Find non-Singular Matrices P & Q such that, $A=\begin{bmatrix}1 & 2 & 3 & 4\\2 & 1 & 4 & 3\\3 & 0 & 5 & -10 \end{bmatrix} $

is reduced to normal form. Also find rank.

$A=\begin{bmatrix}1&2&3&4\\ 2&1&4&3\\ 3&0&5&-10\end{bmatrix} $

$Let\ A=I_3AI_4\\$ \begin{bmatrix}1&2&3&4\\ 2&1&4&3\\ 3&0&5&-10\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix} A \begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} $$

$By\ R_2-2R_1, R_3-2R_1\\ \begin{bmatrix}1&2&3&4\\ 0&-3&-2&-5\\ 0&-6&-4&-22\end{bmatrix} =\begin{bmatrix}1&0&0\\ -2&1&0\\ -3&0&1\end{bmatrix} A \begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} $

$$By \ C_2-2C_1, C_3-3C_1 , C_4-4C_1 \ \begin{bmatrix}1&0&0&0\ 0&-3&-2&-5\ 0&-6&-4&-22\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ -3&0&1\end{bmatrix} A \begin{bmatrix}1&-2&-3&-4\ 0&1&0&0\ 0&0&1&0\ 0&0&0&1\end{bmatrix} $ $By \ R_3-2R_2, \ \begin{bmatrix}1&0&0&0\ 0&-3&-2&-5\ 0&0&0&-12\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix} A \begin{bmatrix}1&-2&-3&-4\ 0&1&0&0\ 0&0&1&0\ 0&0&0&1\end{bmatrix} $ $$By \ - \dfrac{1}{3}C_2, - \dfrac{1}{2}C_3, - \dfrac{1}{5}C_4, \ \begin{bmatrix}1&0&0&0\ 0&1&1&1\ 0&0&0&\dfrac{12}{5}\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix} A \begin{bmatrix}1&\dfrac{2}{3}&\dfrac{3}{2} &\dfrac{4}{5} \ 0&-\dfrac{1}{3}&0&0\ 0&0&-\dfrac{1}{2}&0\ 0&0&0&-\dfrac{1}{5}\end{bmatrix} $ $By \ C_3-C_2,C_4-C_3 \ \begin{bmatrix}1&0&0&0\ 0&1&0&0\ 0&0&0&\dfrac{12}{5}\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix} A \begin{bmatrix}1&\dfrac{2}{3}&\dfrac{5}{6} &-\dfrac{7}{10} \ 0&-\dfrac{1}{3}&\dfrac{1}{3}&0 \ 0&0&-\dfrac{1}{2}&\dfrac{1}{2}\ 0&0&0&-\dfrac{1}{5}\end{bmatrix} $ $ By\ \dfrac{5}{12} C_4 \ \begin{bmatrix}1&0&0&0\ 0&1&0&0\ 0&0&0&1\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix} A \begin{bmatrix}1&\dfrac{2}{3}&\dfrac{5}{6} &-\dfrac{7}{24} \ 0&-\dfrac{1}{3}&\dfrac{1}{3}&0 \ 0&0&-\dfrac{1}{2}&\dfrac{5}{24}\ 0&0&0&-\dfrac{1}{12}\end{bmatrix} $ ---   PLEASE CHOOSE AN APPROPRIATE MODULE FOR THIS QUESTION!             $ By\ C_{34} \ \begin{bmatrix}1&0&0&0\ 0&1&0&0\ 0&0&1&0\end{bmatrix} =\begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix} A \begin{bmatrix}1&\dfrac{2}{3}&-\dfrac{7}{24} &\dfrac{5}{6} \ 0&-\dfrac{1}{3}&0&\dfrac{1}{3}\ 0&0&\dfrac{5}{24}&-\dfrac{1}{2}\ 0&0&-\dfrac{1}{12}&0\end{bmatrix} $ $=\begin{bmatrix}I_3&0\end{bmatrix}=PAQ \ is\ in \ normal \ form.

\therefore Rank\ of\ matrix\ is \ 3\

\ hence\ P= \begin{bmatrix}1&0&0\ -2&1&0\ 1&-2&1\end{bmatrix}, Q= \begin{bmatrix}1&\dfrac{2}{3}&-\dfrac{7}{24} &\dfrac{5}{6} \ 0&-\dfrac{1}{3}&0&\dfrac{1}{3}\ 0&0&\dfrac{5}{24}&-\dfrac{1}{2}\ 0&0&-\dfrac{1}{12}&0\end{bmatrix}

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