Linear Algebra Examples

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}]$

Step 1

Find the eigenvalues.

Tap for more steps...

Step 1.1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{4})$

Step 1.2

The identity matrix or unit matrix of size $4$ is the $4 \times 4$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{4})$ .

Tap for more steps...

Step 1.3.1

Substitute $[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] - λ I_{4})$

Step 1.3.2

Substitute $[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$ for $I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Simplify each term.

Tap for more steps...

Step 1.4.1.1

Multiply $- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.5.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.9.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.10

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.10.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.10.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.11

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.12

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.12.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.12.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.13

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.13.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.13.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.14

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.14.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.14.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.15

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 1.4.1.2.15.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.15.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.16

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 + 0 & 2 + 0 & 0 + 0 \\ 3 + 0 & 2 - λ & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3

Simplify each element.

Tap for more steps...

Step 1.4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 + 0 & 0 + 0 \\ 3 + 0 & 2 - λ & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.2

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 + 0 \\ 3 + 0 & 2 - λ & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.3

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 + 0 & 2 - λ & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.4

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.5

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 + 0 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.7

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 + 0 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.8

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 & 1 - λ & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.9

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 & 1 - λ & 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.10

Add $5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 & 1 - λ & 0 \\ 5 & - 4 + 0 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.11

Add $- 4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 & 1 - λ & 0 \\ 5 & - 4 & 2 + 0 & 3 - λ \end{array}]$

Step 1.4.3.12

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 2 & 0 \\ 3 & 2 - λ & 4 & 0 \\ 2 & - 3 & 1 - λ & 0 \\ 5 & - 4 & 2 & 3 - λ \end{array}]$

Step 1.5

Find the determinant.

Tap for more steps...

Step 1.5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in column $4$ by its cofactor and add.

Tap for more steps...

Step 1.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array} |$

Step 1.5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.5.1.3

The minor for $a_{14}$ is the determinant with row $1$ and column $4$ deleted.

$| \begin{array}{c} 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.4

Multiply element $a_{14}$ by its cofactor.

$0 | \begin{array}{c} 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.5

The minor for $a_{24}$ is the determinant with row $2$ and column $4$ deleted.

$| \begin{array}{c} 4 - λ & 0 & 2 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.6

Multiply element $a_{24}$ by its cofactor.

$0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.7

The minor for $a_{34}$ is the determinant with row $3$ and column $4$ deleted.

$| \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.8

Multiply element $a_{34}$ by its cofactor.

$0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} |$

Step 1.5.1.9

The minor for $a_{44}$ is the determinant with row $4$ and column $4$ deleted.

$| \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.1.10

Multiply element $a_{44}$ by its cofactor.

$(3 - λ) | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.1.11

Add the terms together.

$p (λ) = 0 | \begin{array}{c} 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} | + 0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} | + 0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} | + (3 - λ) | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$ .

$p (λ) = 0 + 0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} | + 0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} | + (3 - λ) | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.3

Multiply $0$ by $| \begin{array}{c} 4 - λ & 0 & 2 \\ 2 & - 3 & 1 - λ \\ 5 & - 4 & 2 \end{array} |$ .

$p (λ) = 0 + 0 + 0 | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} | + (3 - λ) | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.4

Multiply $0$ by $| \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 5 & - 4 & 2 \end{array} |$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) | \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$

Step 1.5.5

Evaluate $| \begin{array}{c} 4 - λ & 0 & 2 \\ 3 & 2 - λ & 4 \\ 2 & - 3 & 1 - λ \end{array} |$ .

Tap for more steps...

Step 1.5.5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

Tap for more steps...

Step 1.5.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 1.5.5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.5.5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 2 - λ & 4 \\ - 3 & 1 - λ \end{array} |$

Step 1.5.5.1.4

Multiply element $a_{11}$ by its cofactor.

$(4 - λ) | \begin{array}{c} 2 - λ & 4 \\ - 3 & 1 - λ \end{array} |$

Step 1.5.5.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 3 & 4 \\ 2 & 1 - λ \end{array} |$

Step 1.5.5.1.6

Multiply element $a_{12}$ by its cofactor.

$0 | \begin{array}{c} 3 & 4 \\ 2 & 1 - λ \end{array} |$

Step 1.5.5.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |$

Step 1.5.5.1.8

Multiply element $a_{13}$ by its cofactor.

$2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |$

Step 1.5.5.1.9

Add the terms together.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) | \begin{array}{c} 2 - λ & 4 \\ - 3 & 1 - λ \end{array} | + 0 | \begin{array}{c} 3 & 4 \\ 2 & 1 - λ \end{array} | + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.2

Multiply $0$ by $| \begin{array}{c} 3 & 4 \\ 2 & 1 - λ \end{array} |$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) | \begin{array}{c} 2 - λ & 4 \\ - 3 & 1 - λ \end{array} | + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3

Evaluate $| \begin{array}{c} 2 - λ & 4 \\ - 3 & 1 - λ \end{array} |$ .

