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Chapter 2: Q3Q (page 93)

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\)

Step by step solution

01

Find the matrix \(3{I_{\bf{2}}} - A\)

The value of \(3{I_2} - A\) can be calculated as follows:

\(\begin{aligned}{c}3{I_2} - A = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}3&0\\0&3\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\end{aligned}\)

02

Find the matrix \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\)

The value of \(\left( {3{I_2}} \right)A\) can be calculated as follows:

\(\begin{aligned}{c}\left( {3{I_2}} \right)A = 3\left( {{I_2}A} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}{1 \times 4 + 0}&{1 \times \left( { - 1} \right) + 0}\\{0 + 1 \times 5}&{0 \times \left( { - 1} \right) + 1 \times \left( { - 2} \right)}\end{aligned}} \right)\\ = 3\left( {\begin{aligned}{*{20}{c}}4&{ - 1}\\5&{ - 2}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\end{aligned}\)

So, \(3{I_2} - A = \left( {\begin{aligned}{*{20}{c}}{ - 1}&1\\{ - 5}&5\end{aligned}} \right)\), and \(\left( {3{I_2}} \right)A = \left( {\begin{aligned}{*{20}{c}}{12}&{ - 3}\\{15}&{ - 6}\end{aligned}} \right)\).

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Most popular questions from this chapter

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

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

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

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 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� 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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