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Chapter 2: Q5Q (page 93)
Use the inverse found in Exercise 1 to solve the system
\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)
The solutions are \({x_1} = 7\) and \({x_2} = - 9\).
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Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let Sand U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that Thas a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].
In exercises 11 and 12, mark each statement True or False. Justify each answer.
a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.
b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.
In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).
34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)
Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).
Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).
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