Question on Matrices: find det(B^T)

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just one last question on matrices, if you don't mind...

Question:

B is a 3*3 matrix det(B)= -3

find det(B^T)

(B^T is B transpose)

My Answer:

have none!

help would be greatly appreciated
 
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What is the relationship between the 2 determinates we are looking at here? Does 'transposing' the matrix effect the determinate? if so, how?
 
Well, to derive it, consider a general 3x3 matrix \left(\begin{array}{ccc}<br /> a&amp;b&amp;c\\d&amp;e&amp;f\\g&amp;h&amp;i \end{array}\right) and expand the determinant

<br /> \left|\begin{array}{ccc}<br /> a&amp;b&amp;c\\d&amp;e&amp;f\\g&amp;h&amp;i \end{array}\right|=<br /> a\left|\begin{array}{cc}e&amp;f\\h&amp;i\end{array}\right| - b\left|\begin{array}{cc}d&amp;f\\g&amp;i\end{array}\right|+c\left|\begin{array}{cc}d&amp;e\\g&amp;h\end{array}\right|=\cdots

Then consider the transposed matrix \left(\begin{array}{ccc}<br /> a&amp;d&amp;g\\b&amp;e&amp;h\\c&amp;f&amp;i \end{array}\right) and expand this in a similar way. Compare the two results.
 
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it's the same!
so the determinant of det(B^T) =det(B)=-3
 
sara_87 said:
it's the same!
so the determinant of det(B^T) =det(B)=-3

Correct. And in reply to your other thread, happy new year to you too!
 
thanx, my new years resolution is not to leave 200 questions till the last minute! it's nearly 2 am I'm going to finish off these ten questions...and go to sleep!
 
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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