## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 1 - Systems of Linear Equations - 1.1 Introduction to Systems of Linear Equations - 1.1 Exercises - Page 12: 88

#### Answer

The condition for the uniqueness of the solution is $$ad-cb\neq0$$

#### Work Step by Step

These are two equations in two variables $x$ and $y$. Let us remind ourselves of the possibilities: Possibility 1: The second equation will be dependent on the first one i. e. we would be able to multiply the first equation by some number different than zero to get the second one. In this case we will have infinitely many solutions. Possibility 2: We can get the left side of the second equation by multiplying the first one by some number different than zero but then their right sides will differ. This means that the equations are inconsistent and we will not have any solutions. Possibility 3: The second equation is independent on the first one, i.e. we cannot obtain its left side by multiplying the left side of the first equation by some number different that zero. In this case we will have a unique solution! So if $k_1a=c$ and $k_2b=d$ then $k_1\neq k_2$ (because we cannot multiply the $a$ and $b$ by the same number to get $c$ and $d$). Multiplying left side of the previous equality with the right side of the second equality and vice versa we get $$k_1ad=k_2bc.$$ Since $k_1\neq k_2$ then we can write $$k_1ad\neq k_1bc$$ $$k_2ad\neq k_2 bc$$ IMPORTANT: Since $k_1$ and $k_2$ are different they cannot be both equal to zero. This means that it would be "legal" to divide at least one of the previous equations ( the first one by $k_1$ or the second one by $k_2$) to get $$ad\neq bc$$ which is often written as $$ad-bc\neq 0.$$

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