Linear Algebra Examples

$[\begin{array}{c} x & y & z \\ 2 x & y & z \\ x & y & z \end{array}]$

Step 1

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

$p (λ) = determinant (A - λ I_{3})$

Step 2

The identity matrix or unit matrix of size $3$ is the $3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 3

Substitute the known values into $p (λ) = determinant (A - λ I_{3})$ .

Tap for more steps...

Step 3.1

Substitute $[\begin{array}{c} x & y & z \\ 2 x & y & z \\ x & y & z \end{array}]$ for $A$ .

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

Step 3.2

Substitute $[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ for $I_{3}$ .

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

Tap for more steps...

Step 4.3.1

Add $y$ and $0$ .

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

Step 4.3.2

Add $z$ and $0$ .

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

Step 4.3.3

Add $2 x$ and $0$ .

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

Step 4.3.4

Add $z$ and $0$ .

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

Step 4.3.5

Add $x$ and $0$ .

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

Step 4.3.6

Add $y$ and $0$ .

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

Step 5

Find the determinant.

Tap for more steps...

Step 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 5.1.1

Consider the corresponding sign chart.

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

Step 5.1.2

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

Step 5.1.3

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

$| \begin{array}{c} y - λ & z \\ y & z - λ \end{array} |$

Step 5.1.4

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

$(x - λ) | \begin{array}{c} y - λ & z \\ y & z - λ \end{array} |$

Step 5.1.5

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

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

Step 5.1.6

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

$- y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} |$

Step 5.1.7

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

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

Step 5.1.8

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

$z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = (x - λ) | \begin{array}{c} y - λ & z \\ y & z - λ \end{array} | - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2

Evaluate $| \begin{array}{c} y - λ & z \\ y & z - λ \end{array} |$ .

Tap for more steps...

Step 5.2.1

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

$p (λ) = (x - λ) ((y - λ) (z - λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2

Simplify the determinant.

Tap for more steps...

Step 5.2.2.1

Simplify each term.

Tap for more steps...

Step 5.2.2.1.1

Expand $(y - λ) (z - λ)$ using the FOIL Method.

Tap for more steps...

Step 5.2.2.1.1.1

Apply the distributive property.

$p (λ) = (x - λ) (y (z - λ) - λ (z - λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.1.2

Apply the distributive property.

$p (λ) = (x - λ) (y z + y (- λ) - λ (z - λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.1.3

Apply the distributive property.

$p (λ) = (x - λ) (y z + y (- λ) - λ z - λ (- λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2

Simplify each term.

Tap for more steps...

Step 5.2.2.1.2.1

Rewrite using the commutative property of multiplication.

$p (λ) = (x - λ) (y z - y λ - λ z - λ (- λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2.2

Rewrite using the commutative property of multiplication.

$p (λ) = (x - λ) (y z - y λ - λ z - 1 \cdot - 1 λ \cdot λ - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2.3

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

Tap for more steps...

Step 5.2.2.1.2.3.1

Move $λ$ .

$p (λ) = (x - λ) (y z - y λ - λ z - 1 \cdot - 1 (λ \cdot λ) - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2.3.2

Multiply $λ$ by $λ$ .

$p (λ) = (x - λ) (y z - y λ - λ z - 1 \cdot - 1 λ^{2} - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2.4

Multiply $- 1$ by $- 1$ .

$p (λ) = (x - λ) (y z - y λ - λ z + 1 λ^{2} - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.1.2.5

Multiply $λ^{2}$ by $1$ .

$p (λ) = (x - λ) (y z - y λ - λ z + λ^{2} - y z) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.2

Combine the opposite terms in $y z - y λ - λ z + λ^{2} - y z$ .

Tap for more steps...

Step 5.2.2.2.1

Subtract $y z$ from $y z$ .

$p (λ) = (x - λ) (- y λ - λ z + λ^{2} + 0) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.2.2

Add $- y λ - λ z + λ^{2}$ and $0$ .

$p (λ) = (x - λ) (- y λ - λ z + λ^{2}) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.2.2.3

Move $λ$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y | \begin{array}{c} 2 x & z \\ x & z - λ \end{array} | + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3

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

Tap for more steps...

Step 5.3.1

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

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (2 x (z - λ) - x z) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3.2

Simplify the determinant.

Tap for more steps...

Step 5.3.2.1

Simplify each term.

Tap for more steps...

