Matrix Identity

**Oiler** · February 25th 2011, 03:43 PM

Hey all,

Given $ B = \left( {\begin{array}{ccc} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} } \right) $ and $ B^{-1} = \left( {\begin{array}{ccc} -1 & 0 & 2 \\ 3 & 1 & -6 \\ 1 & 0 & -1 \\ \end{array} } \right) $
find C, where $BCB^{T} = I$
Thanks.

**Ackbeet** · February 25th 2011, 03:44 PM

What ideas have you had so far?

**Oiler** · February 25th 2011, 04:00 PM

$BCB^{T}=I \Rightarrow BB^{T}C = I \\$
so $C = I(BB^{T})^{-1}\\$
which is $ \[ M = \left( {\begin{array}{ccc} 11 & 3 & -2 \\ 3 & 1 & -6 \\ -21 & -6 & 41 \\ \end{array} } \right) \]$
That seems to work.. not sure if its right though...

**Ackbeet** · February 25th 2011, 04:10 PM

You might just get the right answer for the wrong reasons. *slaps your wrist*

You don't know that $B$ and $C$ commute! You can't just slip $B^{T}$ past the $C$ like that. However, you can left- and right-multiply by inverses, if you know they exist. In this case, you've been given $B^{-1}$ , so you know that exists. How about this first step:

$BCB^{T}=I$

$B^{-1}BCB^{T}=B^{-1}I=B^{-1}$

$CB^{T}=B^{-1}.$

What else could you do?

**Oiler** · February 25th 2011, 04:17 PM

(In regards to my second post - Not ackbeets)
Sorry thats wrong... I think it should be that
The product of $BC$ should give the inverse of $B^{T}$
so $B \cdot C = B^{-1} \Rightarrow C = B^{T-1} \cdot B^{-1}$
Which is

$ M = \left[ {\begin{array}{ccc} 5 & -15 & -3 \\ -15 & 46 & 9 \\ -3 & 9 & 2 \\ \end{array} } \right] $

**Ackbeet** · February 25th 2011, 04:29 PM

A few comments:

1. What is $M?$ No need to introduce another variable if you don't need it.

2. The notation $B^{T-1}$ should really be $(B^{T})^{-1}.$ Otherwise, someone might think you're raising $B$ to the power $T-1.$

3. You've got the order wrong. You need to right-multiply by $(B^{T})^{-1}.$ That is, you've got to do this (picking up from where I left off in my last post):

$CB^{T}=B^{-1}$

$CB^{T}(B^{T})^{-1}=B^{-1}(B^{T})^{-1}$

$C=B^{-1}(B^{T})^{-1}.$

How can you compute the RHS here?

**Oiler** · February 25th 2011, 04:37 PM

Ah, thanks for the pointers. RHS would just be the product of the inverses of B and its Transpose..

**Ackbeet** · February 25th 2011, 04:44 PM

Ah, but you're not directly given $(B^{T})^{-1}.$ How can you get it?

**Oiler** · February 25th 2011, 05:31 PM

Wouldn't that just be:

$ B = \left( {\begin{array}{ccc} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} } \right) $ Then $ (B^{T})^{-1} = \left( {\begin{array}{ccc} 1 & 3 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \\ \end{array} } \right)^{-1} $

**Ackbeet** · February 25th 2011, 05:38 PM

True, but I think you can get what you need faster than that, if you use an identity.

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