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Give a 2x2 matrix with Det(A)=+/-1 that is not orthogonal.

  • Thread starter pyroknife
  • Start date
  • #1
613
3
I attached the problem.
I only need help with part b), I provided part a) just to remind you guys what a orthogonal matrix was.

The only 2x2 matrix I can think of with a determinant of + or - 1 is something like

√1 0
0 1

The determinant of this would be √1 = + or - 1

The second part asks me to show that this matrix is not orthogonal thus A^-1≠A^T

A^T=
√1 0
0 1


To calculate A^-1 I set up the 2x2 matrix with the identify matrix on the right and reduce so the left becomes the I matrix.
√1 0 1 0
0 1 0 1


1 0 1/√1 0
0 1 0 1


But when I do this I get A^-1 =
1/√1 0
0 1


Is this still considered = to A^T =
√1 0
0 1
?
 

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Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
√1 is 1. Try thinking of matrices with nonzero off-diagonal elements.
 
  • #3
2,967
5
A 2x2 orthogonal matrix must satisfy:
[tex]
\left(\begin{array}{cc}
a & b \\

c & d
\end{array} \right) \cdot \left(\begin{array}{cc}
a & c \\

b & d
\end{array}\right) = \left(\begin{array}{cc}
a^2 + b^2 & a \, c + b \, d \\

a \, c + b \, d & c^2 + d^2
\end{array}\right) = \left(\begin{array}{cc}
1 & 0 \\

0 & 1
\end{array}\right)
[/tex]
You can simply make it non-orthogonal if you make the off-diagonal element to be non-zero, for example:
[tex]
a \, c + b \, d = 1
[/tex]
Now, the determinant is:
[tex]
a \, d - b \, c = \pm1
[/tex]
Solve these two equations for a, and b, for example, and you will get a two-parameter family of matrices that are non-orthogonal, but with determinant ±1.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
You can write down a lot of simple examples without thinking half that hard.
 
Last edited:
  • #5
613
3
I'm not sure if I misinterpreted this problem when it says Det(A)=±1, does that mean the + and - have to be both satisfied?

or can I have a matrix that's simply
2 1
1 1
 
  • #6
2,967
5
You can write down a lot of simple examples without thinking half that hard.
yes, you and I can. Please unquote the solution.
 
  • #7
2,967
5
  • #8
613
3
Oh will, I guess I was thinking too hard. Usually when they give ±, I assume that both conditions have to be met.

Can someone verify if I proved part a right? I'm second guessing myself.

Det(A^-1)=Det(A^t)
Since Det(A^t)=Det(A)
Det(A)=Det(A^-1)
Det(A^-1)=1/Det(A)
Thus
Det(A)=1/Det(A)
(Det(A))^2=1
Det(A)=±1
 
  • #9
2,967
5
Oh will, I guess I was thinking too hard. Usually when they give ±, I assume that both conditions have to be met.

Can someone verify if I proved part a right? I'm second guessing myself.

Det(A^-1)=Det(A^t)
Since Det(A^t)=Det(A)
Det(A)=Det(A^-1)
Det(A^-1)=1/Det(A)
Thus
Det(A)=1/Det(A)
(Det(A))^2=1
Det(A)=±1
yes, it is correct.
 
  • #10
Mentallic
Homework Helper
3,798
94
Oh will, I guess I was thinking too hard. Usually when they give ±, I assume that both conditions have to be met.
The [itex]\pm[/itex] operator is called plus or minus, not plus and minus :tongue:
 
  • #11
2,967
5
and imposing the conditions
[tex]
x = 1 \wedge x = -1
[/tex]
leads to a contradiction
[tex]
1 = -1
[/tex]
so, that interpretation doesn't make sense.
 

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