Solving a 2x2 Matrix: Can You Help?

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Homework Help Overview

The discussion revolves around solving a system of equations derived from a 2x2 matrix subtraction problem, where participants are tasked with finding the values of x and y that satisfy the resulting equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various approaches to solving the equations, with some attempting to derive values for x and y directly from the matrix equations. Questions arise regarding the correctness of steps taken and the consistency of the equations derived from the matrix setup.

Discussion Status

The discussion is ongoing, with participants providing hints and feedback on each other's attempts. Some guidance has been offered regarding the need to check all equations for consistency, and there is an acknowledgment of potential errors in the initial problem setup or calculations.

Contextual Notes

Participants note the importance of ensuring that all elements of the matrix are consistent and question whether there may have been a typographical error in the original problem statement, particularly concerning the signs of the constants in the equations.

wat2000
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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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