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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

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)
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