Solve for variable in augmented matrix

In summary, to find the value of h that would make the given augmented matrix consistent, divide the second row by -4, resulting in h = -5.
  • #1
Gurvir
19
0

Homework Statement



Given the augmented matrix for a system of equations below, find m such that the system will have infinitely many solutions.
3 3 -9 15
1 -1 3 -3
-4 -4 12 m

m = ?

Homework Equations


None

The Attempt at a Solution



Well I tried to reduce the matrix to row-echelon form but I'm not sure what to do with the 'm' at all. I am not familiar with variables in matrices. Here's how far I have gotten:

r1/3 and r3/-4

1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4

r2-r1 and r3-r1

1 1 -3 5
0 1 -3 4
0 0 0 m-5/-4

I don't know how to answer for m. I am pretty sure that infinite solutions means that the matrix needs as many variables as equations which I read somewhere else on the forum.
 
Last edited:
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  • #2
Gurvir said:

Homework Statement



Given the augmented matrix for a system of equations below, find m such that the system will have infinitely many solutions.
3 3 -9 15
1 -1 3 -3
-4 -4 12 m

m = ?

Homework Equations


None


The Attempt at a Solution



Well I tried to reduce the matrix to row-echelon form but I'm not sure what to do with the 'm' at all. I am not familiar with variables in matrices. Here's how far I have gotten:

r1/3 and r3/-4

1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4

r2/r1 and r3/r1

1 1 -3 5
0 1 -3 4
0 0 0 m-5/-4

I don't know how to answer for m. I am pretty sure that infinite solutions means that the matrix needs as many variables as equations which I read somewhere else on the forum.
No, it doesn't mean this. If the system represented by your augmented matrix has an infinite number of solutions, there will be fewer equations than variables.

What are you doing here?
Gurvir said:
r2/r1 and r3/r1

1 1 -3 5
0 1 -3 4
0 0 0 m-5/-4
Dividing a row by another row is not one of the three valid row operations. In the matrix above that, you should notice that row 1 and row 3 are almost the same, and will be the same for some value of m.
 
  • #3
Mark44 said:
No, it doesn't mean this. If the system represented by your augmented matrix has an infinite number of solutions, there will be fewer equations than variables.

What are you doing here?
Dividing a row by another row is not one of the three valid row operations. In the matrix above that, you should notice that row 1 and row 3 are almost the same, and will be the same for some value of m.

So wait, you make one row equal to the one with the variable to solve for it? also, dividing is not one of the row operations but i believe multiplying is. where you can multiply 1/-4 which is the same as dividing
 
  • #4
Yes, you can replace a row by a multiple of itself, but you can't divide one row by another, which is what your notation implied that you were doing.
 
  • #5
Mark44 said:
Yes, you can replace a row by a multiple of itself, but you can't divide one row by another, which is what your notation implied that you were doing.

My bad, i didn't mean r2/r1...it was r2-r1 and r3-r1

Anyways, back to my question:
You make one row equal to the one with the variable to solve for it? Or is there some other way
 
  • #6
What does m need to be so that the first and third rows are the same?
1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4
 
  • #7
Mark44 said:
What does m need to be so that the first and third rows are the same?
1 1 -3 5
1 -1 3 -3
1 1 -3 m/-4

Oh! m = -20?

Would this be the same idea?
Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.

5 -3 h
-20 15 5

I can't solve for it, you can never make the first to variables in the matrix equal
You can get:

r2/-5
5 -3 h
4 -3 1

or

r1 * -4
-20 12 -4h
-20 15 5
 
Last edited:
  • #8
Gurvir said:
Oh! m = -20?

Would this be the same idea?
Determine the value of h such that the matrix is the augmented matrix of a consistent linear system.

5 -3 h
-20 15 5

I can't solve for it, you can never make the first to variables in the matrix equal
You can get:

r2/-5
5 -3 h
4 -3 1
You have a mistake in the work above.
Gurvir said:
or

r1 * -4
-20 12 -4h
-20 15 5

In this augmented matrix...

5 -3 h
-20 15 5

it's possible to find a solution for any value of h.
 
  • #9
If you have trouble seeing the "big picture", write it out as an equation.
 
  • #10
flyingpig said:
If you have trouble seeing the "big picture", write it out as an equation.

Sorry I was just brain dead yesterday. You just divide r2 by -4 and then it gives you h. Thanks for your help guys.
 

1. How do I solve for a variable in an augmented matrix?

To solve for a variable in an augmented matrix, you will need to use the row operations of addition, subtraction, and multiplication to eliminate the variable from one of the equations. This will result in a new matrix with the variable isolated on one side and the constants on the other side. You can then use basic algebra to solve for the variable.

2. What is the purpose of using an augmented matrix to solve for a variable?

An augmented matrix is used to solve for a variable in a system of linear equations by organizing the equations and variables into a matrix form. This makes it easier to apply the row operations and solve the system of equations simultaneously, resulting in a simpler and more efficient solution.

3. Can an augmented matrix have more than one solution for a variable?

Yes, an augmented matrix can have infinitely many solutions for a variable. This means that there are multiple values for the variable that satisfy all of the equations in the system. This is known as an inconsistent system of equations.

4. What happens if there is no solution for the variable in an augmented matrix?

If there is no solution for the variable in an augmented matrix, it means that the system of equations is inconsistent and cannot be solved. This could be due to contradictory equations or a lack of information to solve for the variable.

5. Are there any shortcuts for solving for a variable in an augmented matrix?

Yes, there are some shortcuts that can be used to solve for a variable in an augmented matrix, such as using the inverse matrix or Gaussian elimination. However, these methods are more advanced and may require a deeper understanding of matrix algebra.

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