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:
$$\left(\begin{array}{cc} a & b \\ c & d \end{array} \right) \cdot \left(\begin{array}{cc} a & c \\ b & d \end{array}\right) = \left(\begin{array}{cc} a^2 + b^2 & a \, c + b \, d \\ a \, c + b \, d & c^2 + d^2 \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$$
You can simply make it non-orthogonal if you make the off-diagonal element to be non-zero, for example:
$$a \, c + b \, d = 1$$
Now, the determinant is:
$$a \, d - b \, c = \pm1$$
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 :tongue:

 

[Dickfore]

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