Linear Algebra: The transpose of A equals Inverse A, so

jinksys
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If the transpose of A equals the Inverse of A, then det(A)=1.

False. However, I don't follow the logic.

If transA=InverseA, doesn't that mean the matrix is the identity matrix?

The explanation says that det(A)= 1 and -1.
 
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jinksys said:
If the transpose of A equals the Inverse of A, then det(A)=1.

False. However, I don't follow the logic.

If transA=InverseA, doesn't that mean the matrix is the identity matrix?

The explanation says that det(A)= 1 and -1.

If the transpose of A is the inverse of A, it does not have to be the identity matrix. All it says is that the columns of A are an orthonormal basis, as are the rows (check this by matrix multiplication).

Examples are rotation matrices, and reflection matrices (try constructing some 2x2 example to be sure).
 
Try looking at a simple case. How about 1x1 matrices? There is only one unknown -- write down the equations that define what it means for the transpose to equal the inverse.
 
If you start with det(AA^-1)=det(I) and consider that det(A^T)=det(A) you should be able to work this out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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