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Chapter 2: Q2.5-30Q (page 93)

30. Find a different factorization of the A in Exercise 29 i.e., \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{4}} \mathord{\left/

{\vphantom {{\bf{4}} {\bf{3}}}} \right.

\kern-\nulldelimiterspace} {\bf{3}}}}&{ - {\bf{12}}}\\{{{ - {\bf{1}}} \mathord{\left/

{\vphantom {{ - {\bf{1}}} {\bf{4}}}} \right.

\kern-\nulldelimiterspace} {\bf{4}}}}&{\bf{3}}\end{aligned}} \right)\), and thereby design a different ladder network whose transfer matrix is A.

Short Answer

Expert verified

The different factorization of A in Exercise 29 is

\(A = \left( {\begin{aligned}{*{20}{c}}1&{ - {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 4}} \right.

\kern-\nulldelimiterspace} 4}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - 8}\\0&1\end{aligned}} \right)\).

Thus, the ladder network, whose transfer matrix is A, is

Step by step solution

01

Design a ladder network

Construct a ladder network that is different from the network in Exercise 29.

02

Write the transfer matrices

Here, the first circuit is the series circuit with resistance \({R_1}\) , the second circuit is the shunt circuit with resistance \({R_2}\), and the last circuit is the series circuit with resistance \({R_3}\) , whose transfer matrices are \(\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\), and \(\left( {\begin{aligned}{*{20}{c}}1&{ - {R_3}}\\0&1\end{aligned}} \right)\), respectively.

03

Find the transfer matrix for the above ladder network

The transfer matrix for the ladder network is given by in the reverse order.

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}1&{ - {R_3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_3}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - {R_1}}\\0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/

{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right)\end{aligned}\)

04

Set matrix A, which is equal to the transfer matrix in step 2

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/

{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right) = A\\\left( {\begin{aligned}{*{20}{c}}{1 + {{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{ - {R_1} - {{{R_1}{R_3}} \mathord{\left/

{\vphantom {{{R_1}{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} - {R_3}}\\{ - {1 \mathord{\left/

{\vphantom {1 {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}}}&{{{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 1}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}&{ - 12}\\{{{ - 1} \mathord{\left/

{\vphantom {{ - 1} 4}} \right.

\kern-\nulldelimiterspace} 4}}&3\end{aligned}} \right)\end{aligned}\)

This implies that

\(\begin{aligned}{c}{{ - 1} \mathord{\left/

{\vphantom {{ - 1} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} = - {1 \mathord{\left/

{\vphantom {1 4}} \right.

\kern-\nulldelimiterspace} 4}\\{R_2} = 4\end{aligned}\)

\(\begin{aligned}{c}{{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} + 1 = 3\\{{{R_1}} \mathord{\left/

{\vphantom {{{R_1}} 4}} \right.

\kern-\nulldelimiterspace} 4} = 2\\{R_1} = 8\end{aligned}\)

\(\begin{aligned}{c}1 + {{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} {{R_2}}}} \right.

\kern-\nulldelimiterspace} {{R_2}}} = {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}\\{{{R_3}} \mathord{\left/

{\vphantom {{{R_3}} 4}} \right.

\kern-\nulldelimiterspace} 4} = {1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}\\{R_3} = {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}\end{aligned}\)

05

Conclusion

The different factorization of A in Exercise 29 is:

\(A = \left( {\begin{aligned}{*{20}{c}}1&{ - {4 \mathord{\left/

{\vphantom {4 3}} \right.

\kern-\nulldelimiterspace} 3}}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\{ - {1 \mathord{\left/

{\vphantom {1 4}} \right.

\kern-\nulldelimiterspace} 4}}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&{ - 8}\\0&1\end{aligned}} \right)\)

Thus, the ladder network, whose transfer matrix is A, is

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Most popular questions from this chapter

Show that if ABis invertible, so is B.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

38. Use at least three pairs of random \(4 \times 4\) matrices Aand Bto test the equalities \({\left( {A + B} \right)^T} = {A^T} + {B^T}\) and \({\left( {AB} \right)^T} = {A^T}{B^T}\). (See Exercise 37.) Report your conclusions. (Note:Most matrix programs use \(A'\) for \({A^{\bf{T}}}\).

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

4. \[\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

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 \(AB\).

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

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).
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