Chapter 2: Q17Q (page 93)
Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.
Short Answer
\(A = BC{B^{ - 1}}\)
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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 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 \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).
Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]
5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]
Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]
If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.
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