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Chapter 2: Q6Q (page 93)

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{2}}}\\{ - {\bf{3}}}&{\bf{0}}\\{\bf{3}}&{\bf{5}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{3}}\\{\bf{2}}&{ - {\bf{1}}}\end{aligned}} \right)\)

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\)

Step by step solution

01

Find the value of \(A{b_{\bf{1}}}\)

Multiply matrix \(A\) with the first column of matrix \(B\).

\(\begin{aligned}{c}A{b_1} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}\\{\left( { - 3} \right) \times 1 + 0}\\{3 \times 1 + 5 \times 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\{ - 3}\\{13}\end{aligned}} \right)\end{aligned}\)

02

Find the value of \(A{b_{\bf{2}}}\)

Multiply matrix \(A\) with the second column of matrix \(B\).

\(\begin{aligned}{c}A{b_2} = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 3 + 0}\\{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{14}\\{ - 9}\\4\end{aligned}} \right)\end{aligned}\)

03

Write the product \(AB\)

The product \(AB\) can be written as follows:

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

04

Find the product \(AB\) using row-column rule

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}4&{ - 2}\\{ - 3}&0\\3&5\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&3\\2&{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{4 \times 1 + \left( { - 2} \right) \times 2}&{4 \times 3 + \left( { - 2} \right) \times \left( { - 1} \right)}\\{\left( { - 3} \right) \times 1 + 0}&{\left( { - 3} \right) \times 3 + 0}\\{3 \times 1 + 5 \times 2}&{3 \times 3 + 5 \times \left( { - 1} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\end{aligned}\)

So, \(AB = \left( {\begin{aligned}{*{20}{c}}0&{14}\\{ - 3}&{ - 9}\\{13}&4\end{aligned}} \right)\).

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Most popular questions from this chapter

Use the inverse found in Exercise 1 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

Suppose \(CA = {I_n}\)(the \(n \times n\) identity matrix). Show that the equation \(Ax = 0\) has only the trivial solution. Explain why Acannot have more columns than rows.
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