Compute the inverse of a matrix over QQ

Compute the inverses of the following two matrices over \mathbb{Q}.

A = \left[ \begin{array}{ccc} \text{-}7 & \text{-}1 & \text{-}4 \\ 7 & 1 & 3 \\ 1 & 0 & 0 \end{array} \right]
B = \left[ \begin{array}{cccc} 1 & 1 & 0 & 2 \\ 0 & 2 & 1 & \text{-}1 \\ 0 & 2 & 0 & 0 \\ \text{-}1 & 1 & 1 & 0 \end{array} \right]

We will use the strategy suggested by the previous exercise.

Evidently, the reduced row echelon form (over \mathbb{Q}) of [A|I] is \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 & 4 & \text{-}7 \\ 0 & 0 & 1 & \text{-}1 & \text{-}1 & 0 \end{array} \right]. Thus A^{-1} = \left[ \begin{array}{ccc} 0 & 0 & 1 \\ 3 & 4 & \text{-}7 \\ \text{-}1 & \text{-}1 & 0 \end{array} \right].

Likewise, the reduced row echelon form (over \mathbb{Q}) of [B|I] is \left[ \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 1/3 & 2/3 & \text{-}1/2 & \text{-}2/3 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1/2 & 0 \\ 0 & 0 & 1 & 0 & 1/3 & 2/3 & \text{-}1 & 1/3 \\ 0 & 0 & 0 & 1 & 1/3 & \text{-}1/3 & 0 & 1/3 \end{array} \right]. Thus B^{-1} = \left[ \begin{array}{cccc} 1/3 & 2/3 & \text{-}1/3 & \text{-}2/3 \\ 0 & 0 & 1/2 & 0 \\ 1/3 & 2/3 & \text{-}1 & 1/3 \\ 1/3 & \text{-}1/3 & 0 & 1/3 \end{array} \right].

Post a comment or leave a trackback: Trackback URL.

Comments

  • mcoulont  On October 6, 2011 at 6:58 am

    Your B^(-1) seems false (at postion 1;3, it’s -1/2, no ?).
    And your B isn’t exactly like in D&F (although you seem to do the exercise with the original B).

    Greetings.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: