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\textbf{Exercises on multiplication and inverse matrices}
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\begin{problem} Add $AB$ to $AC$ and compare with $A(B+C):$

\vspace{10pt}

$$A = \twobytwo
1 2 
3 4 \hspace{10pt} B = \twobytwo
1 0 
0 0 
\hspace{10pt} C = \twobytwo
0 0 
5 6 $$

\end{problem}

\solution{ We first add $AB$ to $AC:$

$$AB = \twobytwo 1 2 3 4 \twobytwo 1 0 0 0 = \twobytwo 1 0 3 0, \hspace{10pt} AC = \twobytwo 1 2 3 4 \twobytwo 0 0 5 6 = \twobytwo {10} {12} {20} {24} $$

$$\longrightarrow AB+AC = \twobytwo 1 0 3 0 +  \twobytwo {10} {12} {20} {24} = \twobytwo {11} {12} {23} {24} .$$

\medskip

\noindent We then compute $A(B+C):$

$$B+C = \twobytwo 1 0 0 0 + \twobytwo 0 0 5 6 = \twobytwo 1 0 5 6$$

$$\longrightarrow A(B+C) = \twobytwo 1 2 3 4 \twobytwo 1 0 5 6 = \twobytwo  {11} {12} {23} {24}  = AB + AC.$$

\noindent Therefore, $AB + AC = A(B+C).$
}

\begin{problem} 
(2.5 \#24.  {\em Introduction to Linear Algebra:} Strang) Use Gauss-Jordan elimination on $[U \hspace{6pt}I]$ to find the upper triangular $U^{-1}:$

\vspace{10pt}

$$UU^{-1} = I \hspace{6pt} \threebythree 
1 a b
0 1 c
0 0 1
\threebythree
0 0 0 
{x_1} {x_2} {x_3}
0 0 0 
= \threebythree  
1 0 0 
0 1 0 
0 0 1 
.$$
 
\end{problem}

\solution{
Row reduce $[U\hspace{6pt} I]$ to get $[I\hspace{6pt} U^{-1}]$ as follows (here, $R_i$ = row $i$)

$$\left [ \begin{array}{rrrrrrr}
1 & a & b & & 1 & 0 & 0\\
0 & 1 & c & & 0 & 1 & 0\\
0 & 0 & 1 & & 0 & 0&1 \end{array} \right ]  
\longrightarrow \begin{array}{c}
(R_1 = R_1 - a R_2)\\
(R_2 = R_2 - c R_2 )\\ 
\\
\end{array}
\left [ \begin{array}{rrrrrrr}
1 & 0 & b-ac & & 1 & -a & 0\\
0 & 1 & 0 & & 0 & 1 & -c\\
0 & 0 & 1 & & 0 & 0&1 \end{array} \right ]  
  $$

$\longrightarrow \begin{array}{c}
(R_1 = R_1 - (b-ac) R_3)\\
\\
\\
\end{array}
\left [ \begin{array}{rrrrrrr}
1 & 0 & 0 & & 1 & -a & ac-b\\
0 & 1 & 0 & & 0 & 1 & -c\\
0 & 0 & 1 & & 0 & 0&1 \end{array} \right ] = [I \hspace{8pt} L^{-1}]$
}

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