Solving a 2x2 Matrix: Can You Help?

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The discussion revolves around solving a 2x2 matrix equation involving variables x and y. Participants initially attempt to derive equations from the matrix and suggest potential solutions, but inconsistencies arise when checking the equations. The equations x + y = 4 and y - x = 6 are established, but solutions provided do not satisfy all four necessary equations. Ultimately, it is concluded that either there is no solution or a mistake in the problem statement, as the derived values do not consistently satisfy all equations. The conversation emphasizes the importance of ensuring all matrix elements are consistent for a valid solution.
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solve for x and y

|x y|
|-y x|

minus

|-y x|
|x y|
equals

|4 6|
|-4 6| can someone give me a hint?
 
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wat2000 said:
solve for x and y

|x y|
|-y x|

minus

|-y x|
|x y|
equals

|4 6|
|-4 6| can someone give me a hint?

Think about what it means for two matrices to be equal to each other.
 
x+y=4 => y=4-x, then (4-x)-x=6
4-2x=6 => -2x=6/4=3/2 => x=-3/4
so y=4-x=4-(-3/4)=4+3/4=16/4+3/4=19/4
would this be the answer?
 
You are on the right track with that, however be careful when you are solving the actual equations.

"4-2x=6 => -2x=6/4"

That step isn't quite correct.

Another thing that you will need to check is that the calculated values of x and y work for all 4 equations that you could write. Have you missed out a minus sign by any chance on one of the 6's? The reason i ask is because we can write the following two equations:

y - x = 6
x - y = 6

No value of x and y will work with these
 
I tried something different and got x=-1 and y= 5 is that right.
 
That's not what I get. You should get four equations in two unknowns.
Can you confirm that what you wrote is exactly what the problem is?
 
wat2000 said:
I tried something different and got x=-1 and y= 5 is that right.

Yes they would solve the first two equations, however they still do not solve the second two. You need all elements of the matrix to be consistent, so you need solutions to all four equations:

x+y = 4
y-x = 6

You solved those two, but the solution needs to also solve the remaining two:

-y-x = -4
x-y = 6

If you substitute x=-1 and y=5 into the top equation then you get -4 which is ok, but the second equation gives -1-5 = -6 which is inconsistent. Either there is no solution to the whole problem or you have mis-typed the 6 instead of a -6 perhaps?
 
x-(-y)=4
y-x=6
-y-x=-4
x-y=6

4-2x=6
-2x=6-4
-2x=2
x=-1

since y=4-x, then y=4-(-1)
4+1=5
y=5
check: |-1 -5|
|-5-1| minus

|-5 -1|
|-1 5| equals

|4 6|
|-4 6|
Thats how I got it.
 
<br /> <br /> \left[ {\begin{array}{*{20}c}<br /> { - 1} &amp; 5 \\<br /> { - 5} &amp; { - 1} \\<br /> \end{array}} \right] - \left[ {\begin{array}{*{20}c}<br /> { - 5} &amp; { - 1} \\<br /> { - 1} &amp; 5 \\<br /> \end{array}} \right] = \left[ {\begin{array}{*{20}c}<br /> 4 &amp; 6 \\<br /> { - 4} &amp; { - 6} \\<br /> \end{array}} \right] \ne \left[ {\begin{array}{*{20}c}<br /> 4 &amp; 6 \\<br /> { - 4} &amp; 6 \\<br /> \end{array}} \right]
 
  • #10
Where did I mess up?
 
  • #11
Is the question you wrote definitely correct?
 
  • #12
yes. is there just no solution?
 
  • #13
None that i can see.
 
  • #14
ok that's what I thought. Thanks for your help.
 
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