Tap for more steps...

Step 1.5.5.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) ((2 - λ) (1 - λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2

Simplify the determinant.

Tap for more steps...

Step 1.5.5.3.2.1

Simplify each term.

Tap for more steps...

Step 1.5.5.3.2.1.1

Expand $(2 - λ) (1 - λ)$ using the FOIL Method.

Tap for more steps...

Step 1.5.5.3.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 (1 - λ) - λ (1 - λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 \cdot 1 + 2 (- λ) - λ (1 - λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 \cdot 1 + 2 (- λ) - λ \cdot 1 - λ (- λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2

Simplify and combine like terms.

Tap for more steps...

Step 1.5.5.3.2.1.2.1

Simplify each term.

Tap for more steps...

Step 1.5.5.3.2.1.2.1.1

Multiply $2$ by $1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 + 2 (- λ) - λ \cdot 1 - λ (- λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.2

Multiply $- 1$ by $2$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ \cdot 1 - λ (- λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.3

Multiply $- 1$ by $1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ - λ (- λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ - 1 \cdot - 1 λ \cdot λ - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.5

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

Tap for more steps...

Step 1.5.5.3.2.1.2.1.5.1

Move $λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ - 1 \cdot - 1 (λ \cdot λ) - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ - 1 \cdot - 1 λ^{2} - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ + 1 λ^{2} - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 2 λ - λ + λ^{2} - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.2.2

Subtract $λ$ from $- 2 λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 3 λ + λ^{2} - (- 3 \cdot 4)) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.3

Multiply $- (- 3 \cdot 4)$ .

Tap for more steps...

Step 1.5.5.3.2.1.3.1

Multiply $- 3$ by $4$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 3 λ + λ^{2} - - 12) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.1.3.2

Multiply $- 1$ by $- 12$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (2 - 3 λ + λ^{2} + 12) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.2

Add $2$ and $12$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (- 3 λ + λ^{2} + 14) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.3.2.3

Reorder $- 3 λ$ and $λ^{2}$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 | \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |)$

Step 1.5.5.4

Evaluate $| \begin{array}{c} 3 & 2 - λ \\ 2 & - 3 \end{array} |$ .

Tap for more steps...

Step 1.5.5.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (3 \cdot - 3 - 2 (2 - λ)))$

Step 1.5.5.4.2

Simplify the determinant.

Tap for more steps...

Step 1.5.5.4.2.1

Simplify each term.

Tap for more steps...

Step 1.5.5.4.2.1.1

Multiply $3$ by $- 3$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (- 9 - 2 (2 - λ)))$

Step 1.5.5.4.2.1.2

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (- 9 - 2 \cdot 2 - 2 (- λ)))$

Step 1.5.5.4.2.1.3

Multiply $- 2$ by $2$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (- 9 - 4 - 2 (- λ)))$

Step 1.5.5.4.2.1.4

Multiply $- 1$ by $- 2$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (- 9 - 4 + 2 λ))$

Step 1.5.5.4.2.2

Subtract $4$ from $- 9$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (- 13 + 2 λ))$

Step 1.5.5.4.2.3

Reorder $- 13$ and $2 λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 0 + 2 (2 λ - 13))$

Step 1.5.5.5

Simplify the determinant.

Tap for more steps...

Step 1.5.5.5.1

Add $(4 - λ) (λ^{2} - 3 λ + 14)$ and $0$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) ((4 - λ) (λ^{2} - 3 λ + 14) + 2 (2 λ - 13))$

Step 1.5.5.5.2

Simplify each term.

Tap for more steps...

Step 1.5.5.5.2.1

Expand $(4 - λ) (λ^{2} - 3 λ + 14)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} + 4 (- 3 λ) + 4 \cdot 14 - λ \cdot λ^{2} - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2

Simplify each term.

Tap for more steps...

Step 1.5.5.5.2.2.1

Multiply $- 3$ by $4$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 4 \cdot 14 - λ \cdot λ^{2} - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.2

Multiply $4$ by $14$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ \cdot λ^{2} - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.3

Multiply $λ$ by $λ^{2}$ by adding the exponents.

Tap for more steps...

Step 1.5.5.5.2.2.3.1

Move $λ^{2}$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - (λ^{2} λ) - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.3.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 1.5.5.5.2.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - (λ^{2} λ^{1}) - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{2 + 1} - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.3.3

Add $2$ and $1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} - λ (- 3 λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} - 1 \cdot - 3 λ \cdot λ - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.5

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

Tap for more steps...

