Exercise 1.2.5

Question

Start with a matrix $B$. If we want to take combinations of its rows, we premultiply by $A$ to get $A B$. If we want to take combinations of its columns, we postmultiply by $C$ to get $B C .$ For this question we will do both.
\begin{itemize}
\item Row operations then column operations First $A B$ then $(A B) C$
\item Column operations then row operations First $B C$ then $A(B C)$
\end{itemize}
The associative law says that we get the same final result both ways. Verify $(A B) C=A(B C)$ for 
$$A=\left[\begin{array}{ll}1 & a \ 0 & 1\end{array}ight] \quad B=\left[\begin{array}{ll}b_{1} & b_{2} \ b_{3} & b_{4}\end{array}ight] \quad C=\left[\begin{array}{ll}1 & 0 \ c & 1\end{array}ight].$$

Neil Z. · Accepted Answer

$$(AB)C = \begin{bmatrix}b_1+ab_3 & b_2+ab_4\b_3 & b_4\end{bmatrix}\begin{bmatrix}1 &0 \ c& 1\end{bmatrix} = \begin{bmatrix}b_1+ab_3+cb_2+acb_4 & b_2+ab_4\b_3 +cb_4& b_4\end{bmatrix}$$

$$A(BC) = \begin{bmatrix}1 & a\0 & 1\end{bmatrix}\begin{bmatrix}b_1+cb_2 & b_2\b_3+cb_4 & b_4\end{bmatrix} = \begin{bmatrix}b_1+ab_3+cb_2+acb_4 & b_2+ab_4\b_3 +cb_4& b_4\end{bmatrix}$$

So we have verified $(AB)C=A(BC)$.

Exercise 1.2.5

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Exercise 1.2.5

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