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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

Suppose Ais a \(3 \times n\) matrix whose columns span \({\mathbb{R}^3}\). Explain how to construct an \(n \times 3\) matrix Dsuch that \(AD = {I_3}\).

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) 鈥�(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)
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