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 $p (λ) = determinant (A - λ I_{2})$ .

Tap for more steps...

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.

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

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.

Tap for more steps...

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 $- λ \cdot 0$ .

Tap for more steps...

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 $- λ \cdot 0$ .

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

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.

Tap for more steps...

Step 5.2.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1

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

Tap for more steps...

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.

Tap for more steps...

Step 5.2.1.2.1

Simplify each term.

Tap for more steps...

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.

Tap for more steps...

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 $λ^{2}$ .

$p (λ) = λ^{2} - 6 λ - 1$

Step 6

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

$λ^{2} - 6 λ - 1 = 0$

Step 7

Solve for $λ$ .

Tap for more steps...

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 $a = 1$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $λ$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 7.3

Simplify.

Tap for more steps...

Step 7.3.1

Simplify the numerator.

Tap for more steps...

Step 7.3.1.1

Raise $- 6$ to the power of $2$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 7.3.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

Tap for more steps...

Step 7.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Step 7.3.1.2.2

Multiply $- 4$ by $- 1$ .

$λ = \frac{6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Step 7.3.1.3

Add $36$ and $4$ .

$λ = \frac{6 \pm \sqrt{40}}{2 \cdot 1}$

Step 7.3.1.4

Rewrite $40$ as $2^{2} \cdot 10$ .

Tap for more steps...

Step 7.3.1.4.1

Factor $4$ out of $40$ .

$λ = \frac{6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

Step 7.3.1.4.2

Rewrite $4$ as $2^{2}$ .

$λ = \frac{6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Step 7.3.1.5

Pull terms out from under the radical.

$λ = \frac{6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Step 7.3.2

Multiply $2$ by $1$ .

$λ = \frac{6 \pm 2 \sqrt{10}}{2}$

Step 7.3.3

Simplify $\frac{6 \pm 2 \sqrt{10}}{2}$ .

$λ = 3 \pm \sqrt{10}$

Step 7.4

The final answer is the combination of both solutions.

$λ = 3 + \sqrt{10}, 3 - \sqrt{10}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = 3 + \sqrt{10}, 3 - \sqrt{10}$

Decimal Form:

$λ = 6.16227766 \dots, - 0.16227766 \dots$

Enter YOUR Problem