\[ \text {if} A=\begin{bmatrix} 1 &2 &-2 \\-1 &3 &0 \\0 &-2 &1 \end{bmatrix} \] then find two non-singular matrices P&Q such that PAQ is in normal form also find ρ(A) and A-1.

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$A= \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{bmatrix} \\ A= I_3AI_3 \; i.e. \\ \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ By \; R_2 + R_1 \to \begin{bmatrix} 1 & 2 & -2 \\ 0 & 5 & -2 \\ 0 & -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ By C_2 - 2C_1 \to \begin{bmatrix} 1 & 0 & -2 \\ 0 & 5 & -2 \\ 0 & -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ By \; C_3 + 2C_1 \to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & -2 \\ 0 & -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & -2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ By \; C_2 + 2C_3 \to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{bmatrix} \\ By \; R_2 + 2R_3 \to \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 0 & 0 & 1 \\ \end{bmatrix} A \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{bmatrix} \\ \text{which is in normal form} \\ \therefore Rank \; of \; A \; = Q(A)= \; 3 \\ \text{To find A^{-1} we see that,} \\ I = PAQ \\ \therefore AQ= P^{-1} \\ \therefore A^{-1}AQ= A^{-1}P^{-1} \\ \therefore Q= A^{-1}P^{-1} \\ \therefore QP= A^{-1}P^{-1}P = A^{-1} \\ Now, \; QP = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 3 & 2 & 6 \\ 1 &1 & 2 \\ 2 & 2 & 5 \\ \end{bmatrix} \\ \therefore A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 &1 & 2 \\ 2 & 2 & 5 \\ \end{bmatrix}$

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