Practical lesson (part 1). Systems of linear equations
The practical lesson is devoted to systems of linear equations.
Our task is to work out basic concepts that were discussed in the first lecture.
Let's start with the following task.
It is required to find the intersection of two lines given by linear equations.
We already know that each linear equation desummandines a straight line on the plane, if there are two variables in this equation
In order to find the intersection point it is possible, of course, to build these lines on the plane.
But we'll do it differently.
We’ll apply an algebraic way.
Let’s consider a system made up of these equations, and find a solution.
To do this, we add the second to the first one, i.e we perform an elementary transformation.
Then coefficient x will be zeroed out - 4 and 4 will be 0*x.
Further -y+3y=+2y.
On the right side there is four.
The second equation, we write without changes.
Thus, the summand 0*x can be removed because it is zero.
After this transformation, we obtain immediately the value of variable y.
So, y equals two.
We substitute this value into the second equation and find x: 4*x plus 3*2, that is plus 6, equals = 2. So 4*x =-4. Hence x - is -1.
So, this system has a unique solution.
This solution will be a pair of numbers (-1, 2).
Hence the crossing point of these lines has coordinates (-1, 2).
Thus the solution is received.
The common point of these lines is found.
Now let’s consider another task.
There is given a linear equation in the variable x. It is required to examine it, i.e. to determine what values of the variable give the equation one solution, what values give an infinite number, and what values give no solutions.
So a is a parameter that is a fixed number, but we do not know what it is equal to.
Let us remember what cases, generally speaking, are possible if we are given a linear equation with one variable.
The coefficient of the x is marked with letter k.
So, k is multiplied by variable x, a constant summand will be marked b.
Of course, we are accustomed to this simple equation solved by x is equal to b divided by k.
But this is possible only if k is not equal to zero.
That is, if k is not zero, then the solution of the equation is number b / k.
But if k = 0, in this case a study is necessary to see what would happen.
So I suggest to consider for our example a case where a coefficient a x is equal to zero.
Let's write it down (a^2-4) and immediately transform it according to the formula of difference of squares - (a-2)*(a + 2).
Obviously, this coefficient will be equal to zero in one of two cases: if a is 2 or -2.
Let's see what happens when a is two.
We substitute it in our original equation.
It takes the form 0*x=4.
This equation is contradictory.
So there are no solutions, only an empty set.
So, for a which is equal to two we get an empty set of solutions
The second case.
If we take another value - -2.
Again coefficient x is zero, however, the constant summand will also be zero.
We get a zero equation
It is known that its decision is any number, i.e. the whole set of real numbers R.
Well, finally the third case.
When a is not equal to 2 and is not equal to minus 2.
I will write briefly separated by semi-colon.
In this case, we will proceed as we used to do.
We divide by coefficient k and obtain value x.
We can devide because k is not equal to zero coefficient.
Thus, (a + 2) divided by (a ^ 2-4), given that we factored it, we see that one of them can be reduced.
Thus, the equation has the following root, the following decision - x = 1 / (a-2).
So thus we get the answer.
The equation has one solution for all a unequal to numbers 2 and -2.
Here it is the solution.
The equation has infinitely many solutions, when a is -2 – this is R set - infinitely many solutions.
And not just an infinite number, but any number will be the solution.
And finally, the equation has no solution if a is 2.
Thus, if you go back to the original general form of the linear equation with one variable, we can say that if k is not zero, the equation has only one root.
If k is equal to zero, the equation cannot have roots and can have exactly one, sorry, cannot have roots and can have infinitely many solutions.
And another problem in which we consider a system of linear equations.
So the question again is dependent on the parameters.
The question is, under what values a given system of equations has one solution.
It is required to find the solution.
What will we do?
Again, we apply the technique of transformations.
Note that the first summand of these equations is the same.
So we get a zero subtraction summand.
Subtract, for example, the first from the second equation.
We get the following system, which is known to be equivalent to the original one: x + y = a, and the second equation is transformed as follows. I will not write 0 at x, and for y we get a coefficient a-1.
So, (a-1)*y=1-a.
What do we see?
The second equation already contains a variable, and we also can study this equation as in the previous task
We begin with a variable coefficient of - (a-1).
First, we consider the case where the coefficient is equal to one.
More precisely, let's think about what would happen if a is equal to one.
and the second equation is transformed as follows. 0 I will not write at x, while for y we get a coefficient a-1.
And we seek one solution, so we can just assume that a is not equal to one to have this one solution.
So, if a is not equal to one, in this case we multiply by the number inverse of (a-1), that is, divide by (a-1), and we get the following number value.
Note (a-1) and (1-a) are the opposite numbers.
I can write that this way: (1-a), but with a minus sign.
Reducing, we get -1.
But that's not the answer yet, because we have one root for the second equation, and we solve the system
We substitute the value of variable y in the first equation and see what happens.
So, a*x-1 = a.
Lt’s express x.
Again, we do not rush.
Let’s bring to the standard form and again there is a obtain dilemma – the coefficient before x is either equal to zero or not .
If it is zero, our equation, let's write, takes the form 0*x=1, i.e. it will not have a solution, but then the whole system will be inconsistent.
So this case is not suitable for us.
Therefore parameter a should not be equal to zero.
In this case, dividing by a, we obtain the value of variable x.
So, in order that the system has one solution it is necessary that a is not equal to one, and secondly, a should not be zero.
We get the answer to the problem – a is not equal to one, a is not zero, in this case, the system has one solution.
Write it in the form of a pair: the first coordinate (a + 1) and the second one is -1. Well, if the question were wider – to consider all the options, we would say that if a=0, the system is inconsistent, and if a equals one, ... And I suggest you to think on this issue yourself.