# Without graphing , how do you decide whether the following system of linear equations has one solution , infinitely many solutions or no solution?

Mar 31, 2016

A system of $N$ linear equations with $N$ unknown variables that contains no linear dependency between equations (in other words, its determinant is non-zero) will have one and only one solution.

#### Explanation:

Let's consider a system of two linear equations with two unknown variables:
$A x + B y = C$
$D x + E y = F$

If pair $\left(A , B\right)$ is not proportional to pair $\left(D , E\right)$ (that is, there is no such number $k$ that $D = k A$ and $E = k B$, which can be checked by condition $A \cdot E - B \cdot D \ne 0$) then there is one and only one solution:

$x = \frac{C \cdot E - B \cdot F}{A \cdot E - B \cdot D}$, $y = \frac{A \cdot F - C \cdot D}{A \cdot E - B \cdot D}$

Example:
$x + y = 3$
$x - 2 y = - 3$

Solution:
$x = \frac{3 \cdot \left(- 2\right) - 1 \cdot \left(- 3\right)}{1 \cdot \left(- 2\right) - 1 \cdot 1} = 1$
$y = \frac{1 \cdot \left(- 3\right) - 3 \cdot 1}{1 \cdot \left(- 2\right) - 1 \cdot 1} = 2$

If pair $\left(A , B\right)$ is proportional to pair $\left(D , E\right)$ (which means that there is such number $k$ that $D = k A$ and $E = k B$, which can be checked by a condition $A \cdot E - B \cdot D = 0$), there are two cases:

(a) infinite number of solutions if $C$ and $F$ are proportional with the same coefficient as $A$ and $D$, that is $F = k C$, where $k$ is the same coefficient of proportionality;
Example:
$x + y = 3$
$2 x + 2 y = 6$
Here $k = 2$ for all pairs: $D = 2 A$, $E = 2 B$, $F = 2 C$.
The second equation is a trivial consequence of the first (just multiply the first equation by $2$) and, therefore, provides no additional information about unknown, reducing the number of equations, effectively, to 1.

(b) no solutions at all, if $F \ne k C$
Example:
$x + 4 y = 3$
$2 x + 8 y = 5$
In this case equations contradict each other since, by multiplying the first by 2, we derive to equation $2 x + 8 y = 6$, which cannot have common solution with $2 x + 8 y = 5$ since the left parts of these two equations are equal, but the right parts are not.