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Chapter 2: Q2.3-43Q (page 93)

Exercises 42鈥�44 show how to use the condition number of a matrix Ato estimate the accuracy of a computed solution of \(Ax = b\). If the entries of Aand b are accurate to about rsignificant digits and if the condition number of Ais approximately \({\bf{1}}{{\bf{0}}^k}\) (with ka positive integer), then the computed solution of \(Ax = b\) should usually be accurate to at least \(r - k\) significant digits.

43. Repeat Exercise 42 for the matrix in Exercise 10.

Short Answer

Expert verified

The solution has approximately 11 decimal places, and the calculated answer

(\({{\bf{x}}_1}\)) is accurate.

Step by step solution

01

Obtain the condition number of matrix A

Consider matrix A as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}5&3&1&7&9\\6&4&2&8&{ - 8}\\7&5&3&{10}&9\\9&6&4&{ - 9}&{ - 5}\\8&5&2&{11}&4\end{array}} \right]\)

Obtain thecondition number of matrix A by using the MATLAB command shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right];\\ > > {\rm{ C}} = {\rm{cond}}\left( {\rm{A}} \right)\end{array}\]

It gives the output 68622.

Thus, thecondition number of matrix A is 68622.

By comparing with thecondition number of A, that is \({10^k}\), the required condition number is approximately in between \({10^4}\) and \({10^5}\).

02

Obtain the solution by using the MATLAB command

It is discovered that x and\({{\bf{x}}_1}\)agree to at least 11 or 12 significant digits if it run multiple experiments with MATLAB, which properly captures 16 digits.

Obtain a random matrix by using the MATLAB command shown below:

\( > > {\bf{x}} = {\rm{rand}}\left( {5,1} \right)\)

\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{.2190}\\{.0470}\\{.6789}\\{.6793}\\{.9347}\end{array}} \right]\)

Now, compute\({\bf{b}} = A{\bf{x}}\)by using the MATLAB command shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right];\\ > > x = \left[ {.2190{\rm{; }}{\rm{.0470; }}{\rm{.6789; }}{\rm{.6793; }}{\rm{.9347}}} \right]{\rm{;}}\\ > > b = A*x\end{array}\)

The output is \({\bf{b}} = A{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{15.0821}\\{.8165}\\{19.0097}\\{ - 5.8188}\\{14.5557}\end{array}} \right]\).

03

Obtain the MATLAB solution

Compute\({{\bf{x}}_1}\)of\(A{\bf{x}} = {\bf{b}}\)by using the MATLAB command shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{5 3 1 7 9; 6 4 2 8 }} - {\rm{8; 7 5 3 10 9; 9 6 4 }} - {\rm{9 }} - {\rm{5; 8 5 2 11 4}}} \right];\\ > > b = \left[ {15.0821{\rm{; }}{\rm{.8165; 19}}{\rm{.0097; }} - {\rm{5}}{\rm{.8188; 14}}{\rm{.5557}}} \right]{\rm{;}}\\ > > {x_1} = A\backslash b\end{array}\)

The output is\({{\bf{x}}_1} = \left[ {\begin{array}{*{20}{c}}{.2190}\\{.0470}\\{.6789}\\{.6793}\\{.9347}\end{array}} \right]\).

Obtain the difference between x and\({{\bf{x}}_1}\).

\({\bf{x}} - {{\bf{x}}_1} = \left[ {\begin{array}{*{20}{c}}{.3165}\\{ - .6743}\\{.3343}\\{.0158}\\{ - .0005}\end{array}} \right] \times {10^{ - 11}}\)

Thus, the solution has approximately 11 decimal places, and the calculated answer (\({{\bf{x}}_1}\)) is accurate.

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Most popular questions from this chapter

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

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

1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)

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

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.)

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 \(A\).

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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 \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).
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