Short Answer

Consider two positive numbers a and b. Solve the following system:
$| \begin{matrix} a x - b y & = 1 \\ b x + a y & = 0 \end{matrix} |$ .
What are the signs of the solutions x and y? How does x change as b increases?

Therefore,

$[\begin{matrix} \begin{matrix} x \\ y \end{matrix} \end{matrix}] = [\begin{matrix} \frac{a}{a^{2} + b^{2}} \\ \frac{- b}{a^{2} + b^{2}} \end{matrix}]$

Since a and are b positive numbers, it means that x is positive, and y is negative.

Also, as b increases, x decreases.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition.

In matrices, Cramer's rule expresses the solution in terms of the determinants of the coefficient matrix (i.e., for a square matrix) and of matrices obtained from it by replacing one column by the column vector of right-hand-sides of the equations.

Step 2: What are the signs of the solutions x and y?

Using Cramer's rule, we solve:

$\det A = |\begin{matrix} a & - b \\ b & a \end{matrix}| \det A = a^{2} + b^{2} x = \frac{|\begin{matrix} 1 & - b \\ 0 & a \end{matrix}|}{a^{2} + b^{2}} x = \frac{a}{a^{2} + b^{2}} y = \frac{|\begin{matrix} a & 1 \\ b & 0 \end{matrix}|}{a^{2} + b^{2}} y = \frac{- b}{a^{2} + b^{2}}$

Since a and are b positive numbers, it means that x is positive, and y is negative. Also, as b increases, x decreases.

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Linear Algebra With Applications

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Short Answer

Consider two positive numbers a and b. Solve the following system:
$| \begin{matrix} a x - b y & = 1 \\ b x + a y & = 0 \end{matrix} |$ .
What are the signs of the solutions x and y? How does x change as b increases?

Step by Step Solution

Step 1: Definition.

Step 2: What are the signs of the solutions x and y?

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Consider two positive numbers a and b. Solve the following system:
$| \begin{matrix} a x - b y & = 1 \\ b x + a y & = 0 \end{matrix} |$ .
What are the signs of the solutions x and y? How does x change as b increases?