Step 1.5.5.5.2.2.5.1

Move $λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} - 1 \cdot - 3 (λ \cdot λ) - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} - 1 \cdot - 3 λ^{2} - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.6

Multiply $- 1$ by $- 3$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} + 3 λ^{2} - λ \cdot 14 + 2 (2 λ - 13))$

Step 1.5.5.5.2.2.7

Multiply $14$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (4 λ^{2} - 12 λ + 56 - λ^{3} + 3 λ^{2} - 14 λ + 2 (2 λ - 13))$

Step 1.5.5.5.2.3

Add $4 λ^{2}$ and $3 λ^{2}$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 12 λ + 56 - λ^{3} - 14 λ + 2 (2 λ - 13))$

Step 1.5.5.5.2.4

Subtract $14 λ$ from $- 12 λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 26 λ + 56 - λ^{3} + 2 (2 λ - 13))$

Step 1.5.5.5.2.5

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 26 λ + 56 - λ^{3} + 2 (2 λ) + 2 \cdot - 13)$

Step 1.5.5.5.2.6

Multiply $2$ by $2$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 26 λ + 56 - λ^{3} + 4 λ + 2 \cdot - 13)$

Step 1.5.5.5.2.7

Multiply $2$ by $- 13$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 26 λ + 56 - λ^{3} + 4 λ - 26)$

Step 1.5.5.5.3

Add $- 26 λ$ and $4 λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 22 λ + 56 - λ^{3} - 26)$

Step 1.5.5.5.4

Subtract $26$ from $56$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - 22 λ - λ^{3} + 30)$

Step 1.5.5.5.5

Move $- 22 λ$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (7 λ^{2} - λ^{3} - 22 λ + 30)$

Step 1.5.5.5.6

Reorder $7 λ^{2}$ and $- λ^{3}$ .

$p (λ) = 0 + 0 + 0 + (3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$

Step 1.5.6

Simplify the determinant.

Tap for more steps...

Step 1.5.6.1

Combine the opposite terms in $0 + 0 + 0 + (3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$ .

Tap for more steps...

Step 1.5.6.1.1

Add $0$ and $0$ .

$p (λ) = 0 + 0 + (3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$

Step 1.5.6.1.2

Add $0$ and $0$ .

$p (λ) = 0 + (3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$

Step 1.5.6.1.3

Add $0$ and $(3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$ .

$p (λ) = (3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$

Step 1.5.6.2

Expand $(3 - λ) (- λ^{3} + 7 λ^{2} - 22 λ + 30)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 3 (- λ^{3}) + 3 (7 λ^{2}) + 3 (- 22 λ) + 3 \cdot 30 - λ (- λ^{3}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3

Simplify each term.

Tap for more steps...

Step 1.5.6.3.1

Multiply $- 1$ by $3$ .

$p (λ) = - 3 λ^{3} + 3 (7 λ^{2}) + 3 (- 22 λ) + 3 \cdot 30 - λ (- λ^{3}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.2

Multiply $7$ by $3$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} + 3 (- 22 λ) + 3 \cdot 30 - λ (- λ^{3}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.3

Multiply $- 22$ by $3$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 3 \cdot 30 - λ (- λ^{3}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.4

Multiply $3$ by $30$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - λ (- λ^{3}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.5

Rewrite using the commutative property of multiplication.

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - 1 \cdot - 1 λ \cdot λ^{3} - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.6

Multiply $λ$ by $λ^{3}$ by adding the exponents.

Tap for more steps...

Step 1.5.6.3.6.1

Move $λ^{3}$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - 1 \cdot - 1 (λ^{3} λ) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.6.2

Multiply $λ^{3}$ by $λ$ .

Tap for more steps...

Step 1.5.6.3.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.6.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - 1 \cdot - 1 λ^{3 + 1} - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.6.3

Add $3$ and $1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 - 1 \cdot - 1 λ^{4} - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.7

Multiply $- 1$ by $- 1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + 1 λ^{4} - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - λ (7 λ^{2}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.9

Rewrite using the commutative property of multiplication.

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 1 \cdot 7 λ \cdot λ^{2} - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.10

Multiply $λ$ by $λ^{2}$ by adding the exponents.

Tap for more steps...

Step 1.5.6.3.10.1

Move $λ^{2}$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 1 \cdot 7 (λ^{2} λ) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.10.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 1.5.6.3.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 1 \cdot 7 (λ^{2} λ^{1}) - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.10.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 1 \cdot 7 λ^{2 + 1} - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.10.3

Add $2$ and $1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 1 \cdot 7 λ^{3} - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.11

Multiply $- 1$ by $7$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} - λ (- 22 λ) - λ \cdot 30$

Step 1.5.6.3.12

Rewrite using the commutative property of multiplication.

