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Chapter 2: Q26Q (page 93)

Exercises 25 and 26 prove Theorem 4 for \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

26. Show that if \(ad - bc \ne {\bf{0}}\), the formula for \({A^{ - 1}}\) works.

Short Answer

Expert verified

If \(ad - bc \ne 0\), then \({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\).

Step by step solution

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01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed using theaugmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is theidentity matrix. Matrix Ahas an inverse only if \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\) is row equivalent to \(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\).

02

Obtain the inverse of matrix A

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\).

Write the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)as shown below:

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}a&b&1&0\\c&d&0&1\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

Multiply row one by\(\frac{1}{a}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\c&d&0&1\end{aligned}} \right)\)

Use the\({x_1}\)term in the first equation to eliminate the\(c{x_1}\)term from the second equation. Add\( - c\)times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{d - \frac{{bc}}{a}}&{ - c/a}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right)\)

Use the\(\frac{{b{x_2}}}{a}\)term in the first equation to eliminate the\(\left( {\frac{{ad - bc}}{a}} \right){x_2}\)term from the second equation. Add\( - \left( {\frac{b}{{ad - bc}}} \right)\)times row two to row one.

\(\left( {\begin{aligned}{*{20}{c}}1&{b/a}&{1/a}&0\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\0&{\frac{{ad - bc}}{a}}&{ - c/a}&1\end{aligned}} \right)\)

Multiply row two by\(\left( {\frac{a}{{ad - bc}}} \right)\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\0&1&{\frac{{ - c}}{{ad - bc}}}&{\frac{a}{{ad - bc}}}\end{aligned}} \right)\)

By comparing with\(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\), you get\({A^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{\frac{d}{{ad - bc}}}&{ - \frac{b}{{ad - bc}}}\\{ - \frac{c}{{ad - bc}}}&{\frac{a}{{ad - bc}}}\end{aligned}} \right)\), or\({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\).

Thus, theinverseof matrix is \({A^{ - 1}} = \frac{1}{{ad - bc}}\left( {\begin{aligned}{*{20}{c}}d&{ - b}\\{ - c}&a\end{aligned}} \right)\) if \(ad - bc \ne 0\).

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Most popular questions from this chapter

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?

Show that block upper triangular matrix \(A\) in Example 5is invertible if and only if both \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible. [Hint: If \({A_{{\bf{11}}}}\) and \({A_{{\bf{12}}}}\) are invertible, the formula for \({A^{ - {\bf{1}}}}\) given in Example 5 actually works as the inverse of \(A\).] This fact about \(A\) is an important part of several computer algorithims that estimates eigenvalues of matrices. Eigenvalues are discussed in chapter 5.

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

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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