Question

For a system of linear equations in three variables the minimum number of equations required to get unique solution is _________

1 point

1

2

3

4​

Answer 1

Answer:

Tank you send me the difference between interpreter and

Step-by-step explanation:

Tank you send me the difference

Answer 2

Answer: 3

Step-by-step explanation:

First let's look at one variable:  x.  If you had one linear equation  

involving x, you'd know what it is:

                           4x = 12        ->      clearly x = 3

If you had another equation in that system, such as -2x = 10, this

would not give a consistent solution for x.  The variable x cannot

simultaneously equal 3 and -5.

consider a system of two variables:  x,y.  Two equations are  

necessary to come to a unique solution:

x + y = 6 

2x - y = 0

In this case, you can easily solve the two equations in two unknowns for x and y, and find that x = 2, y = 4.

In general, more equations added to this system will make the solution  

disappear, as the system becomes "overconstrained":

 x + y = 6   solving the first two equations together implies x = 2.

2x - y = 0   solving the last two equations together implies x = -1.

x - y = 1

Clearly, the added equation does not help.  There are too many  

constraints on the variables x and y.

On the other hand, fewer equations don't provide enough constraints

for a unique solution.  If we have only the first equation alone  

(i.e., x + y = 6), this tells us something about the relationship

between x and y, and allows x = 2 (and y = 4) as a possible solution,  

but it is not a unique solution.  In fact, there would be infinite

possible solutions to that equation by itself.

A similar type of reasoning applies as we increase the number of

variables to three, four, and so on.  We will need one additional

equation for each additional variable.

Remark I:  it is not always true that more equations than variables leads to no solution.  Sometimes, we can get lucky:

x + y = 6

2x - y = 0

 4x - y = 4  this added equation is "consistent" with x = 2, y = 4.

 However, this situation does not usually happen.

Remark II:  it is not always true that having exactly N linear

equations for N variables always leads to a unique solution.  It

is possible to have either no solution (inconsistent) or an infinite number of solutions (underconstrained):

Examples with N = 3:

x + y + z = 8                 x + y + z = 8

x + y     = 4              2x + 2y + 2z = 16

2x + 2y   = 5                         z = 5

This system is             This system is underconstrained.

inconsistent.              We know the exact value of only z.

          So minimum number of equations required is 3