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} - 1 \cdot - 22 λ \cdot λ - λ \cdot 30$

Step 1.5.6.3.13

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

Tap for more steps...

Step 1.5.6.3.13.1

Move $λ$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} - 1 \cdot - 22 (λ \cdot λ) - λ \cdot 30$

Step 1.5.6.3.13.2

Multiply $λ$ by $λ$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} - 1 \cdot - 22 λ^{2} - λ \cdot 30$

Step 1.5.6.3.14

Multiply $- 1$ by $- 22$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} + 22 λ^{2} - λ \cdot 30$

Step 1.5.6.3.15

Multiply $30$ by $- 1$ .

$p (λ) = - 3 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} - 7 λ^{3} + 22 λ^{2} - 30 λ$

Step 1.5.6.4

Subtract $7 λ^{3}$ from $- 3 λ^{3}$ .

$p (λ) = - 10 λ^{3} + 21 λ^{2} - 66 λ + 90 + λ^{4} + 22 λ^{2} - 30 λ$

Step 1.5.6.5

Add $21 λ^{2}$ and $22 λ^{2}$ .

$p (λ) = - 10 λ^{3} + 43 λ^{2} - 66 λ + 90 + λ^{4} - 30 λ$

Step 1.5.6.6

Subtract $30 λ$ from $- 66 λ$ .

$p (λ) = - 10 λ^{3} + 43 λ^{2} - 96 λ + 90 + λ^{4}$

Step 1.5.6.7

Move $90$ .

$p (λ) = - 10 λ^{3} + 43 λ^{2} - 96 λ + λ^{4} + 90$

Step 1.5.6.8

Move $- 96 λ$ .

$p (λ) = - 10 λ^{3} + 43 λ^{2} + λ^{4} - 96 λ + 90$

Step 1.5.6.9

Move $43 λ^{2}$ .

$p (λ) = - 10 λ^{3} + λ^{4} + 43 λ^{2} - 96 λ + 90$

Step 1.5.6.10

Reorder $- 10 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90$

Step 1.6

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

$λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90 = 0$

Step 1.7

Solve for $λ$ .

Tap for more steps...

Step 1.7.1

Factor the left side of the equation.

Tap for more steps...

Step 1.7.1.1

Factor $λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90$ using the rational roots test.

Tap for more steps...

Step 1.7.1.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 90, \pm 2, \pm 45, \pm 3, \pm 30, \pm 5, \pm 18, \pm 6, \pm 15, \pm 9, \pm 10$

$q = \pm 1$

Step 1.7.1.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 90, \pm 2, \pm 45, \pm 3, \pm 30, \pm 5, \pm 18, \pm 6, \pm 15, \pm 9, \pm 10$

Step 1.7.1.1.3

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Tap for more steps...

Step 1.7.1.1.3.1

Substitute $3$ into the polynomial.

$3^{4} - 10 \cdot 3^{3} + 43 \cdot 3^{2} - 96 \cdot 3 + 90$

Step 1.7.1.1.3.2

Raise $3$ to the power of $4$ .

$81 - 10 \cdot 3^{3} + 43 \cdot 3^{2} - 96 \cdot 3 + 90$

Step 1.7.1.1.3.3

Raise $3$ to the power of $3$ .

$81 - 10 \cdot 27 + 43 \cdot 3^{2} - 96 \cdot 3 + 90$

Step 1.7.1.1.3.4

Multiply $- 10$ by $27$ .

$81 - 270 + 43 \cdot 3^{2} - 96 \cdot 3 + 90$

Step 1.7.1.1.3.5

Subtract $270$ from $81$ .

$- 189 + 43 \cdot 3^{2} - 96 \cdot 3 + 90$

Step 1.7.1.1.3.6

Raise $3$ to the power of $2$ .

$- 189 + 43 \cdot 9 - 96 \cdot 3 + 90$

Step 1.7.1.1.3.7

Multiply $43$ by $9$ .

$- 189 + 387 - 96 \cdot 3 + 90$

Step 1.7.1.1.3.8

Add $- 189$ and $387$ .

$198 - 96 \cdot 3 + 90$

Step 1.7.1.1.3.9

Multiply $- 96$ by $3$ .

$198 - 288 + 90$

Step 1.7.1.1.3.10

Subtract $288$ from $198$ .

$- 90 + 90$

Step 1.7.1.1.3.11

Add $- 90$ and $90$ .

$0$

Step 1.7.1.1.4

Since $3$ is a known root, divide the polynomial by $λ - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90}{λ - 3}$

Step 1.7.1.1.5

Divide $λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90$ by $λ - 3$ .

Tap for more steps...