Step 5.3.2.1.1

Apply the distributive property.

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (2 x z + 2 x (- λ) - x z) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3.2.1.2

Rewrite using the commutative property of multiplication.

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (2 x z + 2 \cdot - 1 x λ - x z) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (2 x z - 2 x λ - x z) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3.2.2

Subtract $x z$ from $2 x z$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (1 x z - 2 x λ) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.3.2.3

Multiply $x$ by $1$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z | \begin{array}{c} 2 x & y - λ \\ x & y \end{array} |$

Step 5.4

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

Tap for more steps...

Step 5.4.1

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

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (2 x y - x (y - λ))$

Step 5.4.2

Simplify the determinant.

Tap for more steps...

Step 5.4.2.1

Simplify each term.

Tap for more steps...

Step 5.4.2.1.1

Apply the distributive property.

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (2 x y - x y - x (- λ))$

Step 5.4.2.1.2

Rewrite using the commutative property of multiplication.

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (2 x y - x y - 1 \cdot - 1 x λ)$

Step 5.4.2.1.3

Simplify each term.

Tap for more steps...

Step 5.4.2.1.3.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (2 x y - x y + 1 x λ)$

Step 5.4.2.1.3.2

Multiply $x$ by $1$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (2 x y - x y + x λ)$

Step 5.4.2.2

Subtract $x y$ from $2 x y$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (1 x y + x λ)$

Step 5.4.2.3

Multiply $x$ by $1$ .

$p (λ) = (x - λ) (- y λ - 1 z λ + λ^{2}) - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5

Simplify the determinant.

Tap for more steps...

Step 5.5.1

Simplify each term.

Tap for more steps...

Step 5.5.1.1

Rewrite $- 1 z$ as $- z$ .

$p (λ) = (x - λ) (- y λ - z λ + λ^{2}) - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.2

Expand $(x - λ) (- y λ - z λ + λ^{2})$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = x (- y λ) + x (- z λ) + x λ^{2} - λ (- y λ) - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3

Simplify each term.

Tap for more steps...

Step 5.5.1.3.1

Rewrite using the commutative property of multiplication.

$p (λ) = - x y λ + x (- z λ) + x λ^{2} - λ (- y λ) - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.2

Rewrite using the commutative property of multiplication.

$p (λ) = - x y λ - x z λ + x λ^{2} - λ (- y λ) - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.3

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

Tap for more steps...

Step 5.5.1.3.3.1

Move $λ$ .

$p (λ) = - x y λ - x z λ + x λ^{2} - (λ \cdot λ) (- y) - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.3.2

Multiply $λ$ by $λ$ .

$p (λ) = - x y λ - x z λ + x λ^{2} - λ^{2} (- y) - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = - x y λ - x z λ + x λ^{2} - 1 \cdot - 1 λ^{2} y - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.5

Multiply $- 1$ by $- 1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + 1 λ^{2} y - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.6

Multiply $λ^{2}$ by $1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y - λ (- z λ) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.7

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

Tap for more steps...

Step 5.5.1.3.7.1

Move $λ$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y - (λ \cdot λ) (- z) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.7.2

Multiply $λ$ by $λ$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y - λ^{2} (- z) - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.8

Rewrite using the commutative property of multiplication.

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y - 1 \cdot - 1 λ^{2} z - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.9

Multiply $- 1$ by $- 1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + 1 λ^{2} z - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.10

Multiply $λ^{2}$ by $1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ \cdot λ^{2} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.11

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

Tap for more steps...

Step 5.5.1.3.11.1

Move $λ^{2}$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - (λ^{2} λ) - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.11.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.5.1.3.11.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - (λ^{2} λ^{1}) - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.11.2.2

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

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{2 + 1} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.3.11.3

Add $2$ and $1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - y (x z - 2 x λ) + z (x y + x λ)$

Step 5.5.1.4

Apply the distributive property.

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - y (x z) - y (- 2 x λ) + z (x y + x λ)$

Step 5.5.1.5

Multiply $- 2$ by $- 1$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - y x z + 2 y (x λ) + z (x y + x λ)$

Step 5.5.1.6

Apply the distributive property.

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - y x z + 2 y x λ + z x y + z x λ$

Step 5.5.2

Combine the opposite terms in $- x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - y x z + 2 y x λ + z x y + z x λ$ .

Tap for more steps...

