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

[pyroknife]

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
?

 

[Dick]

√1 is 1. Try thinking of matrices with nonzero off-diagonal elements.

 

[Dickfore]

A 2x2 orthogonal matrix must satisfy:
<br /> \left(\begin{array}{cc}<br /> a &amp; b \\<br /> <br /> c &amp; d<br /> \end{array} \right) \cdot \left(\begin{array}{cc}<br /> a &amp; c \\<br /> <br /> b &amp; d<br /> \end{array}\right) = \left(\begin{array}{cc}<br /> a^2 + b^2 &amp; a \, c + b \, d \\<br /> <br /> a \, c + b \, d &amp; c^2 + d^2<br /> \end{array}\right) = \left(\begin{array}{cc}<br /> 1 &amp; 0 \\<br /> <br /> 0 &amp; 1<br /> \end{array}\right)<br />
You can simply make it non-orthogonal if you make the off-diagonal element to be non-zero, for example:
<br /> a \, c + b \, d = 1<br />
Now, the determinant is:
<br /> a \, d - b \, c = \pm1<br />
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.

 

[Dick]

You can write down a lot of simple examples without thinking half that hard.

 

[pyroknife]

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

 

[Dickfore]

You can write down a lot of simple examples without thinking half that hard.

yes, you and I can. Please unquote the solution.

 

[Dickfore]

can I have a matrix that's simply
2 1
1 1

yes.

 

[pyroknife]

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

 

[Dickfore]

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.

 

[Mentallic]

Oh will, I guess I was thinking too hard. Usually when they give ±, I assume that both conditions have to be met.

The \pm operator is called plus or minus, not plus and minus

 

[Dickfore]

and imposing the conditions
<br /> x = 1 \wedge x = -1<br />
leads to a contradiction
<br /> 1 = -1<br />
so, that interpretation doesn't make sense.