Step 1.7.1.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

Step 1.7.1.1.5.2

Divide the highest order term in the dividend $λ^{4}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

Step 1.7.1.1.5.3

Multiply the new quotient term by the divisor.

$λ^{3}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

Step 1.7.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $λ^{4} - 3 λ^{3}$

$λ^{3}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

Step 1.7.1.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

Step 1.7.1.1.5.6

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

Step 1.7.1.1.5.7

Divide the highest order term in the dividend $- 7 λ^{3}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$7 λ^{2}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

Step 1.7.1.1.5.8

Multiply the new quotient term by the divisor.

$λ^{3}$

$7 λ^{2}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

Step 1.7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 λ^{3} + 21 λ^{2}$

$λ^{3}$

$7 λ^{2}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

Step 1.7.1.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$7 λ^{2}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

Step 1.7.1.1.5.11

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$7 λ^{2}$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

Step 1.7.1.1.5.12

Divide the highest order term in the dividend $22 λ^{2}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$7 λ^{2}$

$22 λ$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

Step 1.7.1.1.5.13

Multiply the new quotient term by the divisor.

$λ^{3}$

$7 λ^{2}$

$22 λ$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

Step 1.7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $22 λ^{2} - 66 λ$

$λ^{3}$

$7 λ^{2}$

$22 λ$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

Step 1.7.1.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$7 λ^{2}$

$22 λ$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

Step 1.7.1.1.5.16

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$7 λ^{2}$

$22 λ$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

$90$

Step 1.7.1.1.5.17

Divide the highest order term in the dividend $- 30 λ$ by the highest order term in divisor $λ$ .

$λ^{3}$

$7 λ^{2}$

$22 λ$

$30$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

$90$

Step 1.7.1.1.5.18

Multiply the new quotient term by the divisor.

$λ^{3}$

$7 λ^{2}$

$22 λ$

$30$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

$90$

$30 λ$

$90$

Step 1.7.1.1.5.19

The expression needs to be subtracted from the dividend, so change all the signs in $- 30 λ + 90$

$λ^{3}$

$7 λ^{2}$

$22 λ$

$30$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

$90$

$30 λ$

$90$

Step 1.7.1.1.5.20

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$7 λ^{2}$

$22 λ$

$30$

$λ$

$3$

$λ^{4}$

$10 λ^{3}$

$43 λ^{2}$

$96 λ$

$90$

$λ^{4}$

$3 λ^{3}$

$7 λ^{3}$

$43 λ^{2}$

$7 λ^{3}$

$21 λ^{2}$

$22 λ^{2}$

$96 λ$

$22 λ^{2}$

$66 λ$

$30 λ$

$90$

$30 λ$

$90$

$0$

Step 1.7.1.1.5.21

Since the remander is $0$ , the final answer is the quotient.

$λ^{3} - 7 λ^{2} + 22 λ - 30$

Step 1.7.1.1.6

Write $λ^{4} - 10 λ^{3} + 43 λ^{2} - 96 λ + 90$ as a set of factors.

$(λ - 3) (λ^{3} - 7 λ^{2} + 22 λ - 30) = 0$

Step 1.7.1.2

Factor $λ^{3} - 7 λ^{2} + 22 λ - 30$ using the rational roots test.

Tap for more steps...

Step 1.7.1.2.1

Factor $λ^{3} - 7 λ^{2} + 22 λ - 30$ using the rational roots test.

Tap for more steps...

Step 1.7.1.2.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 30, \pm 2, \pm 15, \pm 3, \pm 10, \pm 5, \pm 6$

$q = \pm 1$

Step 1.7.1.2.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 30, \pm 2, \pm 15, \pm 3, \pm 10, \pm 5, \pm 6$

Step 1.7.1.2.1.3

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Tap for more steps...

Step 1.7.1.2.1.3.1

Substitute $3$ into the polynomial.

$3^{3} - 7 \cdot 3^{2} + 22 \cdot 3 - 30$

Step 1.7.1.2.1.3.2

Raise $3$ to the power of $3$ .

$27 - 7 \cdot 3^{2} + 22 \cdot 3 - 30$

Step 1.7.1.2.1.3.3

Raise $3$ to the power of $2$ .

$27 - 7 \cdot 9 + 22 \cdot 3 - 30$

Step 1.7.1.2.1.3.4

Multiply $- 7$ by $9$ .

$27 - 63 + 22 \cdot 3 - 30$

Step 1.7.1.2.1.3.5

Subtract $63$ from $27$ .

$- 36 + 22 \cdot 3 - 30$

Step 1.7.1.2.1.3.6

Multiply $22$ by $3$ .

$- 36 + 66 - 30$

Step 1.7.1.2.1.3.7

Add $- 36$ and $66$ .

$30 - 30$

Step 1.7.1.2.1.3.8

Subtract $30$ from $30$ .

