MHB Finding Unique solution for system of linear equations

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To determine when the given matrix has a unique solution or infinitely many solutions, the key is to analyze the determinant. A unique solution exists when the determinant is non-zero. The values of k that make the determinant zero need to be identified, as they indicate potential cases for infinite solutions or no solutions. If the determinant is zero, further substitution is necessary to check for repeated or parallel equations. Understanding these concepts will clarify the conditions for unique and infinite solutions in the system of linear equations.
diggybob
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Hey guys, I am a little bit stuck on a recent math question and i was wondering if i could get some help about the best way to go about doing it

i have a matrice which is

1 2 -1 / -3
0 1 (-k-3) /-5
0 0 (k^2-2k) /(5k+11)

and i need to find when it has a unique solution, and infinitely many. Now i don't think i can have infinitely many because from what i understand the points at 3,3 and 3,4 both need to =0, and i can't find an x that does that. I am not sure how to go about finding the unique solution either.
i also can't get the matrix latex to work at all, sorry
 
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diggybob said:
Hey guys, I am a little bit stuck on a recent math question and i was wondering if i could get some help about the best way to go about doing it

i have a matrice which is

1 2 -1 / -3
0 1 (-k-3) /-5
0 0 (k^2-2k) /(5k+11)

and i need to find when it has a unique solution, and infinitely many. Now i don't think i can have infinitely many because from what i understand the points at 3,3 and 3,4 both need to =0, and i can't find an x that does that. I am not sure how to go about finding the unique solution either.
i also can't get the matrix latex to work at all, sorry

First of all, the singular of "matrices" is "matrix", not "matrice".

You will have unique solution wherever the determinant nonzero.

Once you have the values of k where the determinant is 0, then you will need to substitute them into determine if you have repeated equations (which would mean infinite solutions) or parallel equations (which would mean no solutions)...
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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