My Math Forum Can anyone enlighten me yet?

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 February 22nd, 2009, 11:44 AM #1 Newbie   Joined: Feb 2009 Posts: 11 Thanks: 0 Can anyone enlighten me yet? Is anyone able to help me understand this determinant problem yet? I understand 2nd degree and I thought I understood how to cross multiply 3rd degree ones but with there being letters involved, it stumped me. Is all you do is look at the first determinant and then because the second determinant has a row interchange, then all you do is deal with the numbers of the second one and just change the sign of the answer, because of the interchange? I know the answer is "6" but I just want to know for sure how it is worked. Thanks for any understanding. "Help with 3rd order Determinants by JudaStar » Fri Feb 20, 2009 9:49 pm I have gone over and over 3rd order determinants and their properties in my College Algebra book and I cannot understand. I understand 2nd order ones. Here is the determinant problem. It's going to be hard to post it without the vertical straight lines but I'm going to try. I will use enclosures instead...just pretend they are the vertical straight lines. I don't even understand how the 1st one equals 3 much less what the second one equals. The problem says: "Use properties of determinants to find the value of the second determinant, if the value of the first is known." [x y z] [u v w] [u v w]=3 [2 -6 -2]=? [1 -3 -1] [x y z]JudaStar It would not post the way I had it with the determinants apart side by side. It did not separate them so I am going to post one on top and the other one under it. In the book they are side by side. [x y z] [u v w]=3 [1 -3 -1] [x y z] [2 -6 -2]=? [u v w]
 March 5th, 2009, 02:03 PM #2 Senior Member   Joined: Feb 2009 Posts: 172 Thanks: 5 Re: Can anyone enlighten me yet? When you exchanged Line 2 and 3 the new determinant is the previous multiplied by -1 and when you multiplied the new Line 3 (Old Line 2) by 2 your determinant is multplied by 2. Hence the new determinant is -6.

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