$0$

Step 1.7.1.2.1.4

Since $3$ is a known root, divide the polynomial by $λ - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{3} - 7 λ^{2} + 22 λ - 30}{λ - 3}$

Step 1.7.1.2.1.5

Divide $λ^{3} - 7 λ^{2} + 22 λ - 30$ by $λ - 3$ .

Tap for more steps...

Step 1.7.1.2.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$3$

$λ^{3}$

$7 λ^{2}$

$22 λ$

$30$

Step 1.7.1.2.1.5.2

Divide the highest order term in the dividend $λ^{3}$ by the highest order term in divisor $λ$ .

					$λ^{2}$
	$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$

Step 1.7.1.2.1.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			+	$λ^{3}$	-	$3 λ^{2}$

Step 1.7.1.2.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $λ^{3} - 3 λ^{2}$

				$λ^{2}$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$

Step 1.7.1.2.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$

Step 1.7.1.2.1.5.6

Pull the next terms from the original dividend down into the current dividend.

				$λ^{2}$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$

Step 1.7.1.2.1.5.7

Divide the highest order term in the dividend $- 4 λ^{2}$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$

Step 1.7.1.2.1.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					-	$4 λ^{2}$	+	$12 λ$

Step 1.7.1.2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 λ^{2} + 12 λ$

				$λ^{2}$	-	$4 λ$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$

Step 1.7.1.2.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$	-	$4 λ$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$

Step 1.7.1.2.1.5.11

Pull the next terms from the original dividend down into the current dividend.

				$λ^{2}$	-	$4 λ$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$	-	$30$

Step 1.7.1.2.1.5.12

Divide the highest order term in the dividend $10 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$	+	$10$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$	-	$30$

Step 1.7.1.2.1.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$	+	$10$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$	-	$30$
							+	$10 λ$	-	$30$

Step 1.7.1.2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $10 λ - 30$

				$λ^{2}$	-	$4 λ$	+	$10$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$	-	$30$
							-	$10 λ$	+	$30$

Step 1.7.1.2.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$	-	$4 λ$	+	$10$
$λ$	-	$3$		$λ^{3}$	-	$7 λ^{2}$	+	$22 λ$	-	$30$
			-	$λ^{3}$	+	$3 λ^{2}$
					-	$4 λ^{2}$	+	$22 λ$
					+	$4 λ^{2}$	-	$12 λ$
							+	$10 λ$	-	$30$
							-	$10 λ$	+	$30$
										$0$

Step 1.7.1.2.1.5.16

Since the remander is $0$ , the final answer is the quotient.

$λ^{2} - 4 λ + 10$

Step 1.7.1.2.1.6

Write $λ^{3} - 7 λ^{2} + 22 λ - 30$ as a set of factors.

$(λ - 3) ((λ - 3) (λ^{2} - 4 λ + 10)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$(λ - 3) (λ - 3) (λ^{2} - 4 λ + 10) = 0$

Step 1.7.1.3

Combine like factors.

Tap for more steps...

Step 1.7.1.3.1

Raise $λ - 3$ to the power of $1$ .

$(λ - 3) (λ - 3) (λ^{2} - 4 λ + 10) = 0$

Step 1.7.1.3.2

Raise $λ - 3$ to the power of $1$ .

$(λ - 3) (λ - 3) (λ^{2} - 4 λ + 10) = 0$

Step 1.7.1.3.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(λ - 3)}^{1 + 1} (λ^{2} - 4 λ + 10) = 0$

Step 1.7.1.3.4

Add $1$ and $1$ .

${(λ - 3)}^{2} (λ^{2} - 4 λ + 10) = 0$

Step 1.7.2

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

${(λ - 3)}^{2} = 0$

$λ^{2} - 4 λ + 10 = 0$

Step 1.7.3

Set ${(λ - 3)}^{2}$ equal to $0$ and solve for $λ$ .

Tap for more steps...

Step 1.7.3.1

Set ${(λ - 3)}^{2}$ equal to $0$ .

${(λ - 3)}^{2} = 0$

Step 1.7.3.2

Solve ${(λ - 3)}^{2} = 0$ for $λ$ .

Tap for more steps...

Step 1.7.3.2.1

Set the $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 1.7.3.2.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 1.7.4

Set $λ^{2} - 4 λ + 10$ equal to $0$ and solve for $λ$ .

Tap for more steps...

Step 1.7.4.1

Set $λ^{2} - 4 λ + 10$ equal to $0$ .

$λ^{2} - 4 λ + 10 = 0$

Step 1.7.4.2

Solve $λ^{2} - 4 λ + 10 = 0$ for $λ$ .

Tap for more steps...

Step 1.7.4.2.1

Use the quadratic formula to find the solutions.

