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Either my brain is seriously fried up right now or the computer is wrong.

If I have a matrix $\begin{bmatrix} 4 & -2\\ 2 & -1 \\ 0 & 0 \end{bmatrix}$ multiply by its transpose $\begin{bmatrix} 4 &2 &0 \\ -2&-1 &0 \end{bmatrix}$, I should get a $3 \times 3$ $\begin{bmatrix} 20 &10 &0 \\ 10 &5 &0 \\ 0&0 &0 \end{bmatrix}$

For some reason Maple is giving me a $2 \times 2$ $\begin{bmatrix} 20 &10 \\ 10 &5 \\ \end{bmatrix}$

Why did they delete the last row and column of 0s? You can't do that

Arturo Magidin
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Hed
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    This is maybe just a Maple convention: you should change your title to indicate that it's something happening in Maple. Also, don't assume your questions are too dumb for this this forum :) It's not like MathOverflow... – William Apr 10 '12 at 02:35
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    Are you doing $A A^T$ or $A^T A$? – lhf Apr 10 '12 at 02:46
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    Never mind, I am the one who is wrong...I multiplied it the other way – Hed Apr 10 '12 at 02:46
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    Can you edit the question so that its title is useful in some way? – Mariano Suárez-Álvarez Apr 10 '12 at 02:52
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    Note, though, that if you multiplied them the other way, you should have gotten $$\left(\begin{array}{rr}20&-10\-10&5\end{array}\right)$$and not what you say you got. – Arturo Magidin Apr 10 '12 at 03:06

2 Answers2

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You have multiplied the second matrix with the first matrix. Sice matrix multiplication is not commute the answer is wrong. And you have also ignored the negative signs. They should not be ignored.

chndn
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0

Do you try the result with one row of os or try some square matrix with one row of os?

Maybe is the storage of maple?

Jimmy Wang
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