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Chapter 2: Q16Q (page 93)
Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)
It is proved that Ais invertible.
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Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.
Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).
In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.
27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).
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 \(B \ne C\).
Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.
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