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

Step 1.7.4.2.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = 10$ into the quadratic formula and solve for $λ$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 1.7.4.2.3

Simplify.

Tap for more steps...

Step 1.7.4.2.3.1

Simplify the numerator.

Tap for more steps...

Step 1.7.4.2.3.1.1

Raise $- 4$ to the power of $2$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Step 1.7.4.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 10$ .

Tap for more steps...

Step 1.7.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 10}}{2 \cdot 1}$

Step 1.7.4.2.3.1.2.2

Multiply $- 4$ by $10$ .

$λ = \frac{4 \pm \sqrt{16 - 40}}{2 \cdot 1}$

Step 1.7.4.2.3.1.3

Subtract $40$ from $16$ .

$λ = \frac{4 \pm \sqrt{- 24}}{2 \cdot 1}$

Step 1.7.4.2.3.1.4

Rewrite $- 24$ as $- 1 (24)$ .

$λ = \frac{4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 1}$

Step 1.7.4.2.3.1.5

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$λ = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 1}$

Step 1.7.4.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$λ = \frac{4 \pm i \cdot \sqrt{24}}{2 \cdot 1}$

Step 1.7.4.2.3.1.7

Rewrite $24$ as $2^{2} \cdot 6$ .

Tap for more steps...

Step 1.7.4.2.3.1.7.1

Factor $4$ out of $24$ .

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

Step 1.7.4.2.3.1.7.2

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

$λ = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Step 1.7.4.2.3.1.8

Pull terms out from under the radical.

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

Step 1.7.4.2.3.1.9

Move $2$ to the left of $i$ .

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

Step 1.7.4.2.3.2

Multiply $2$ by $1$ .

$λ = \frac{4 \pm 2 i \sqrt{6}}{2}$

Step 1.7.4.2.3.3

Simplify $\frac{4 \pm 2 i \sqrt{6}}{2}$ .

$λ = 2 \pm i \sqrt{6}$

Step 1.7.4.2.4

The final answer is the combination of both solutions.

$λ = 2 + i \sqrt{6}, 2 - i \sqrt{6}$

Step 1.7.5

The final solution is all the values that make ${(λ - 3)}^{2} (λ^{2} - 4 λ + 10) = 0$ true.

$λ = 3, 2 + i \sqrt{6}, 2 - i \sqrt{6}$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{4})$

Step 3

Find the eigenvector using the eigenvalue $λ = 3$ .

Tap for more steps...

Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] - 3 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Multiply $- 3$ by each element of the matrix.

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 3.2.1.2.1

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.10

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.11

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.12

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.13

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.14

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.15

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & - 3 \cdot 1 \end{array}]$

Step 3.2.1.2.16

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & - 3 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 4 - 3 & 0 + 0 & 2 + 0 & 0 + 0 \\ 3 + 0 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3

Simplify each element.

Tap for more steps...

Step 3.2.3.1

Subtract $3$ from $4$ .

$[\begin{array}{c} 1 & 0 + 0 & 2 + 0 & 0 + 0 \\ 3 + 0 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.2

Add $0$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 + 0 & 0 + 0 \\ 3 + 0 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.3

Add $2$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 + 0 \\ 3 + 0 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 + 0 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.5

Add $3$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & 2 - 3 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.6

Subtract $3$ from $2$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 + 0 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.7

Add $4$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 + 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 + 0 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.9

Add $2$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 + 0 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.10

Add $- 3$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & 1 - 3 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.11

Subtract $3$ from $1$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 + 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.12

Add $0$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 \\ 5 + 0 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.13

Add $5$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 \\ 5 & - 4 + 0 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.14

Add $- 4$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 \\ 5 & - 4 & 2 + 0 & 3 - 3 \end{array}]$

Step 3.2.3.15

Add $2$ and $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 \\ 5 & - 4 & 2 & 3 - 3 \end{array}]$

Step 3.2.3.16

Subtract $3$ from $3$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 \\ 3 & - 1 & 4 & 0 \\ 2 & - 3 & - 2 & 0 \\ 5 & - 4 & 2 & 0 \end{array}]$

Step 3.3

Find the null space when $λ = 3$ .

Tap for more steps...

Step 3.3.1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 3 & - 1 & 4 & 0 & 0 \\ 2 & - 3 & - 2 & 0 & 0 \\ 5 & - 4 & 2 & 0 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

Tap for more steps...

Step 3.3.2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 3.3.2.1.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 3 - 3 \cdot 1 & - 1 - 3 \cdot 0 & 4 - 3 \cdot 2 & 0 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ 2 & - 3 & - 2 & 0 & 0 \\ 5 & - 4 & 2 & 0 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & - 1 & - 2 & 0 & 0 \\ 2 & - 3 & - 2 & 0 & 0 \\ 5 & - 4 & 2 & 0 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

Tap for more steps...

