Chapter 2: Q17Q (page 93)
Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.
\(A = BC{B^{ - 1}}\)
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Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.
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.)
Use the inverse found in Exercise 1 to solve the system
\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)
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 \(AB\).
\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)
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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)
\(A + 2B\), \(3C - E\), \(CB\), \(EB\).
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