Linear Algebra Examples

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

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 $p (λ) = determinant (A - λ I_{3})$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $2$ and $0$ .

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

Step 4.3.2

Add $1$ and $0$ .

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

Step 4.3.3

Add $1$ and $0$ .

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

Step 4.3.4

Subtract $λ$ from $0$ .

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

Step 4.3.5

Add $0$ and $0$ .

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

Step 4.3.6

Add $0$ and $0$ .

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

Step 4.3.7

Add $2$ and $0$ .

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

Step 5

Find the determinant.

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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 column $1$ by its cofactor and add.

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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.

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

Step 5.1.4

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

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

Step 5.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

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

Step 5.1.6

Multiply element $a_{21}$ by its cofactor.

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

Step 5.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

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

Step 5.1.8

Multiply element $a_{31}$ by its cofactor.

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

Step 5.1.9

Add the terms together.

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

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Apply the distributive property.

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

Step 5.3.2.1.2

Multiply $- 1$ by $1$ .

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

Step 5.3.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.3.2.1.4

Simplify each term.

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Step 5.3.2.1.4.1

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

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Step 5.3.2.1.4.1.1

Move $λ$ .

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

Step 5.3.2.1.4.1.2

Multiply $λ$ by $λ$ .

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

Step 5.3.2.1.4.2

Multiply $- 1$ by $- 1$ .

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

Step 5.3.2.1.4.3

Multiply $λ^{2}$ by $1$ .

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

Step 5.3.2.1.5

Multiply $- 2$ by $0$ .

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

Step 5.3.2.2

Add $- λ + λ^{2}$ and $0$ .

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

Step 5.3.2.3

Reorder $- λ$ and $λ^{2}$ .

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

Step 5.4

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

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Step 5.4.1

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

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 (1 - λ) - 2 \cdot 1) + 0$

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Apply the distributive property.

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 \cdot 1 + 2 (- λ) - 2 \cdot 1) + 0$

Step 5.4.2.1.2

Multiply $2$ by $1$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 + 2 (- λ) - 2 \cdot 1) + 0$

Step 5.4.2.1.3

Multiply $- 1$ by $2$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 - 2 λ - 2 \cdot 1) + 0$

Step 5.4.2.1.4

Multiply $- 2$ by $1$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (2 - 2 λ - 2) + 0$

Step 5.4.2.2

Combine the opposite terms in $2 - 2 λ - 2$ .

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Step 5.4.2.2.1

Subtract $2$ from $2$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (- 2 λ + 0) + 0$

Step 5.4.2.2.2

Add $- 2 λ$ and $0$ .

$p (λ) = (2 - λ) (λ^{2} - λ) - 1 (- 2 λ) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Add $(2 - λ) (λ^{2} - λ) - 1 (- 2 λ)$ and $0$ .

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

Step 5.5.2

Simplify each term.

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Step 5.5.2.1

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

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Step 5.5.2.1.1

Apply the distributive property.

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

Step 5.5.2.1.2

Apply the distributive property.

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

Step 5.5.2.1.3

Apply the distributive property.

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

Step 5.5.2.2

Simplify and combine like terms.

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Step 5.5.2.2.1

Simplify each term.

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Step 5.5.2.2.1.1

Multiply $- 1$ by $2$ .

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

Step 5.5.2.2.1.2

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

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Step 5.5.2.2.1.2.1

Move $λ^{2}$ .

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

Step 5.5.2.2.1.2.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.2.2.1.2.2.1

Raise $λ$ to the power of $1$ .

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

Step 5.5.2.2.1.2.2.2

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

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

Step 5.5.2.2.1.2.3

Add $2$ and $1$ .

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

Step 5.5.2.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.5.2.2.1.4

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

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Step 5.5.2.2.1.4.1

Move $λ$ .

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

Step 5.5.2.2.1.4.2

Multiply $λ$ by $λ$ .

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

Step 5.5.2.2.1.5

Multiply $- 1$ by $- 1$ .

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

Step 5.5.2.2.1.6

Multiply $λ^{2}$ by $1$ .

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

Step 5.5.2.2.2

Add $2 λ^{2}$ and $λ^{2}$ .

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

Step 5.5.2.3

Multiply $- 2$ by $- 1$ .

$p (λ) = 3 λ^{2} - 2 λ - λ^{3} + 2 λ$

Step 5.5.3

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

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Step 5.5.3.1

Add $- 2 λ$ and $2 λ$ .

$p (λ) = 3 λ^{2} - λ^{3} + 0$

Step 5.5.3.2

Add $3 λ^{2} - λ^{3}$ and $0$ .

$p (λ) = 3 λ^{2} - λ^{3}$

Step 5.5.4

Reorder $3 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 3 λ^{2}$

Step 6

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

$- λ^{3} + 3 λ^{2} = 0$

Step 7

Solve for $λ$ .

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Step 7.1

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

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Step 7.1.1

Factor $- λ^{2}$ out of $- λ^{3}$ .

$- λ^{2} λ + 3 λ^{2} = 0$

Step 7.1.2

Factor $- λ^{2}$ out of $3 λ^{2}$ .

$- λ^{2} λ - λ^{2} \cdot - 3 = 0$

Step 7.1.3

Factor $- λ^{2}$ out of $- λ^{2} (λ) - λ^{2} (- 3)$ .

$- λ^{2} (λ - 3) = 0$

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 $0$ .

$λ^{2} = 0$

$λ - 3 = 0$

Step 7.3

Set $λ^{2}$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $λ^{2}$ equal to $0$ .

$λ^{2} = 0$

Step 7.3.2

Solve $λ^{2} = 0$ for $λ$ .

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Step 7.3.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$λ = \pm \sqrt{0}$

Step 7.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 7.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$λ = \pm \sqrt{0^{2}}$

Step 7.3.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$λ = \pm 0$

Step 7.3.2.2.3

Plus or minus $0$ is $0$ .

$λ = 0$

Step 7.4

Set $λ - 3$ equal to $0$ and solve for $λ$ .

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Step 7.4.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 7.4.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 7.5

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

$λ = 0, 3$

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