On http://people.revoledu.com/kardi/tutorial/LinearAlgebra/VectorCrossProduct.html, the equation in the property starting \"The magnitude of vector cross...\" is missing norm-bars. In other words, a x b => ||a x b||. There is also a mysterious \'m\' or \'n\' in the RHS of the equation.

Thanks Mark for the report. The vector n is the cross product result explained in the figure right below the properties. To make it clearer, I fixed the equations that you have pointed out. Thanks also for the tips on Sphinx.

Page: http://people.revoledu.com/kardi/tutorial/LinearAlgebra/VectorTripleCrossProduct.html Properties: \"Similarly, ...\" (and the preceding property) -- use either (a^T c) or (a . c), but not (a^T . c). Also, have you considered converting this website to use Sphinx? It makes fantastic websites, and allows you to write your math inline in the rst source files as LaTeX formulas. -- http://sphinx.pocoo.org/

I use the SVD decomposition in order to inverse my matrix but it dont work. If i put my matrix in your web site ... it works ... How do you do to inverse a matyrix with the svd decomposition... Thanks a Lot... Your site is a beatiful land !!!

I AM DOING SOME RESEARCH IN COMBINATORICS. IF I HAVE A 7BY7 MATRIX WITH 3 DOWN THE MAIN DIAGONAL AND 1 EVERYWHERE ELSE. THE SQUARE ROOT WILL BE A 7BY7 SYMMETRIC MATRIX WITH THREE 1'S IN EACH ROW AND COLUMN.ALSO THE DIRECT PRODUCT OF EACH PAIR OF ROWS AND EACH PAIROF COLUMNS WILL EQUAL 1.IF I CAN SEE HOW TO DO THIS SIMPLEST OF ALL EXAMPLES THEN I SHOULD BE ABLE TO WORK ON AN 81 BY 81 MATRIX WITH 16 DOWN THE MAIN DIAGONAL AND 3 EVERWHERE ELSE.I WOULD BE LOOKING FOR THE SQUARE ROOT OF THIS MATRIX WHICH WOULD HAVE SIXTEEN 1'S IN EACH ROW AND COLUMN. ALSO THE DIRECT PRODUCT OF EACH PAIR OF ROWS AND COLUMNS WOULD BE THREE.THIS MATRIX WOULD ALSO HAVE TO SYMMETRIC. NOTE;WE NEED A=ATRANSPOSE TO MAKE THIS WORK.

I'm really happy and greatful because this site has help improve my maths

I tried this matrix: 2,1,4,1,0,0; 3,5,1,0,1,0; 2,0,6,0,0,1; 0,0,0,0,0,0; 0,0,0,0,0,0; 0,0,0,0,0,0; mistake happens at the end: Add: R(2) = R(2) + 1_3/7 * R(3) : 1,0,0,7_1/2,-1_1/2,-4_3/4; 0,1,0,-3,1,2_1/2; --> mistake at (-3) supposed to be (-4) 0,0,1,-2_1/2,1/2,1_3/4; 0,0,0,0,0,0; 0,0,0,0,0,0; 0,0,0,0,0,0;