Step 5.5.2.1

Reorder the factors in the terms $- y x z$ and $z x y$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - x y z + 2 y x λ + x y z + z x λ$

Step 5.5.2.2

Add $- x y z$ and $x y z$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ + 0 + z x λ$

Step 5.5.2.3

Add $- x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ$ and $0$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ + z x λ$

Step 5.5.2.4

Reorder the factors in the terms $- x z λ$ and $z x λ$ .

$p (λ) = - x y λ - x z λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ + x z λ$

Step 5.5.2.5

Add $- x z λ$ and $x z λ$ .

$p (λ) = - x y λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ + 0$

Step 5.5.2.6

Add $- x y λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ$ and $0$ .

$p (λ) = - x y λ + x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 2 y x λ$

Step 5.5.3

Add $- x y λ$ and $2 y x λ$ .

Tap for more steps...

Step 5.5.3.1

Move $y$ .

$p (λ) = x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} - x y λ + 2 x y λ$

Step 5.5.3.2

Add $- x y λ$ and $2 x y λ$ .

$p (λ) = x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + 1 x y λ$

Step 5.5.4

Multiply $x$ by $1$ .

$p (λ) = x λ^{2} + λ^{2} y + λ^{2} z - λ^{3} + x y λ$

Step 5.5.5

Reorder $λ^{2}$ and $y$ .

$p (λ) = x λ^{2} + y λ^{2} + λ^{2} z - λ^{3} + x y λ$

Step 5.5.6

Reorder $λ^{2}$ and $z$ .

$p (λ) = x λ^{2} + y λ^{2} + z λ^{2} - λ^{3} + x y λ$

Step 5.5.7

Move $- λ^{3}$ .

$p (λ) = x λ^{2} + y λ^{2} + z λ^{2} + x y λ - λ^{3}$

Step 5.5.8

Move $z λ^{2}$ .

$p (λ) = x λ^{2} + y λ^{2} + x y λ + z λ^{2} - λ^{3}$

Step 5.5.9

Move $y λ^{2}$ .

$p (λ) = x λ^{2} + x y λ + y λ^{2} + z λ^{2} - λ^{3}$

Step 5.5.10

Reorder $x λ^{2}$ and $x y λ$ .

$p (λ) = x y λ + x λ^{2} + y λ^{2} + z λ^{2} - λ^{3}$

Step 6

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

$x y λ + x λ^{2} + y λ^{2} + z λ^{2} - λ^{3} = 0$

Step 7

Solve for $λ$ .

Tap for more steps...

Step 7.1

Factor $λ$ out of $x y λ + x λ^{2} + y λ^{2} + z λ^{2} - λ^{3}$ .

Tap for more steps...

Step 7.1.1

Factor $λ$ out of $x y λ$ .

$λ (x y) + x λ^{2} + y λ^{2} + z λ^{2} - λ^{3} = 0$

Step 7.1.2

Factor $λ$ out of $x λ^{2}$ .

$λ (x y) + λ (x λ) + y λ^{2} + z λ^{2} - λ^{3} = 0$

Step 7.1.3

Factor $λ$ out of $y λ^{2}$ .

$λ (x y) + λ (x λ) + λ (y λ) + z λ^{2} - λ^{3} = 0$

Step 7.1.4

Factor $λ$ out of $z λ^{2}$ .

$λ (x y) + λ (x λ) + λ (y λ) + λ (z λ) - λ^{3} = 0$

Step 7.1.5

Factor $λ$ out of $- λ^{3}$ .

$λ (x y) + λ (x λ) + λ (y λ) + λ (z λ) + λ (- λ^{2}) = 0$

Step 7.1.6

Factor $λ$ out of $λ (x y) + λ (x λ)$ .

$λ (x y + x λ) + λ (y λ) + λ (z λ) + λ (- λ^{2}) = 0$

Step 7.1.7

Factor $λ$ out of $λ (x y + x λ) + λ (y λ)$ .

$λ (x y + x λ + y λ) + λ (z λ) + λ (- λ^{2}) = 0$

Step 7.1.8

Factor $λ$ out of $λ (x y + x λ + y λ) + λ (z λ)$ .

$λ (x y + x λ + y λ + z λ) + λ (- λ^{2}) = 0$

Step 7.1.9

Factor $λ$ out of $λ (x y + x λ + y λ + z λ) + λ (- λ^{2})$ .

