Linear algebra/System of linear equations

In this course we will investigate solutions to linear equations, vectors and vector spaces and connections between them.

Of course you're familiar with basic linear equations from Algebra. They consist of equations such as

3x+2y=7\,

.

and you're hopefully familiar with all the usual problems that go along with this sort of equation. Such as how to find the slope, how to place the equation in point-slope form, standard form, or slope-intercept form.

All of these ideas are useful, but we will, at first, be interested in systems of linear equations. A system of equations is some collection of equations that are all expected to hold at the same time. Let us start with an example of such a system.

\left\{{\begin{aligned}3x-y&=14\\2x+y&=11\end{aligned}}\right.

.

Usually the first questions to ask is are there any solutions to a system of equations. By this we mean is there a pair of numbers so that both equations hold when you plug in this pair. For example, in the above system, if you take $x=5$ and $y=1$ then you can check that both equations hold for this pair of numbers:

\left\{{\begin{aligned}3\cdot 5-1&=14\\2\cdot 5+1&=11\end{aligned}}\right.

.

Notice that we are asking for the same pair of numbers to satisfy both equations. The first thing to realize is that just because you ask for two equations to hold at the same time doesn't mean it is always possible. Consider the system:

\left\{{\begin{aligned}x+y&=2\\x+y&=3\end{aligned}}\right.

Clearly there is a problem. If $x+y=2$ and $x+y=3$ then you would have $2=x+y=3$ , and so $2=3$ . Which is absurd. Just because we would like there to be a pair of numbers that satisfy both equations doesn't mean that we will get what we like. One of the main goals of this course will be to understand when systems of equations have solutions, how many solutions they have, and how to find them.

In the first example above there is an easy way to find that $x=5$ and $y=1$ is the solution directly from the equations. If you add these two equations together, you can see that the y's cancel each other out. When this happens, you will get $5x=25$ , or $x=5$ . Substituting back into the above, we find that $y=1$ . Note that this is the only solution to the system of equations.

Frequently one encounters systems of equations with more then just two variables. For example one may be interested in finding a solution to the system:

\left\{{\begin{aligned}3x+2y-z&=2\\2x-3y+z&=5\\x+y+z&=11\end{aligned}}\right.

.

Solving Linear Systems Algebraically

One was mentioned above, but there are other ways to solve a system of linear equations without graphing.

Variable Elimination

2x+y=11\,

-4x+3y=13\,

Multiply 1st row by 2 add to 2nd row

4x+2y=22\,

-4x+3y=13\,

Yields

5y=35\,

=>

y=7

Multiply 1st row by -3 add to 2nd row

-6x-3y=-33\,

-4x+3y=13\,

Yields

-10x=-20\,

=>

x=2

Substitution

If you get a system of equations that looks like this:

2x+y=11\,

-4x+3y=13\,

You can switch around some terms in the first to get this:

y=-2x+11\,

Then you can substitute that into the bottom one so that it looks like this:

-4x+3(-2x+11)=13\,

-4x-6x+33=13\,

-10x+33=13\,

-10x=-20\,

x=2\,

Then, you can substitute 2 into an x from either equation and solve for y. It's usually easier to substitute it in the one that had the single y. In this case, after substituting 2 for x, you would find that y = 7.

Determinant

If you get a system of equations that looks like this:

2x+y=11\,

-4x+3y=13\,

Solve for y

x+{\frac {1}{2}}y={\frac {11}{2}}\,

x+{\frac {3}{-4}}y={\frac {13}{-4}}\,

y({\frac {1}{2}}-{\frac {3}{4}})={\frac {11}{2}}-{\frac {13}{-4}}

y={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{4}}}}

Solve for x

2x+y=11\,

{\frac {-4}{3}}x+y={\frac {13}{3}}\,

x(2-{\frac {-4}{3}})=11-{\frac {13}{3}}

x={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{4}}}}

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