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If $A=\begin{vmatrix} 1&2&-2\\ -1&3&0\\ 0&-2&1\\ \end{vmatrix}$ then find two non singular matrices P & Q. Such that PAQ is in normal form. Also find P(A) and $A^{-1}$
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$$A=\begin{vmatrix} 1&2&-2\ -1&3&0\ 0&-2&1\ \end{vmatrix}$$

It can also be written as in PAQ form

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

Now $R_2+R_1$

$\begin{vmatrix} 1&2&-2\\ 0&5&-2\\ 0&-2&1\\ \end{vmatrix}= \begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{vmatrix}A\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

Now $C_2-2C_1$

$\begin{vmatrix} 1&0&-2\\ 0&5&-2\\ 0&-2&1\\ \end{vmatrix}= \begin{vmatrix} 1&0&0\\ 1&1&0\\ 0&0&1\\ \end{vmatrix}A\begin{vmatrix} 1&-2&0\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

Now $C_3+2C_1$

$\begin{vmatrix} 1&0&0\\ 0&5&-2\\ 0&-2&1\\ \end{vmatrix}= \begin{vmatrix} 1&0&0\\ 1&1&0\\ 0&0&1\\ \end{vmatrix}A\begin{vmatrix} 1&-2&2\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

Now $R_2+2R_3$

$\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&-2&1\\ \end{vmatrix}= \begin{vmatrix} 1&0&0\\ 1&1&2\\ 0&0&1\\ \end{vmatrix}A\begin{vmatrix} 1&-2&2\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

Now $R_3+2R_2$

$\begin{vmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{vmatrix}= \begin{vmatrix} 1&0&0\\ 1&1&2\\ 2&2&5\\ \end{vmatrix}A\begin{vmatrix} 1&-2&2\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

∴ Comparing it with PAQ

We get

$P=\begin{vmatrix} 1&0&0\\1&1&2\\2&2&5\\ \end{vmatrix} $ & $Q=\begin{vmatrix} 1&-2&2\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} $

∴Rank of A i.e P(A)=No.of Non Zero rows=3

$∴P(A)=3 \\ ∵PAQ=I \\ ∴A=\dfrac 1{PQ} \\ or A^{-1}=Q.P$

$\therefore A^{-1}= \begin{vmatrix} 1&-2&2\\ 0&1&0\\ 0&0&1\\ \end{vmatrix} \times \begin{vmatrix} 1&0&0\\1&1&2\\2&2&5\\ \end{vmatrix} \\ \therefore A^{-1} = \begin{vmatrix} 3&2&6\\1&1&2\\2&2&5\\ \end{vmatrix}$

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