Compute the products in Exercises 1–4 using (a) the definition, as

in Example 1, and (b) the row–vector rule for computing \(A{\bf{x}}\). If a product is undefined, explain why.

3. \(\left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\)

According to the definition, the weights in a linear combination of matrix A columns are represented by the entries in vector x.

Moreover, the product by using the row-vector rule is defined as shown below:

\(\begin{aligned}{c}A{\bf{x}} &= \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\\ &= {x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\end{aligned}\)

The number of columns in matrix \(A\)should be equal to the number of entries in vector x so that \[A{\bf{x}}\] can be defined.

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\).

It is observed that the number of columns in matrix \(A\) is 2.

Consider matrix \(x = \left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\).

It is observed that the number of entries in vector x is 2.

Since the number of columns in matrix \(A\) is equal to the number of entries in vector x; so \(A{\bf{x}}\) is defined.

Therefore, the product of \(\left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\) is defined.

a.

Compute the product of \(\left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right]\) using the definition as shown below:

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right] &= 2\left[ {\begin{array}{*{20}{c}}6\\{ - 4}\\7\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}5\\{ - 3}\\6\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12}\\{ - 8}\\{14}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{15}\\{ - 9}\\{18}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12 - 15}\\{ - 8 + 9}\\{14 - 18}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\end{aligned}\)

b.

Now, compute the value of \(A{\bf{x}}\) by using the row-vector rule as shown below:

\(\begin{aligned}{c}\left[ {\begin{array}{*{20}{c}}6&5\\{ - 4}&{ - 3}\\7&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\{ - 3}\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}{6 \cdot 2 + 5 \cdot \left( { - 3} \right)}\\{\left( { - 4} \right) \cdot 2 + \left( { - 3} \right) \cdot \left( { - 3} \right)}\\{7 \cdot 2 + 6 \cdot \left( { - 3} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{12 - 15}\\{ - 8 + 9}\\{14 - 18}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\end{aligned}\)

Thus, the product is \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\{ - 4}\end{array}} \right]\).