Linear Algebra Examples

$[\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}]$

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

$[\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}]$ for

$A$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] - λ I_{2})$

Step 3.2

Substitute

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

$I_{2}$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 \\ 3 & 5 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 + 0 \\ 3 + 0 & 5 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add

$2$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 \\ 3 + 0 & 5 - λ \end{array}]$

Step 4.3.2

Add

$3$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 \\ 3 & 5 - λ \end{array}]$

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

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

$p (λ) = (1 - λ) (5 - λ) - 3 \cdot 2$

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

$(1 - λ) (5 - λ)$ using the FOIL Method.

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Step 5.2.1.1.1

Apply the distributive property.

$p (λ) = 1 (5 - λ) - λ (5 - λ) - 3 \cdot 2$

Step 5.2.1.1.2

Apply the distributive property.

$p (λ) = 1 \cdot 5 + 1 (- λ) - λ (5 - λ) - 3 \cdot 2$

Step 5.2.1.1.3

Apply the distributive property.

$p (λ) = 1 \cdot 5 + 1 (- λ) - λ \cdot 5 - λ (- λ) - 3 \cdot 2$

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

$1$ .

$p (λ) = 5 + 1 (- λ) - λ \cdot 5 - λ (- λ) - 3 \cdot 2$

Step 5.2.1.2.1.2

Multiply

$- λ$ by

$1$ .

$p (λ) = 5 - λ - λ \cdot 5 - λ (- λ) - 3 \cdot 2$

Step 5.2.1.2.1.3

Multiply

$5$ by

$- 1$ .

$p (λ) = 5 - λ - 5 λ - λ (- λ) - 3 \cdot 2$

Step 5.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 5 - λ - 5 λ - 1 \cdot - 1 λ \cdot λ - 3 \cdot 2$

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

$λ$ .

$p (λ) = 5 - λ - 5 λ - 1 \cdot - 1 (λ \cdot λ) - 3 \cdot 2$

Step 5.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = 5 - λ - 5 λ - 1 \cdot - 1 λ^{2} - 3 \cdot 2$

Step 5.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

$p (λ) = 5 - λ - 5 λ + 1 λ^{2} - 3 \cdot 2$

Step 5.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = 5 - λ - 5 λ + λ^{2} - 3 \cdot 2$

Step 5.2.1.2.2

Subtract

$5 λ$ from

$- λ$ .

$p (λ) = 5 - 6 λ + λ^{2} - 3 \cdot 2$

Step 5.2.1.3

Multiply

$- 3$ by

$2$ .

$p (λ) = 5 - 6 λ + λ^{2} - 6$

Step 5.2.2

Subtract

$6$ from

$5$ .

$p (λ) = - 6 λ + λ^{2} - 1$

Step 5.2.3

Reorder

$- 6 λ$ and

$λ^{2}$ .

$p (λ) = λ^{2} - 6 λ - 1$

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