 5.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 10 - 5 λ - 2 λ + 1 λ^{2} - (- 2 \cdot - 2)$

Step 5.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 10 - 5 λ - 2 λ + λ^{2} - (- 2 \cdot - 2)$

Step 5.2.1.2.2

Subtract $2 λ$ from $- 5 λ$ .

$p (λ) = 10 - 7 λ + λ^{2} - (- 2 \cdot - 2)$

Step 5.2.1.3

Multiply $- (- 2 \cdot - 2)$ .

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Step 5.2.1.3.1

Multiply $- 2$ by $- 2$ .

$p (λ) = 10 - 7 λ + λ^{2} - 1 \cdot 4$

Step 5.2.1.3.2

Multiply $- 1$ by $4$ .

$p (λ) = 10 - 7 λ + λ^{2} - 4$

Step 5.2.2

Subtract $4$ from $10$ .

$p (λ) = - 7 λ + λ^{2} + 6$

Step 5.2.3

Reorder $- 7 λ$ and $λ^{2}$ .

$p (λ) = λ^{2} - 7 λ + 6$

Step 6

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

$λ^{2} - 7 λ + 6 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor $λ^{2} - 7 λ + 6$ using the AC method.

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Step 7.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 7$ .

$- 6, - 1$

Step 7.1.2

Write the factored form using these integers.

$(λ - 6) (λ - 1) = 0$

Step 7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$λ - 6 = 0$

$λ - 1 = 0$

Step 7.3

Set $λ - 6$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $λ - 6$ equal to $0$ .

$λ - 6 = 0$

Step 7.3.2

Add $6$ to both sides of the equation.

$λ = 6$

Step 7.4

Set $λ - 1$ equal to $0$ and solve for $λ$ .

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Step 7.4.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 7.4.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 7.5

The final solution is all the values that make $(λ - 6) (λ - 1) = 0$ true.

$λ = 6, 1$