Algebra Examples

$[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$

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

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

Substitute the known values in the formula.

$p (λ) = determinant ([\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] - λ [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}])$

Subtract the eigenvalue times the identity matrix from the original matrix.

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

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$ .

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Simplify element $0, 0$ by multiplying $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Simplify element $0, 1$ by multiplying $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ - λ \cdot 0 & - λ \cdot 1 \end{array}]$

Simplify element $1, 0$ by multiplying $- λ \cdot 0$ .

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \cdot 1 \end{array}]$

Simplify element $1, 1$ by multiplying $- λ \cdot 1$ .

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$

Combine the similar matrices with each others.

$p (λ) = [\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}] + [\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$

Simplify each element of the matrix $[\begin{array}{c} - λ + 3 & 0 + 2 \\ 0 + 4 & - λ + 6 \end{array}]$ .

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Combine the same size matrices $[\begin{array}{c} 3 & 2 \\ 4 & 6 \end{array}]$ and $[\begin{array}{c} - λ & 0 \\ 0 & - λ \end{array}]$ by adding the corresponding elements of each.

$p (λ) = [\begin{array}{c} - λ + 3 & 0 + 2 \\ 0 + 4 & - λ + 6 \end{array}]$

Simplify element $0, 1$ of the matrix.

$p (λ) = [\begin{array}{c} - λ + 3 & 2 \\ 0 + 4 & - λ + 6 \end{array}]$

Simplify element $1, 0$ of the matrix.

$p (λ) = [\begin{array}{c} - λ + 3 & 2 \\ 4 & - λ + 6 \end{array}]$

The determinant of $[\begin{array}{c} - λ + 3 & 2 \\ 4 & - λ + 6 \end{array}]$ is $λ^{2} - 9 λ + 10$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} - L + 3 & 2 \\ 4 & - L + 6 \end{array}] = | \begin{array}{c} - L + 3 & 2 \\ 4 & - L + 6 \end{array} |$

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

$p (λ) = (- λ + 3) (- λ + 6) - 4 \cdot 2$

Simplify the determinant.

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Simplify each term.

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Expand $(- λ + 3) (- λ + 6)$ using the FOIL Method.

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Apply the distributive property.

$p (λ) = - λ (- λ + 6) + 3 (- λ + 6) - 4 \cdot 2$

Apply the distributive property.

$p (λ) = - λ (- λ) - λ \cdot 6 + 3 (- λ + 6) - 4 \cdot 2$

Apply the distributive property.

$p (λ) = - λ (- λ) - λ \cdot 6 + 3 (- λ) + 3 \cdot 6 - 4 \cdot 2$

Simplify and combine like terms.

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Simplify each term.

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Multiply $λ$ by $λ$ .

$p (λ) = - 1 \cdot (- 1 λ^{2}) - λ \cdot 6 + 3 (- λ) + 3 \cdot 6 - 4 \cdot 2$

Multiply $- 1$ by $- 1$ .

$p (λ) = 1 λ^{2} - λ \cdot 6 + 3 (- λ) + 3 \cdot 6 - 4 \cdot 2$

Multiply $λ^{2}$ by $1$ .

$p (λ) = λ^{2} - λ \cdot 6 + 3 (- λ) + 3 \cdot 6 - 4 \cdot 2$

Multiply $6$ by $- 1$ .

$p (λ) = λ^{2} - 6 λ + 3 (- λ) + 3 \cdot 6 - 4 \cdot 2$

Multiply $- 1$ by $3$ .

$p (λ) = λ^{2} - 6 λ - 3 λ + 3 \cdot 6 - 4 \cdot 2$

Multiply $3$ by $6$ .

$p (λ) = λ^{2} - 6 λ - 3 λ + 18 - 4 \cdot 2$

Subtract $3 λ$ from $- 6 λ$ .

$p (λ) = λ^{2} - 9 λ + 18 - 4 \cdot 2$

Multiply $- 4$ by $2$ .

$p (λ) = λ^{2} - 9 λ + 18 - 8$

Subtract $8$ from $18$ .

$p (λ) = λ^{2} - 9 λ + 10$

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