Solving a Matrix: How to Go Further?

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In summary, the given matrix A can be transformed into a reduced echelon form by swapping the second and third rows, resulting in a leading entry of 1 in the second row and a leading entry of (h - 2)/(2h - 6) in the third row. To find values of h that would make A invertible, the third row's leading entry must be nonzero, so h cannot be equal to 3. Other possible values for h can be determined by continuing to reduce the matrix and solving for h.
  • #1
Boom101
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A =
| 1 2 0 |
| 0 2h-6 h-2 |
| 0 1 1 |

How do I go further? 2h - 6 = 1? Or should it be (2h-6 times a constant x) = 1, and then write the answer in terms of x? I thought of also using x(2h-6) = 1, isolate x and replace it in x(h-2) = 0 to put the answer in terms of h. But I don't think it should be in terms of any variables. Help please!
 
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  • #2
Boom101 said:

Homework Statement



A =
| 1 3 1 |
| 2 2h h |
|-3 -8 -2 |

Find all values of h so A is invertible. The thing is I cannot use determinants.

Homework Equations



Elementary Row Operations

The Attempt at a Solution



Now I attempted to make it reduced echelon form and got

A =
| 1 2 0 |
| 0 2h-6 h-2 |
| 0 1 1 |

But how do I go further? 2h - 6 = 1? Or should it be (2h-6 times a constant x) = 1, and then write the answer in terms of x? I thought of also using x(2h-6) = 1, isolate x and replace it in x(h-2) = 0 to put the answer in terms of h. But I don't think it should be in terms of any variables. Help please!
To get a 1 entry in the 2nd row, you need to divide by 2h - 6. Does that suggest a value that h cannot be?
 
  • #3
Mark44 said:
To get a 1 entry in the 2nd row, you need to divide by 2h - 6. Does that suggest a value that h cannot be?

Careful. h=3 doesn't make the matrix singular. Exchange the second and third rows and keep reducing.
 
  • #4
Ok so I get h-2 / 2h-6 I still don't understand how to finish.
 
  • #5
After swapping the 2nd and 3rd rows, as Dick suggested, you should have this:
[tex]\left[ \begin{array}{c c c}
1 & 2 & 0 \\
0 & 1 & 1 \\
0 & 2h - 6 & h - 2 \end{array}
\right][/tex]

Use the new 2nd row to eliminate the leading entry in the 3rd row. You should not get (h - 2)/(2h - 6).
As Dick suggested, swap the 2nd and 3rd rows.
 
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  • #6
Thanks everyone for the help. How do I find values other than h = 3?
 
  • #7
Boom101 said:
Thanks everyone for the help. How do I find values other than h = 3?

You'd better listen more closely to the help. h=3 is wrong. Swap the rows and finish the reduction. Then tell me what you get.
 

1. What is a matrix and why is it important to solve?

A matrix is a rectangular array of numbers or symbols that are arranged in rows and columns. It is important to solve because it is a powerful tool used in various fields of mathematics, science, and engineering to represent and manipulate data and equations.

2. How do you solve a matrix?

To solve a matrix, you can use various methods such as Gaussian elimination, Cramer's rule, and matrix inversion. These methods involve performing mathematical operations on the matrix to reduce it to a simpler form and find the solution.

3. What are the applications of solving a matrix?

Solving a matrix has various applications in fields such as computer graphics, economics, physics, and engineering. It is used to solve systems of linear equations, find optimal solutions in linear programming, and perform transformations in computer graphics.

4. What are the common mistakes to avoid when solving a matrix?

Some common mistakes to avoid when solving a matrix include not paying attention to the order of operations, not properly simplifying the matrix, and making errors while performing calculations. It is important to double-check your work and use a calculator if needed.

5. Are there any tips for efficiently solving a matrix?

Some tips for efficiently solving a matrix include organizing your work and breaking down the problem into smaller steps, understanding the properties and rules of matrix operations, and practicing regularly to improve your skills. It is also helpful to seek guidance from a teacher or tutor if you are struggling.

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