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)