# How to find the inverse from this?

One problem I stumbled upon sounds like this:

$$P=\left[ \begin{array}{ c c } 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{array} \right]$$

B=[A|P]

If

$$B=\left[ \begin{array}{ c c } 1&0&0&x_1&x_2&x_3 \\ 0&1&0&x_4&x_5&x_6 \\ 0&0&1&x_7&x_8&x_9 \end{array} \right]$$

1. What is $A^{-1}$?
2. What do the last three columns of B represent?

Thank you.

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What does $\left[\,A\vert P\,\right]$ mean? – user7530 Nov 27 '11 at 11:19
Hint: look at $\mathbf P^\top\mathbf P$. – J. M. Nov 27 '11 at 11:19
@7530: it's a block matrix. – J. M. Nov 27 '11 at 11:20
I recently wrote up something related to this. Hope it helps. – user13838 Nov 27 '11 at 11:26
@J.M. I must still be missing something... if B is just a block matrix then clearly $A=I$, the right columns are the columns of $P$ and there's nothing to do in this problem? – user7530 Nov 27 '11 at 11:38

Note that $P$ is an elementary matrix. Moreover:

1) Multiplication of $A$ by $P$ on the right interchanges columns one and two of $A$.

2) Multiplication of $A$ by $P$ on the left interchanges rows one and two of $A$.

3) $P^{-1}=P$.

Now suppose $B=[\,A|P\,]$ and that

$B$ is similar to $\left[ \begin{array}{ c c } 1&0&0&x_1&x_2&x_3 \\ 0&1&0&x_4&x_5&x_6 \\ 0&0&1&x_7&x_8&x_9 \end{array} \right]$ (I assume this is what the OP meant).

Then $[AP|PP]=[\,AP\,|I\,\,]$ is similar to

$\left[ \begin{array}{ c c } 0&1&0&x_2&x_1&x_3 \\ 1&0&0&x_5&x_4&x_6 \\ 0&0&1&x_8&x_7&x_9 \end{array} \right]\sim \left[ \begin{array}{ c c } 1&0&0&x_5&x_4&x_6 \\ 0&1&0&x_2&x_1&x_3 \\ 0&0&1&x_8&x_7&x_9 \end{array} \right]$

So $(AP)^{-1} =\left[ \matrix{ x_5&x_4&x_6 \\ x_2&x_1&x_3 \\ x_8&x_7&x_9 }\right]$.

Now $PA^{-1}=P^{-1}A^{-1}=(AP)^{-1}$; whence $A^{-1}=P(AP)^{-1}$, so: $$A^{-1} =P\left[ \matrix{ x_5&x_4&x_6 \\ x_2&x_1&x_3 \\ x_8&x_7&x_9 }\right]= \left[ \matrix{ x_2&x_1&x_3 \\ x_5&x_4&x_6 \\ x_8&x_7&x_9 }\right].$$

The last three columns of $B$ give the matrix $P$, which is described above.

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If I understand your question correctly, you are given that $[A|P]\sim[I|C]$, i.e. $[I|C]$ can be obtained from $[A|P]$ using row operations. The matrices $P$ and $C$ are given and you want to know what $A^{-1}$ is.

Performing a row operation is the same thing as multiplying both matrices by an elementary matrix from the left - see e.g. this question: elementary matrices and row operations.

So basically you know that $$[I|C]=[EA|EP]$$ where $E=E_n E_{n-1}\dots E_1$ is the product of the elementary matrices used in the row reduction.

Thus you know that $EA=I$, which means $E=A^{-1}$, i.e. $E$ is the matrix you want to get.

So you are asked to find $E$ such that $EP=C$. This means that $E=CP^{-1}$.

For the given matrix $P$ it's even simpler. You can notice that $PP=I$. This means that $P^{-1}=P$ and $E=CP$.

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