|a-b b-c| = |2 2 | |c-d d-a| |-6 2| My attempt: This leads to (a-b = 2), (b-c =

Question

|a-b b-c| = |2 2 |

|c-d d-a| |-6 2|

My attempt:

This leads to (a-b = 2), (b-c = 2), (c-d = -6),(d-a = 2). But this is where I get stuck. I know the method to find the values of the variables when there are three of them; but when there are more than that the same method does not work. The method that I tried (the one for 3 variables) is to solve for one variable in terms of another, then substitute that into another one of the equations.

So, for example: from the first equation we get: (a = b + 2). The right hand side of that can be substituted in where ever we see "a" in another equation, such as the fourth one. Which gives: d-(b + 2) = 2. And we continue like this until we get the values of all the variables. Unfortunately, this method does not seem to work here. Any idea what to do?

How would I do that?

Explanation / Answer

Yes, adding "a- b= 2" and "b- c= 2" gives a- c= 4. Adding that to "c- d= -6", a- d= -2. Finally, adding that to "d- a= 2", both d and a cancel leaving 0= 0, an equation that is true for all a, b, c, and d. That is, this system of equations has an infinte number of equation. You can say a little bit more. Taking "d" to be any number, both "a- d= -2" and "d- a= 2" gives a= d- 2. Then "a- c= 4" becomes (d- 2)- c= 4 so that c= d- 6. And "a- b= 2" becomes d- 2- b= 2 so b= d- 4. That is the "solution set" for this system of equations is one dimensional. Using "t" as a parameter, all solutions are given by a= t- 2, b= t- 4, c= t- 6, d= t.

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