Question

Consider two positive numbers a and b. Solve the following system:

$\left|\begin{array}{cc}ax-by& =1\\ bx+ay& =0\end{array}\right|$.

What are the signs of the solutions x and y? How does x change as bincreases?

Short answer

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.

Step-by-step solution

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:

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

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