Linear Algebra Examples

$[\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}]$

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 $[\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 0 & 0 \\ - 3 & 9 & 18 \\ 2 & - 3 & - 6 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 + 0 & 0 + 0 \\ - 3 + 0 & 9 - λ & 18 + 0 \\ 2 + 0 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3

Simplify each element.

Tap for more steps...

Step 4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 + 0 \\ - 3 + 0 & 9 - λ & 18 + 0 \\ 2 + 0 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 \\ - 3 + 0 & 9 - λ & 18 + 0 \\ 2 + 0 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3.3

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 \\ - 3 & 9 - λ & 18 + 0 \\ 2 + 0 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3.4

Add $18$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 \\ - 3 & 9 - λ & 18 \\ 2 + 0 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3.5

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 \\ - 3 & 9 - λ & 18 \\ 2 & - 3 + 0 & - 6 - λ \end{array}]$

Step 4.3.6

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 0 & 0 \\ - 3 & 9 - λ & 18 \\ 2 & - 3 & - 6 - λ \end{array}]$

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.

$| \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} |$

Step 5.1.4

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

$(3 - λ) | \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} - 3 & 18 \\ 2 & - 6 - λ \end{array} |$

Step 5.1.6

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

$0 | \begin{array}{c} - 3 & 18 \\ 2 & - 6 - λ \end{array} |$

Step 5.1.7

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

$| \begin{array}{c} - 3 & 9 - λ \\ 2 & - 3 \end{array} |$

Step 5.1.8

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

$0 | \begin{array}{c} - 3 & 9 - λ \\ 2 & - 3 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = (3 - λ) | \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} | + 0 | \begin{array}{c} - 3 & 18 \\ 2 & - 6 - λ \end{array} | + 0 | \begin{array}{c} - 3 & 9 - λ \\ 2 & - 3 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} - 3 & 18 \\ 2 & - 6 - λ \end{array} |$ .

$p (λ) = (3 - λ) | \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} | + 0 + 0 | \begin{array}{c} - 3 & 9 - λ \\ 2 & - 3 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} - 3 & 9 - λ \\ 2 & - 3 \end{array} |$ .

$p (λ) = (3 - λ) | \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} | + 0 + 0$

Step 5.4

Evaluate $| \begin{array}{c} 9 - λ & 18 \\ - 3 & - 6 - λ \end{array} |$ .

Tap for more steps...

Step 5.4.1

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

$p (λ) = (3 - λ) ((9 - λ) (- 6 - λ) - (- 3 \cdot 18)) + 0 + 0$

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

Expand $(9 - λ) (- 6 - λ)$ using the FOIL Method.

Tap for more steps...

Step 5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (3 - λ) (9 (- 6 - λ) - λ (- 6 - λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (3 - λ) (9 \cdot - 6 + 9 (- λ) - λ (- 6 - λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (3 - λ) (9 \cdot - 6 + 9 (- λ) - λ \cdot - 6 - λ (- λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2

Simplify and combine like terms.

Tap for more steps...

Step 5.4.2.1.2.1

Simplify each term.

Tap for more steps...

Step 5.4.2.1.2.1.1

Multiply $9$ by $- 6$ .

$p (λ) = (3 - λ) (- 54 + 9 (- λ) - λ \cdot - 6 - λ (- λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.2

Multiply $- 1$ by $9$ .

$p (λ) = (3 - λ) (- 54 - 9 λ - λ \cdot - 6 - λ (- λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.3

Multiply $- 6$ by $- 1$ .

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ - λ (- λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ - 1 \cdot - 1 λ \cdot λ - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.5

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

Tap for more steps...

Step 5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ - 1 \cdot - 1 (λ \cdot λ) - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ - 1 \cdot - 1 λ^{2} - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ + 1 λ^{2} - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (3 - λ) (- 54 - 9 λ + 6 λ + λ^{2} - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.2.2

Add $- 9 λ$ and $6 λ$ .

$p (λ) = (3 - λ) (- 54 - 3 λ + λ^{2} - (- 3 \cdot 18)) + 0 + 0$

Step 5.4.2.1.3

Multiply $- (- 3 \cdot 18)$ .

Tap for more steps...

Step 5.4.2.1.3.1

Multiply $- 3$ by $18$ .

$p (λ) = (3 - λ) (- 54 - 3 λ + λ^{2} - - 54) + 0 + 0$

Step 5.4.2.1.3.2

Multiply $- 1$ by $- 54$ .

$p (λ) = (3 - λ) (- 54 - 3 λ + λ^{2} + 54) + 0 + 0$

Step 5.4.2.2

Combine the opposite terms in $- 54 - 3 λ + λ^{2} + 54$ .

Tap for more steps...

Step 5.4.2.2.1

Add $- 54$ and $54$ .

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

Step 5.4.2.2.2

Add $- 3 λ + λ^{2}$ and $0$ .

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

Step 5.4.2.3

Reorder $- 3 λ$ and $λ^{2}$ .

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

Step 5.5

Simplify the determinant.

Tap for more steps...

Step 5.5.1

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

Tap for more steps...