Step 3.3.2.2.1

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & - 1 & - 2 & 0 & 0 \\ 2 - 2 \cdot 1 & - 3 - 2 \cdot 0 & - 2 - 2 \cdot 2 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 \\ 5 & - 4 & 2 & 0 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & - 1 & - 2 & 0 & 0 \\ 0 & - 3 & - 6 & 0 & 0 \\ 5 & - 4 & 2 & 0 & 0 \end{array}]$

Step 3.3.2.3

Perform the row operation $R_{4} = R_{4} - 5 R_{1}$ to make the entry at $4, 1$ a $0$ .

Tap for more steps...

Step 3.3.2.3.1

Perform the row operation $R_{4} = R_{4} - 5 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & - 1 & - 2 & 0 & 0 \\ 0 & - 3 & - 6 & 0 & 0 \\ 5 - 5 \cdot 1 & - 4 - 5 \cdot 0 & 2 - 5 \cdot 2 & 0 - 5 \cdot 0 & 0 - 5 \cdot 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & - 1 & - 2 & 0 & 0 \\ 0 & - 3 & - 6 & 0 & 0 \\ 0 & - 4 & - 8 & 0 & 0 \end{array}]$

Step 3.3.2.4

Multiply each element of $R_{2}$ by $- 1$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 3.3.2.4.1

Multiply each element of $R_{2}$ by $- 1$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ - 0 & - - 1 & - - 2 & - 0 & - 0 \\ 0 & - 3 & - 6 & 0 & 0 \\ 0 & - 4 & - 8 & 0 & 0 \end{array}]$

Step 3.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & - 3 & - 6 & 0 & 0 \\ 0 & - 4 & - 8 & 0 & 0 \end{array}]$

Step 3.3.2.5

Perform the row operation $R_{3} = R_{3} + 3 R_{2}$ to make the entry at $3, 2$ a $0$ .

Tap for more steps...

Step 3.3.2.5.1

Perform the row operation $R_{3} = R_{3} + 3 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 6 + 3 \cdot 2 & 0 + 3 \cdot 0 & 0 + 3 \cdot 0 \\ 0 & - 4 & - 8 & 0 & 0 \end{array}]$

Step 3.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & - 4 & - 8 & 0 & 0 \end{array}]$

Step 3.3.2.6

Perform the row operation $R_{4} = R_{4} + 4 R_{2}$ to make the entry at $4, 2$ a $0$ .

Tap for more steps...

Step 3.3.2.6.1

Perform the row operation $R_{4} = R_{4} + 4 R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 8 + 4 \cdot 2 & 0 + 4 \cdot 0 & 0 + 4 \cdot 0 \end{array}]$

Step 3.3.2.6.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 0 & 2 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.3

Use the result matrix to declare the final solution to the system of equations.

$x_{1} + 2 x_{3} = 0$

$x_{2} + 2 x_{3} = 0$

$0 = 0$

Step 3.3.4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - 2 x_{3} \\ - 2 x_{3} \\ x_{3} \\ x_{4} \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{3} [\begin{array}{c} - 2 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${x_{3} [\begin{array}{c} - 2 \\ - 2 \\ 1 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4} \in R}$

Step 3.3.7

The solution is the set of vectors created from the free variables of the system.

${[\begin{array}{c} - 2 \\ - 2 \\ 1 \\ 0 \end{array}], [\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue $λ = 2 + i \sqrt{6}$ .

Tap for more steps...

Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] - (2 + i \sqrt{6}) [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 4.2

Simplify.

Tap for more steps...

Step 4.2.1

Simplify each term.

Tap for more steps...

Step 4.2.1.1

Apply the distributive property.

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + (- 1 \cdot 2 - (i \sqrt{6})) [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 4.2.1.2

Multiply $- 1$ by $2$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + (- 2 - (i \sqrt{6})) [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 4.2.1.3

Multiply $- 2 - i \sqrt{6}$ by each element of the matrix.

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} (- 2 - i \sqrt{6}) \cdot 1 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 \end{array}]$

Step 4.2.1.4

Simplify each element in the matrix.

Tap for more steps...

Step 4.2.1.4.1

Multiply $- 2 - i \sqrt{6}$ by $1$ .

$[\begin{array}{c} 4 & 0 & 2 & 0 \\ 3 & 2 & 4 & 0 \\ 2 & - 3 & 1 & 0 \\ 5 & - 4 & 2 & 3 \end{array}] + [\begin{array}{c} - 2 - i \sqrt{6} & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 & (- 2 - i \sqrt{6}) \cdot 0 \\ (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 0 & (- 2 - i \sqrt{6}) \cdot 1 \end{array}]$

Step 4.2.1.4.2