Q2] b) Reduce the following matrix to its normal form and hence find its rank.

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A =$ \begin{bmatrix} 3 & -2 & 0 & 1 \\ 0 & 2 & 2 & 7 \\ 1 & -2 & -3 & 2 \\ 0 & 1 & 2 & 1 \end{bmatrix} $

$2R_3-R_1 R_1- R_3 R_1- R_2 R_2- R_4$

$ \begin{bmatrix} 1 & 2 & 6 & -3 \\ 0 & 2 & 2 & 7 \\ 1 & -2 & -3 & 2 \\ 0 & 1 & 2 & 1 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 2 & 6 & -3 \\ 0 & 2 & 2 & 7 \\ 0 & 4 & 9 & -5 \\ 0 & 1 & 2 & 1 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 0 & 4 & -10 \\ 0 & 2 & 2 & 7 \\ 0 & 4 & 9 & -5 \\ 0 & 1 & 2 & 1 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 0 & 4 & -10 \\ 0 & 1 & 0 & 6 \\ 0 & 4 & 9 & -5 \\ 0 & 1 & 2 & 1 \end{bmatrix} $

$R_2- R_4 4R_2- R_3 R_1+ \frac{2R_4 R_3}{(-9)}$

=>$ \begin{bmatrix} 1 & 0 & 4 & -10 \\ 0 & 1 & 0 & 6 \\ 0 & 4 & 9 & -5 \\ 0 & 0 & -2 & 5 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 0 & 4 & -10 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & -9 & 29 \\ 0 & 0 & -2 & 5 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & -9 & 29 \\ 0 & 0 & -2 & 5 \end{bmatrix} $ =>$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & \frac{-29}{9} \\ 0 & 0 & -2 & 5 \end{bmatrix} $

$2R_3+R_4 6C_2- C_4 \frac{29}{9}C_3+ \frac{C_4 R_4}{(-13/9)}$

$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & \frac{-29}{9} \\ 0 & 0 & 0 & \frac{-13}{9} \end{bmatrix} $=>$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{-29}{9} \\ 0 & 0 & 0 & \frac{-13}{9} \end{bmatrix} $=>$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & \frac{-13}{9} \end{bmatrix} $=>$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $

Hence the given matrix is converted to its normal form

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