Solving a Matrix: How to Go Further?

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Homework Help Overview

The discussion revolves around a matrix problem where participants are exploring how to determine values of the variable h that affect the invertibility of the matrix. The context includes performing row operations and understanding the implications of certain values of h on the matrix's properties.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to manipulate the matrix into reduced echelon form and are questioning how to proceed with isolating variables and interpreting results. There are discussions about the implications of specific values of h, particularly regarding the matrix becoming singular.

Discussion Status

The conversation is active, with participants providing suggestions on row operations and questioning the validity of certain assumptions about h. There is a focus on clarifying misunderstandings and exploring different approaches to the problem without reaching a definitive conclusion.

Contextual Notes

Participants note that the use of determinants is not allowed, which adds a layer of complexity to the problem. There is also a mention of needing to avoid specific values of h that would lead to division by zero during row operations.

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!
 
Last edited:
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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?
 
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.
 
Ok so I get h-2 / 2h-6 I still don't understand how to finish.
 
After swapping the 2nd and 3rd rows, as Dick suggested, you should have this:
[tex]\left[ \begin{array}{c c c}<br /> 1 & 2 & 0 \\<br /> 0 & 1 & 1 \\<br /> 0 & 2h - 6 & h - 2 \end{array}<br /> \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.
 
Last edited:
Thanks everyone for the help. How do I find values other than h = 3?
 
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.
 

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