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Linear Algebra and its Applications

David C. Lay, Steven R. Lay and Judi J. McDonald

$Math Studyset Vaia Explanations$ Math

5th Edition · 483 pages

ISBN: 978-03219822384

Chapter 2: Q2.4-12Q (page 93)

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.

Short Answer

Expert verified

a. True

b. False

Step by step solution

01

Analyze the expression \[Ax\]

For the product of matrix \(A\) and vector \({\bf{x}}\), the row-column rule for multiplication of block matrices provides the most general way.

So, product \(A{\bf{x}}\) is a special case of multiplication.

02

Analyze the expression \(BA\)

Partition matrices can be multiplied by the usual row-column rule. If the block entries are scalars when \(AB\) exists, the column partition of \(A\) matches the row partition of \(B\).

For the given matrices, the column partition is possible for \(B\), and only row operations are possible for matrix \(A\).

Therefore, only product \(BA\) is possible, \(AB\) is not possible.

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Most popular questions from this chapter

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

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 \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)

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