Find two matrices P,Q such that PAQ is in normal form?

Matrix A is given as $$\pmatrix{2 & 1 & -3 & -6 \\ 3 & -3 & 1 & 2 \\ 1 & 1 & 1 & 2 \\}$$ To find 2 matrices P and Q such that P.A.Q is in normal form. the method that must be employed is equating A = P.A.Q -->eqn 1 where initially P as identity matrix is set to order 3x3 and Q as identity is set to order 4x4 depending upon the order of matrix A.It seems this is the correct method.But according to Mathematica matrices with orders 3x3 and 3x4 cannot be multiplied.But according to the method given in "eqn 1" above the a on the LHS has to be converted to Echelon form(using row transformations only) and the same steps have to be applied to P on the RHS,keeping A(on RHS side) and Q as it is.After the echelon form is obtained,the value of P is also obtained.Now to obtain Q,we have to further convert the A on LHS side to Normal form (applying column transformations only) at the same time these transformations are to be applied to Q on RHS while keeping P and A(RHS side) as it is.After A on LHS is converted to Normal form we get the value of Q and our problem is complete.

Now my question is "Is this method correct ?" and if yes, then "Is this answers correct which i obtained following the same method ? the answers is $$p=\pmatrix{1 & 0 & 0 \\ 0 & 1 & -3 \\ 6 & 1 & 9 \\}$$ $$Q=\pmatrix{1\over 2& 1\over 2& 5\over 56& 0 \\ 0& -1\over 6& -1\over 84& 0 \\ 0& 0& 1\over 28& -2 \\ 0& 0& 0& 1 \\}$$ If this method is wrong, then please be show me the correct method or point to a webpage that has the correct method.(i could not solve this directly in Mathematica you can additionally to do that)

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1. I find it hard to believe that Mathematica says $3$-by-$3$ and $3$-by-$4$ matrices can't be multiplied. 2. What do you mean by "normal form? –  Gerry Myerson Nov 29 '12 at 4:34
when i enter the three matrices and try to take the product Mathematica says –  Joao Nov 29 '12 at 6:27
when i enter the three matrices and try to take the product Mathematica says "Thread::tdlen: Objects of unequal length in {{1,0,0},{0,1,-3},{6,1,9}} {{2,1,-3,-6},<<1>>,{1,1,1,2}} {{1/2,1/2,5/56,0},{0,-(1/6),-(1/84),0},{0,0,1/28,-2},{0,0,0,1}} cannot be combined. >>" foll matrices are said to be in Normal form [(100),(010),(001)] OR [(1000),(0100),(0010),(0001)] OR [(10),(01)] Definition:A matrix A of order "n" and rank "r" is said to be in Normal form if it looks like one of the below matrices [Ir] OR [(Ir,0),(0,0)] OR [ir o] etc –  Joao Nov 29 '12 at 6:35
In above comment Ir represents identity matrix of order "r" –  Joao Nov 29 '12 at 6:38
I don't know what you mean by {{1,0,0},{0,1,-3},{6,1,9}}, nor what syntax Mathematica is expecting. Your <<1>> looks suspicious - is Mathematica supposed to know that stands for the row $3\ -3\ 1\ 2$? Then you use some completely different conventions in defining normal form. Is it the same as what's often called reduced row-echelon form? –  Gerry Myerson Nov 29 '12 at 22:33

1 Answer

The only thing to be kept in mind regarding the normal form of matrices is that the matrix [A](m x n) is to be converted in the form: [A](m x n) = [P](m x m) [A](m x n) [Q](n x n)
Starting from [A](m x n) = [I](m x m) [A](m x n) [I](n x n), all the elementary row transformations should be performed on [I](m x m) and all the elementary column transformations should be performed on [I](n x n). [I](m x m) means identity matrix of order (m x m) The correct normal form for the given matrix needs to be verified because I think that the answer is incorrect (Multiply your matrices to see.) The correct answer is:

P = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -\frac{1}{2} & \frac{3}{2}\\ \frac{1}{14} & -\frac{5}{28} & \frac{11}{28} \end{bmatrix}

Q = \begin{bmatrix} 1 & -1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & -3 & -2 \\ 0 & 0 & 0 & 1 \end{bmatrix}

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Please use LaTeX to format this answer. –  Alizter Oct 25 '14 at 20:51