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Question 198943: Hi. I am trying to make a determinant problem for someone I am tutoring. I would like something similar to the following problem that requires factoring in the result, the answer to this problem is 5/3 and -1. I am not sure how to reverse a problem to make the answer something similar to this example. Thanks to anyone who can help.
Here is the example I am trying to simulate:
= -3
Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website! Hi. I am trying to make a determinant problem for someone I am tutoring. I would like something similar to the following problem that requires factoring in the result, the answer to this problem is 5/3 and -1. I am not sure how to reverse a problem to make the answer something similar to this example. Thanks to anyone who can help.
Here is the example I am trying to simulate:
= -3
Here's how to make as many of them as you like.
1. Choose any integer that you want to be one of
the values of x, the solution to the problem.
I'll arbitrarily choose x=2
2. Choose any two multiples of that number, positive or negative.
I'll choose two arbitrary multiples of 2, as 6 and 8.
3. Make up any determinant that contains those two multiples such
that neither is in the same row or column as the other. Evaluate
the determinant.
= 14
Notice I put the 6 in the 3rd row 2nd column position
and the 8 in the 1st row 3rd column position, and I
evaluated the determinant as 14.
Since x=2, now all I have to do is replace the 6 by 2x and
the 8 by 4x
= 14
If you evaluate the determinant you get:
Divide through by 8
So there are two solutions,
and
You can even have more fun with that determinant because
it has a 4 and a -2. The 4 is 2x and the -2 is -x.
So you can replace them too. Like this:
= 14
However make sure you don't have x's so as to give a
third degree equation. Always leave one row or column
without any unknowns. I chose to leave the first column
without unknowns, Notice if had put an x where the
2 is in the 2nd row 1st column, it would have resulted in
a cubic equation.
But as it is, the equation comes out
So there are two solutions here, and
It's interesting that even though was the same in
both problems, the other solution was not the same as the
other solution in the first problem.
Edwin
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