Find the Characteristic Equation [[2,1],[4,0]]

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

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.

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

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

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

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

Simplify each element in the matrix.

Rearrange $- λ \cdot 1$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 1$ .

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

Add the corresponding elements.

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

Simplify each element of the matrix $[\begin{array}{c} 2 - λ & 1 + 0 \\ 4 + 0 & 0 - λ \end{array}]$ .

Simplify $1 + 0$ .

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

Simplify $4 + 0$ .

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

Simplify $0 - λ$ .

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

Find the determinant of $[\begin{array}{c} 2 - λ & 1 \\ 4 & - λ \end{array}]$ .

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

$determinant [\begin{array}{c} 2 - λ & 1 \\ 4 & - λ \end{array}] = | \begin{array}{c} 2 - λ & 1 \\ 4 & - λ \end{array} |$

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

$p (λ) = (2 - λ) (- λ) - 4 \cdot 1$

Simplify each term.

Apply the distributive property.

$p (λ) = 2 (- λ) - λ (- λ) - 4 \cdot 1$

Multiply $- 1$ by $2$ .

$p (λ) = - 2 λ - λ (- λ) - 4 \cdot 1$

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ - 1 \cdot (- 1 λ \cdot λ) - 4 \cdot 1$

Simplify each term.

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

Move $λ$ .

$p (λ) = - 2 λ - 1 \cdot (- 1 (λ \cdot λ)) - 4 \cdot 1$

Multiply $λ$ by $λ$ .

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

Multiply $- 1$ by $- 1$ .

$p (λ) = - 2 λ + 1 λ^{2} - 4 \cdot 1$

Multiply $λ^{2}$ by $1$ .

$p (λ) = - 2 λ + λ^{2} - 4 \cdot 1$

Multiply $- 4$ by $1$ .

$p (λ) = - 2 λ + λ^{2} - 4$

Reorder the polynomial.

$p (λ) = λ^{2} - 2 λ - 4$

Find the Characteristic Equation [[2,1],[4,0]]

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