Is the given system of equations solvable with back substitution?

Correct. In summary, the conversation discusses the solution to a system of equations with 5 variables and 3 equations. The final row of the matrix represents an equation involving all 5 variables. The conversation also considers the possibilities for the solution, which can be a unique solution, multiple solutions, or no solution. The correct answer is B, as there are two variables that can be assigned values.
  • #1
TyErd
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Homework Statement



I have attached the question

Homework Equations





The Attempt at a Solution



I think the answer is no solution because there is 5 variables but only 3 equations. Is that correct?
 

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  • #2
TyErd said:

Homework Statement



I have attached the question

Homework Equations





The Attempt at a Solution



I think the answer is no solution because there is 5 variables but only 3 equations. Is that correct?

No, that is not the reason. Look at the final row; it is shorthand for an equation involving x1, x2, x3, x4, x5. What is the equation?

RGV
 
  • #3
do you mean 0x1 + 0x2 + 0x3 + 0x4 + 0x5 = 4?
 
  • #4
if the third is also an equation that means there must be an answer right? So because we have to assign variables and we have 5unknowns and 3 equations that must mean two values will be variables right?? so the answer has to be B yeah??
 
  • #5
TyErd said:
do you mean 0x1 + 0x2 + 0x3 + 0x4 + 0x5 = 4?
For what values of the variables x1, x1, x2, x3, and x4 will this be a true statement?


TyErd said:
if the third is also an equation that means there must be an answer right?
Not necessarily. There are three possibilities for a system of equations (which are here represented by an augmented matrix):
1) a unique solution
2) multiple solutions
3) no solution.
TyErd said:
So because we have to assign variables and we have 5unknowns and 3 equations that must mean two values will be variables right?? so the answer has to be B yeah??
 

FAQ: Is the given system of equations solvable with back substitution?

What is the purpose of "Matrices-back substitution" in scientific research?

Matrices-back substitution is a technique used to solve systems of linear equations, which are commonly encountered in scientific research. It allows for the efficient and accurate calculation of unknown variables in a system of equations by manipulating matrices.

How does "Matrices-back substitution" differ from other methods of solving systems of equations?

Unlike other methods such as Gaussian elimination, matrices-back substitution is specifically used for solving systems of equations that have already been converted into matrix form. It involves working backwards from the last equation in the system to find the values of the unknown variables.

What are the key steps involved in using "Matrices-back substitution" to solve a system of equations?

The first step is to convert the system of equations into matrix form. Then, the matrix is manipulated using elementary row operations to reduce it to upper triangular form. The next step is to use back substitution to find the values of the unknown variables. Finally, the solutions are checked by plugging them back into the original equations.

Can "Matrices-back substitution" be used for systems of equations with any number of unknown variables?

Yes, matrices-back substitution can be used for systems of equations with any number of unknown variables. However, as the number of variables increases, the calculations can become more complex and time-consuming, making it less efficient compared to other methods.

What are the potential limitations of using "Matrices-back substitution"?

One limitation is that it can only be used for systems of equations that have already been converted into matrix form. This means that the researcher must have some knowledge of matrix algebra. Additionally, as the number of equations and unknown variables increases, the calculations can become more difficult and prone to errors.

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