- #1
What is the Correct Determinant for a Misprinted Matrix?
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Homework Help Overview
The discussion revolves around determining the correct determinant for a misprinted matrix, specifically comparing two matrices, A and B, and exploring the implications of row and column operations on their determinants.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the effects of row and column operations on determinants, question the numerical discrepancies in their calculations, and explore the possibility of misprints in the problem setup.
Discussion Status
There are various attempts to compute the determinants of matrices A and B, with some participants expressing frustration over incorrect numerical results. Suggestions regarding potential misprints in the problem have been raised, indicating an ongoing exploration of the issue without a clear consensus.
Contextual Notes
Participants note that the problem may involve fixed values for the unknown elements, and there is a suggestion that the instructor may have made an error in the problem's formulation.
- #2
Hurkyl
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What have you tried?
- #3
Jameson
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I assume you're familiar with solving a 3x3 Matrix?
- #4
HallsofIvy
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It looks to me like B can be derived from A by "row operations".
(and a few "column" operations- but they work the same way.)
Do you know how row-column operations on a matrix affect its determinant?
Swapping two rows (columns) multiplies the determinant by -1.
Multiplying a row (column) by a number multiplies the determinant by that number.
Adding a multiple of one row (column) to another doesn't change the determinant.
(and a few "column" operations- but they work the same way.)
Do you know how row-column operations on a matrix affect its determinant?
Swapping two rows (columns) multiplies the determinant by -1.
Multiplying a row (column) by a number multiplies the determinant by that number.
Adding a multiple of one row (column) to another doesn't change the determinant.
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- #5
plucker_08
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i've tried the problem...but, the numerical coefficients of b is always wrong...
- #6
CarlB
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Here's my two cents, assuming that I haven't made any arithmetic errors.
Put c=e=g=1, and the rest zero, and compute the determinants for the two matrices. I got |A|=-1, |B|=-12.
Put b=i=1, and the rest zero, and again compute the determinants for the two matrices. I got |A|= -2, |B|= -8.
Consequently, if you're looking for |B| to be a multiple of |A| you are going to be disappointed. So I think that the suggestions given so far for solving this problem will, as the OP says, come up short.
The restriction that |A|=1 is only one equation restricting 7 unknowns. This could get ugly.
Carl
Put c=e=g=1, and the rest zero, and compute the determinants for the two matrices. I got |A|=-1, |B|=-12.
Put b=i=1, and the rest zero, and again compute the determinants for the two matrices. I got |A|= -2, |B|= -8.
Consequently, if you're looking for |B| to be a multiple of |A| you are going to be disappointed. So I think that the suggestions given so far for solving this problem will, as the OP says, come up short.
The restriction that |A|=1 is only one equation restricting 7 unknowns. This could get ugly.
Carl
- #7
plucker_08
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the unknown values of the elements and its determinants are fixed values...
thanks in advance!
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- #8
CarlB
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plucker_08 said:
Of course everyone knows this already. Perhaps my post was too subtle.
Here's some guesses. You're taking a Linear Algebra class. You still don't have a solution to the problem. The instructor wrote the problem; it wasn't taken from a book. Instead, he made the problem up on his own (and thought that the answer was 12 or 44). Your instructor will admit the error sometime after the homework is due when some of the students complain that they can't work it either.
Carl
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- #9
HallsofIvy
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Possibly a misprint: if the lower left corner of B were 4h/2+ 2b, then it would be doable (and, unless I have misplaced an operation, the answer is -1/3). The instructer may have forgotten to divide the lower left corner by 3 when he (I think) multiplied the bottom row by 3.