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Chapter 2: Q5Q (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 \(AB\).

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

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\)

Step by step solution

01

Find the value of \(A{b_{\bf{1}}}\)

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

\(\begin{aligned}{c}A{b_1} = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\left( { - 1} \right) \times 3 + 2 \times \left( { - 2} \right)}\\{5 \times 3 + 4 \times \left( { - 2} \right)}\\{2 \times 3 + \left( { - 3} \right) \times \left( { - 2} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}\\7\\{12}\end{aligned}} \right)\end{aligned}\)

02

Find the value of \(A{b_{\bf{2}}}\)

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

\(\begin{aligned}{c}A{b_2} = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 2}\\1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\left( { - 1} \right) \times \left( { - 2} \right) + 2 \times 1}\\{5 \times \left( { - 2} \right) + 4 \times 1}\\{2 \times \left( { - 2} \right) + \left( { - 3} \right) \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}4\\{ - 6}\\{ - 7}\end{aligned}} \right)\end{aligned}\)

03

Write the product \(AB\)

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

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\end{aligned}\)

04

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

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2\\5&4\\2&{ - 3}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3&{ - 2}\\{ - 2}&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 1 \times \left( 3 \right) + 2 \times \left( { - 2} \right)}&{ - 1 \times \left( { - 2} \right) + 2 \times 1}\\{5 \times 3 + 4 \times \left( { - 2} \right)}&{5 \times \left( { - 2} \right) + 4 \times 1}\\{2 \times 3 + \left( { - 3} \right) \times \left( { - 2} \right)}&{2 \times \left( { - 2} \right) + \left( { - 3} \right) \times 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\end{aligned}\)

So, \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 7}&4\\7&{ - 6}\\{12}&{ - 7}\end{aligned}} \right)\).

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Most popular questions from this chapter

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)

Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.
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