Step 5.5.1.1

Add $(3 - λ) (λ^{2} - 3 λ)$ and $0$ .

$p (λ) = (3 - λ) (λ^{2} - 3 λ) + 0$

Step 5.5.1.2

Add $(3 - λ) (λ^{2} - 3 λ)$ and $0$ .

$p (λ) = (3 - λ) (λ^{2} - 3 λ)$

Step 5.5.2

Expand $(3 - λ) (λ^{2} - 3 λ)$ using the FOIL Method.

Tap for more steps...

Step 5.5.2.1

Apply the distributive property.

$p (λ) = 3 (λ^{2} - 3 λ) - λ (λ^{2} - 3 λ)$

Step 5.5.2.2

Apply the distributive property.

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

Step 5.5.2.3

Apply the distributive property.

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

Step 5.5.3

Simplify and combine like terms.

Tap for more steps...

Step 5.5.3.1

Simplify each term.

Tap for more steps...

Step 5.5.3.1.1

Multiply $- 3$ by $3$ .

$p (λ) = 3 λ^{2} - 9 λ - λ \cdot λ^{2} - λ (- 3 λ)$

Step 5.5.3.1.2

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

Tap for more steps...

Step 5.5.3.1.2.1

Move $λ^{2}$ .

$p (λ) = 3 λ^{2} - 9 λ - (λ^{2} λ) - λ (- 3 λ)$

Step 5.5.3.1.2.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.5.3.1.2.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 3 λ^{2} - 9 λ - (λ^{2} λ^{1}) - λ (- 3 λ)$

Step 5.5.3.1.2.2.2

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

$p (λ) = 3 λ^{2} - 9 λ - λ^{2 + 1} - λ (- 3 λ)$

Step 5.5.3.1.2.3

Add $2$ and $1$ .

$p (λ) = 3 λ^{2} - 9 λ - λ^{3} - λ (- 3 λ)$

Step 5.5.3.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{2} - 9 λ - λ^{3} - 1 \cdot - 3 λ \cdot λ$

Step 5.5.3.1.4

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

Tap for more steps...

Step 5.5.3.1.4.1

Move $λ$ .

$p (λ) = 3 λ^{2} - 9 λ - λ^{3} - 1 \cdot - 3 (λ \cdot λ)$

Step 5.5.3.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = 3 λ^{2} - 9 λ - λ^{3} - 1 \cdot - 3 λ^{2}$

Step 5.5.3.1.5

Multiply $- 1$ by $- 3$ .

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

Step 5.5.3.2

Add $3 λ^{2}$ and $3 λ^{2}$ .

$p (λ) = 6 λ^{2} - 9 λ - λ^{3}$

Step 5.5.4

Move $- 9 λ$ .

$p (λ) = 6 λ^{2} - λ^{3} - 9 λ$

Step 5.5.5

Reorder $6 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 9 λ$

Step 6

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

$- λ^{3} + 6 λ^{2} - 9 λ = 0$

Step 7

Solve for $λ$ .

Tap for more steps...

Step 7.1

Factor the left side of the equation.

Tap for more steps...

Step 7.1.1

Factor $- λ$ out of $- λ^{3} + 6 λ^{2} - 9 λ$ .

Tap for more steps...

Step 7.1.1.1

Factor $- λ$ out of $- λ^{3}$ .

$- λ \cdot λ^{2} + 6 λ^{2} - 9 λ = 0$

Step 7.1.1.2

Factor $- λ$ out of $6 λ^{2}$ .

$- λ \cdot λ^{2} - λ (- 6 λ) - 9 λ = 0$

Step 7.1.1.3

Factor $- λ$ out of $- 9 λ$ .

$- λ \cdot λ^{2} - λ (- 6 λ) - λ \cdot 9 = 0$

Step 7.1.1.4

Factor $- λ$ out of $- λ (λ^{2}) - λ (- 6 λ)$ .

$- λ (λ^{2} - 6 λ) - λ \cdot 9 = 0$

Step 7.1.1.5

Factor $- λ$ out of $- λ (λ^{2} - 6 λ) - λ (9)$ .

$- λ (λ^{2} - 6 λ + 9) = 0$

Step 7.1.2

Factor using the perfect square rule.

Tap for more steps...

Step 7.1.2.1

Rewrite $9$ as $3^{2}$ .

$- λ (λ^{2} - 6 λ + 3^{2}) = 0$

Step 7.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 λ = 2 \cdot λ \cdot 3$

Step 7.1.2.3

Rewrite the polynomial.

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

Step 7.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = λ$ and $b = 3$ .

$- λ {(λ - 3)}^{2} = 0$

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 $0$ .

$λ = 0$

${(λ - 3)}^{2} = 0$

Step 7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 7.4

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

Tap for more steps...

Step 7.4.1

Set ${(λ - 3)}^{2}$ equal to $0$ .

${(λ - 3)}^{2} = 0$

Step 7.4.2

Solve ${(λ - 3)}^{2} = 0$ for $λ$ .

Tap for more steps...

Step 7.4.2.1

Set the $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 7.4.2.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 7.5

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

$λ = 0, 3$