1. Systems of linear equations with a parameter

Systems of linear equations with a parameter are solved by the same basic methods as ordinary systems of equations: the substitution method, the method of adding equations, and the graphical method. Knowledge of the graphical interpretation of linear systems makes it easy to answer the question about the number of roots and their existence.

Example 1.

Find all values ​​for parameter a for which the system of equations has no solutions.

(x + (a 2 – 3)y = a,
(x + y = 2.

Solution.

Let's look at several ways to solve this task.

1 way. We use the property: the system has no solutions if the ratio of the coefficients in front of x is equal to the ratio of the coefficients in front of y, but not equal to the ratio of the free terms (a/a 1 = b/b 1 ≠ c/c 1). Then we have:

1/1 = (a 2 – 3)/1 ≠ a/2 or system

(and 2 – 3 = 1,
(a ≠ 2.

From the first equation a 2 = 4, therefore, taking into account the condition that a ≠ 2, we get the answer.

Answer: a = -2.

Method 2. We solve by substitution method.

(2 – y + (a 2 – 3)y = a,
(x = 2 – y,

((a 2 – 3)y – y = a – 2,
(x = 2 – y.

After taking the common factor y out of brackets in the first equation, we get:

((a 2 – 4)y = a – 2,
(x = 2 – y.

The system has no solutions if the first equation has no solutions, that is

(and 2 – 4 = 0,
(a – 2 ≠ 0.

Obviously, a = ±2, but taking into account the second condition, the answer only comes with a minus answer.

Answer: a = -2.

Example 2.

Find all values ​​for parameter a for which the system of equations has an infinite number of solutions.

(8x + ay = 2,
(ax + 2y = 1.

Solution.

According to the property, if the ratio of the coefficients of x and y is the same, and is equal to the ratio of the free members of the system, then it has an infinite number of solutions (i.e. a/a 1 = b/b 1 = c/c 1). Therefore 8/a = a/2 = 2/1. Solving each of the resulting equations, we find that a = 4 is the answer in this example.

Answer: a = 4.

2. Systems of rational equations with a parameter

Example 3.

(3|x| + y = 2,
(|x| + 2y = a.

Solution.

Let's multiply the first equation of the system by 2:

(6|x| + 2y = 4,
(|x| + 2y = a.

Subtracting the second equation from the first, we get 5|x| = 4 – a. This equation will have a unique solution for a = 4. In other cases, this equation will have two solutions (for a< 4) или ни одного (при а > 4).

Answer: a = 4.

Example 4.

Find all values ​​of the parameter a for which the system of equations has a unique solution.

(x + y = a,
(y – x 2 = 1.

Solution.

We will solve this system using the graphical method. Thus, the graph of the second equation of the system is a parabola raised along the Oy axis upward by one unit segment. The first equation specifies a set of lines parallel to the line y = -x (picture 1). It is clearly seen from the figure that the system has a solution if the straight line y = -x + a is tangent to the parabola at a point with coordinates (-0.5, 1.25). Substituting these coordinates into the straight line equation instead of x and y, we find the value of parameter a:

1.25 = 0.5 + a;

Answer: a = 0.75.

Example 5.

Using the substitution method, find out at what value of the parameter a, the system has a unique solution.

(ax – y = a + 1,
(ax + (a + 2)y = 2.

Solution.

From the first equation we express y and substitute it into the second:

(y = ax – a – 1,
(ax + (a + 2)(ax – a – 1) = 2.

Let us reduce the second equation to the form kx = b, which will have a unique solution for k ≠ 0. We have:

ax + a 2 x – a 2 – a + 2ax – 2a – 2 = 2;

a 2 x + 3ax = 2 + a 2 + 3a + 2.

We represent the square trinomial a 2 + 3a + 2 as a product of brackets

(a + 2)(a + 1), and on the left we take x out of brackets:

(a 2 + 3a)x = 2 + (a + 2)(a + 1).

