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Found in: Page 307

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

Therefore,

$\left[\begin{array}{c}\begin{array}{c}x\\ y\end{array}\end{array}\right]=\left[\begin{array}{c}\frac{a}{{a}^{2}+{b}^{2}}\\ \frac{-b}{{a}^{2}+{b}^{2}}\end{array}\right]$

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 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:

$\mathrm{det}\mathrm{A}=\left|\begin{array}{cc}\mathrm{a}& -\mathrm{b}\\ \mathrm{b}& \mathrm{a}\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{det}\mathrm{A}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{x}=\frac{\left|\begin{array}{cc}1& -\mathrm{b}\\ 0& \mathrm{a}\end{array}\right|}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{x}=\frac{\mathrm{a}}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\frac{\left|\begin{array}{cc}\mathrm{a}& 1\\ \mathrm{b}& 0\end{array}\right|}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\frac{-\mathrm{b}}{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}$

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