Exercise 2.5.23

Question

Follow the 3 by 3 text example but with plus signs in $A$. Eliminate above and below the pivots to reduce $\left[\boldsymbol{A}\boldsymbol{I}ight]$ to $\left[\boldsymbol{I}\boldsymbol{A}^{-1}ight]$:
$$ \left[\boldsymbol{A}\boldsymbol{I}ight] =\left[\begin{array}{rrrrrr}
2 & 1 & 0 & 1 & 0 & 0 \
1 & 2 & 1 & 0 & 1 & 0 \
0 & 1 & 2 & 0 & 0 & 1
\end{array}ight].$$

Prakash Anand · Accepted Answer

We have
$$
\begin{array}{llll}
\left[\boldsymbol{A}\boldsymbol{I}\right] &=\left[\begin{array}{rrr|rrr}
2 & 1 & 0 & 1 & 0 & 0 \
1 & 2 & 1 & 0 & 1 & 0 \
0 & 1 & 2 & 0 & 0 & 1
\end{array}\right]\[10pt]

&\rightarrow\left[\begin{array}{rrr|rrr}
2 & 1 & 0 & 1 & 0 & 0 \
0 & 3 / 2 & 1 & -1 / 2 & 1 & 0 \
0 & 1 & 2 & 0 & 0 & 1
\end{array}\right]\[10pt]

&\rightarrow\left[\begin{array}{rrr|rrr}
2 & 1 & 0     &   1 & 0 & 0 \
0 & 3 / 2 & 1 &  -1 / 2 & 1 & 0 \
0 & 0 & 4 / 3 &  1 / 3 & -2 / 3 & 1
\end{array}\right]\[10pt]

&\rightarrow \left[\begin{array}{rrr|rrr}
2 & 1 & 0     &   1 & 0 & 0 \
0 & 3 / 2 & 0 & -3 / 4 & 3 / 2 & -3 / 4 \
0 & 0 & 4 / 3 & 1 / 3 & -2 / 3 & 1
\end{array}\right]\[10pt]

&\rightarrow \left[\begin{array}{rrr|rrr}
2 & 0 & 0 & 3 / 2 & -1 & 1 / 2 \
0 & 3 / 2 & 0 & -3 / 4 & 3 / 2 & -3 / 4 \
0 & 0 & 4 / 3 & 1 / 3 & -2 / 3 & 1
\end{array}\right]\[10pt]

&\rightarrow\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 3 / 4 & -1 / 2 & 1 / 4 \
0 & 1 & 0 & -1 / 2 & 1 & -1 / 2 \
0 & 0 & 1 & 1 / 4 & -1 / 2 & 3 / 4
\end{array}\right]\[10pt]

&=\left[\boldsymbol{I}\boldsymbol{A}^{-1}\right].
\end{array}
$$

Exercise 2.5.23

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Exercise 2.5.23

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