Obviously, a 2 + 3a should not be equal to zero, therefore,

a 2 + 3a ≠ 0, a(a + 3) ≠ 0, which means a ≠ 0 and ≠ -3.

Answer: a ≠ 0; ≠ -3.

Example 6.

Using the graphical solution method, determine at what value of parameter a the system has a unique solution.

(x 2 + y 2 = 9,
(y – |x| = a.

Solution.

Based on the condition, we construct a circle with a center at the origin and a radius of 3 unit segments; this is what is specified by the first equation of the system

x 2 + y 2 = 9. The second equation of the system (y = |x| + a) is a broken line. By using figure 2 We consider all possible cases of its location relative to the circle. It is easy to see that a = 3.

Answer: a = 3.

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The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. In mathematics, there are problems in which it is necessary to search for solutions to linear and quadratic equations in general form or to search for the number of roots that an equation has depending on the value of a parameter. All these tasks have parameters.

Consider the following equations as an illustrative example:

\[y = kx,\] where \ are variables, \ is a parameter;

\[y = kx + b,\] where \ are variables, \ is a parameter;

\[аx^2 + bх + с = 0,\] where \ is a variable, \[а, b, с\] is a parameter.

Solving an equation with a parameter means, as a rule, solving an infinite set of equations.

However, following a certain algorithm, you can easily solve the following equations:

1. Determine the “control” values ​​of the parameter.

2. Solve the original equation for [\x\] with the parameter values ​​defined in the first paragraph.

3. Solve the original equation for [\x\] for parameter values ​​different from those chosen in the first paragraph.

Let's say we're given the following equation:

\[\mid 6 - x \mid = a.\]

Having analyzed the initial data, it is clear that a \[\ge 0.\]

According to the modulus rule \ we express \

Answer: \where\

Where can I solve an equation with a parameter online?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

Note. In the example given, the calculation of all determinants ended with a representation in the form of a product of factors, one of which (13) was reduced during division. This situation is very common. Therefore, there is no need to rush to multiply the factors, although most often they do not cancel.

Problem 4.4. Solve systems of equations using Cramer's rule:

Solving the above problems shows that Cramer's formulas represent a unified and convenient method for finding solutions to systems of linear equations.

Note: The use of Cramer's formulas is greatly simplified if you need to find only one of the unknowns: in this case, you only need to count two determinants.

2.4.4. Systems of equations with parameters

Above, systems of linear algebraic equations with fixed coefficients for unknowns and the right-hand sides of the equations were considered throughout. In practical problems, very often these coefficients and the values ​​of the right-hand sides are not known accurately. Therefore, it is necessary to analyze the influence of such parameters on the solution of systems.

Example 4.5. Investigate the dependence of the solution to a system of equations

3 x + 8 y = a5 x + 9 y = b

from parameters a and b.

Here, only the right-hand sides of the equations depend on the parameters. Because the

To find a solution, you can use Cramer's formulas. We have:

y(a, b)

x(a, b)

Thus, each pair of parameters a and b corresponds to a unique pair of numbers x and y that satisfies the given system of equations. This means that the solution to the system of equations is an ordered pair and two functions of two variables (parameters a and b). Both functions are defined for any values ​​of these parameters and depend linearly on the independent variables a and b. In addition, x is monotonically increasing

and explore the dependence of their solution on parameters a and b. Recommendation. Plot the resulting solutions x (a, b) and y (a, b)

as functions of the variable parameters a and b. Explain why in all problems the solutions depend linearly on the parameters a and b.

Example 4.6. Investigate the dependence of the solution to a system of equations

This determinant is not equal to zero only when a ≠ − 5. Therefore, Cramer’s formulas can be used only when a ≠ − 5. In this case:

Let us consider separately the case a = − 5. Then the original system is:

− 2 x −10 y = 5 x +5 y = b

− 5 − c x = c , y = 2

Of course, there is arbitrariness in choosing the value of any of the unknowns, and the solution can also be written in the form:

x = − 5 2 − 5 c , y = c

Thus, the dependence on the parameter of the coefficients for the unknowns of the original system can give rise to the absence of a solution or the presence of an infinite number of solutions. The discovered fact is a generalization of what was previously known for one equation ax = b and for systems of two linear equations with two unknowns.

Remark 1. The introduction of the constant c into the solution of a system of equations resembles arbitrariness in the choice of the integration constant.

Note 2. The considered example shows that, as for one equation, for linear algebraic systems with a large number of equations and unknowns, only three different cases are possible: a single solution, no solution, or infinitely many solutions.

Problem 4.6. Explore solutions to the system of equations:

Problem 4.7. Come up with your own system of two algebraic equations with two unknowns and two parameters and study it depending on the values ​​of the parameters.

Questions for self-control

1) What is the minor of a determinant element?

2) What is the difference between algebraic complement and minor element of a determinant?

3) What is an adjoint matrix?

4) How to find the adjoint matrix for a given matrix?

5) What is the order of the adjoint matrix?

6) In what case does the inverse matrix not exist?

7) Which matrix is ​​called non-singular?

8) Under what conditions can Cramer's formulas be used?

9) What is the solution to a system of linear algebraic equations?

10) What determinants are included in Cramer's formulas?

11) When do determinants depend on parameters?

12) Can the product of the adjoint matrix and the original matrix be a scalar matrix?

13) How does the rearrangement of factors affect the result when multiplying the adjoint and original matrix?

14) What are Cramer's formulas?

15) Under what conditions can a solution to a system of linear algebraic equations be found using Cramer’s rule (formulas)?

Let's solve a system of equations with a parameter (A. Larin, option 98)

Find all values ​​of the parameter, for each of which the system

has exactly one solution.

Let's take a closer look at the system. In the first equation of the system, the left side is , and the right side does not depend on the parameter. That is, we can consider this equation as the equation of the function

and we can plot this function.

Second equation of the system

depends on the parameter, and by selecting a complete square on the left side of the equation, we obtain the equation of a circle.

So it makes sense to plot graphs of each equation and see at what value of the parameter these graphs have one intersection point.

Let's start with the first equation. First, let's open the modules. To do this, we equate each submodular expression to zero in order to find the points at which the sign changes.

The first submodular expression changes sign at , the second - at .

Let's plot these points on the coordinate line and find the signs of each submodular expression on each interval:

Note that for and the equation does not make sense, so we puncture these points.

Now let's expand the modules on each interval. (Remember: if a submodular expression is greater than or equal to zero, then we expand the module with the same sign, and if less than zero, then with the opposite sign.)

Both submodular expressions are negative, therefore, we expand both modules with the opposite sign:

That is, when the original function has the form

On this interval, the first submodular expression is negative, and the second is positive, therefore we obtain:

- the function does not exist on this interval.

3. title="x>2">!}

On this interval, both submodular expressions are positive; we expand both modules with the same sign. We get:

That is, with title="x>2"> исходная функция имеет вид !}

So, we got the graph of the function

Now let's look at the second equation:

Let’s select a complete square on the left side of the equation; to do this, add the number 4 to both sides of the equation:

For a specific value of the parameter, the graph of this equation is a circle with a center at a point with coordinates, the radius of which is 5. For different values, we have a series of circles:

We will move the circle from bottom to top until it touches the left side of the graph of the first function. In the picture this circle is red. The center of this circle is the point, its coordinates are (-2;-3). Further, when moving upward, the circle has one intersection point with the left side of the function graph, that is, the system has a unique solution.

We continue to move the circle up until it touches the right side of the graph of the first function. This will happen when the center of the circle is at the point with coordinates (-2;0) - in the figure this circle is blue.

When moving further upward, the circle will intersect both the left and right parts of the graph of the first function, that is, the circle will have two points of intersection with the graph of the first function, and the system will have two solutions. This situation continues until the center of the circle is at the point with coordinates (-2; 5) - this circle is green. At this point the circle touches the left side of the graph and intersects the right. That is, the system has one solution.

So, the system has a unique solution when (-3;0]}