Linear Algebra Examples

${[\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}]}^{- 1}$

Step 1

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

$p (λ) = determinant (A - λ I_{1})$

Step 2

The identity matrix or unit matrix of size $1$ is the $1 \times 1$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 3

Substitute the known values into $p (λ) = determinant (A - λ I_{1})$ .

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Step 3.1

Substitute $[\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}] - λ I_{1})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}] - λ [\begin{array}{c} 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}] + [\begin{array}{c} - λ \cdot 1 \end{array}])$

Step 4.1.2

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} \end{array}] + [\begin{array}{c} - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

To write $- λ$ as a fraction with a common denominator, multiply by $\frac{x y}{x y}$ .

$p (λ) = determinant [\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} - λ \cdot \frac{x y}{x y} \end{array}]$

Step 4.3.2

Combine $- λ$ and $\frac{x y}{x y}$ .

$p (λ) = determinant [\begin{array}{c} \frac{\sqrt{y} - \sqrt{x}}{x y} + \frac{- λ (x y)}{x y} \end{array}]$

Step 4.3.3

Combine the numerators over the common denominator.

$p (λ) = determinant [\begin{array}{c} \frac{\sqrt{y} - \sqrt{x} - λ x y}{x y} \end{array}]$

Step 5

The determinant of a $1 \times 1$ matrix is the element itself.

$p (λ) = \frac{\sqrt{y} - \sqrt{x} - λ x y}{x y}$