$λ (x y + x λ + y λ + z λ - λ^{2}) = 0$

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 $0$ .

$λ = 0$

$x y + x λ + y λ + z λ - λ^{2} = 0$

Step 7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 7.4

Set $x y + x λ + y λ + z λ - λ^{2}$ equal to $0$ and solve for $λ$ .

Tap for more steps...

Step 7.4.1

Set $x y + x λ + y λ + z λ - λ^{2}$ equal to $0$ .

$x y + x λ + y λ + z λ - λ^{2} = 0$

Step 7.4.2

Solve $x y + x λ + y λ + z λ - λ^{2} = 0$ for $λ$ .

Tap for more steps...

Step 7.4.2.1

Use the quadratic formula to find the solutions.

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

Step 7.4.2.2

Substitute the values $a = - 1$ , $b = x + y + z$ , and $c = x y$ into the quadratic formula and solve for $λ$ .

$\frac{- (x + y + z) \pm \sqrt{{(x + y + z)}^{2} - 4 \cdot (- 1 \cdot (x y))}}{2 \cdot - 1}$

Step 7.4.2.3

Simplify.

Tap for more steps...

Step 7.4.2.3.1

Simplify the numerator.

Tap for more steps...

Step 7.4.2.3.1.1

Apply the distributive property.

$λ = \frac{- x - y - z \pm \sqrt{{(x + y + z)}^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.2

Rewrite ${(x + y + z)}^{2}$ as $(x + y + z) (x + y + z)$ .

$λ = \frac{- x - y - z \pm \sqrt{(x + y + z) (x + y + z) - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.3

Expand $(x + y + z) (x + y + z)$ by multiplying each term in the first expression by each term in the second expression.

$λ = \frac{- x - y - z \pm \sqrt{x \cdot x + x y + x z + y x + y \cdot y + y z + z x + z y + z \cdot z - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.4

Simplify each term.

Tap for more steps...

Step 7.4.2.3.1.4.1

Multiply $x$ by $x$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + x y + x z + y x + y \cdot y + y z + z x + z y + z \cdot z - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.4.2

Multiply $y$ by $y$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + x y + x z + y x + y^{2} + y z + z x + z y + z \cdot z - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.4.3

Multiply $z$ by $z$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + x y + x z + y x + y^{2} + y z + z x + z y + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.5

Add $x y$ and $y x$ .

Tap for more steps...

Step 7.4.2.3.1.5.1

Reorder $y$ and $x$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + x y + x y + x z + y^{2} + y z + z x + z y + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.5.2

Add $x y$ and $x y$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + x z + y^{2} + y z + z x + z y + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.6

Add $x z$ and $z x$ .

Tap for more steps...

Step 7.4.2.3.1.6.1

Reorder $z$ and $x$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + y^{2} + y z + x z + x z + z y + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.6.2

Add $x z$ and $x z$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + y^{2} + y z + 2 x z + z y + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.7

Add $y z$ and $z y$ .

Tap for more steps...

Step 7.4.2.3.1.7.1

Reorder $z$ and $y$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + y^{2} + y z + y z + 2 x z + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.7.2

Add $y z$ and $y z$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + y^{2} + 2 y z + 2 x z + z^{2} - 4 \cdot - 1 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.8

Multiply $- 4$ by $- 1$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + 2 x y + y^{2} + 2 y z + 2 x z + z^{2} + 4 \cdot (x y)}}{2 \cdot - 1}$

Step 7.4.2.3.1.9

Add $2 x y$ and $4 x y$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2 \cdot - 1}$

Step 7.4.2.3.2

Multiply $2$ by $- 1$ .

$λ = \frac{- x - y - z \pm \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{- 2}$

Step 7.4.2.3.3

Simplify $\frac{- x - y - z \pm \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{- 2}$ .

$λ = \frac{x + y + z \pm \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

Step 7.4.2.4

The final answer is the combination of both solutions.

$λ = \frac{x + y + z + \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z - \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z + \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z - \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z + \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z - \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

Step 7.5

The final solution is all the values that make $λ (x y + x λ + y λ + z λ - λ^{2}) = 0$ true.

$λ = 0$

$λ = \frac{x + y + z + \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z - \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = 0$

$λ = \frac{x + y + z + \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$

$λ = \frac{x + y + z - \sqrt{x^{2} + y^{2} + 2 y z + 2 x z + z^{2} + 6 x y}}{2}$