Question

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.

4. \(\left[ {\begin{array}{*{20}{c}}8&3&{ - 4}\\5&1&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right]\)

Short answer

The product is defined and it is \(\left[ {\begin{array}{*{20}{c}}7\\8\end{array}} \right]\).

Step-by-step solution

Write the condition for the product of a vector and a matrix

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

Also, 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.

Obtain the number of columns in matrix A

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}8&3&{ - 4}\\5&1&2\end{array}} \right]\).

It is observed that the number of columns in matrix \(A\) is 3.

Obtain the number of entries in vector x

Consider matrix \(x = \left[ {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right]\).

It is observed that the number of entries in vector x is 3.

Check if \(Ax\) is defined or not

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 \(\left[ {\begin{array}{*{20}{c}}8&3&{ - 4}\\5&1&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right]\) is defined.

Obtain the product of the matrix and the vector

a.

Compute the product of \(\left[ {\begin{array}{*{20}{c}}8&3&{ - 4}\\5&1&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right]\) using the definition as shown below:

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

b.

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

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

Thus, the product is \(\left[ {\begin{array}{*{20}{c}}7\\8\end{array}} \right]\).