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Chapter 3: Q17E (page 165)

Question: 17. Show that if A is \({\bf{2}} \times {\bf{2}}\), then Theorem 8 gives the same formula for \({A^{ - {\bf{1}}}}\) as that given by theorem 4 in Section 2.2.

Short Answer

Expert verified

Hence, \({A^{ - 1}} = \frac{1}{{\det A}}{\rm{adj}}\,A = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)\) for \(ad - bc \ne 0\), and \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Theorem 8 gives the same formula for \({A^{ - 1}}\) as that given by Theorem 4 in Section 2.2.

Step by step solution

01

First, find the determinant of A

Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Then,

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right|\\ = ad - bc\end{array}\)

Note that \({A^{ - 1}}\) exists if \(\det A \ne 0\) , i.e., \(ad - bc \ne 0\).

02

Find the cofactors of A

The four cofactorsare:

\(\begin{array}{c}{C_{11}} = {\left( { - 1} \right)^2}\left| d \right|\\ = d\end{array}\)

\(\begin{array}{c}{C_{12}} = {\left( { - 1} \right)^3}\left| c \right|\\ = - c\end{array}\)

\(\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| b \right|\\ = - b\end{array}\)

\(\begin{array}{c}{C_{22}} = {\left( { - 1} \right)^4}\left| a \right|\\ = a\end{array}\)

03

Find the adjugate matrix of A

\(\begin{array}{c}{\rm{adj}}\,A = {\left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right)^T}\\ = {\left( {\begin{array}{*{20}{c}}d&{ - c}\\{ - b}&a\end{array}} \right)^T}\\{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)\end{array}\)

04

Use Theorem 8 to find the inverse of A

By Theorem 8,

\(\begin{array}{c}{A^{ - 1}} = \frac{1}{{\det A}}{\rm{adj}}\,A\\ = \frac{1}{{ad - bc}}\left( {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right)\end{array}\)

\(ad - bc \ne 0\).

Hence, Theorem 8 gives the same formula for \({A^{ - 1}}\) as that given by Theorem 4 in Section 2.2.

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Most popular questions from this chapter

Combine the methods of row reduction and cofactor expansion to compute the determinant in Exercise 12.

12. \(\left| {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}&{\bf{3}}&{\bf{0}}\\{\bf{3}}&{\bf{4}}&{\bf{3}}&{\bf{0}}\\{{\bf{11}}}&{\bf{4}}&{\bf{6}}&{\bf{6}}\\{\bf{4}}&{\bf{2}}&{\bf{4}}&{\bf{3}}\end{aligned}} \right|\)

The expansion of a \({\bf{3}} \times {\bf{3}}\) determinant can be remembered by the following device. Write a second type of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals.

Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises 15-18. Warning: This trick does not generalize in any reasonable way to \({\bf{4}} \times {\bf{4}}\) or larger matrices.

15. \(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{4}}\\{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\end{array}} \right|\)

Compute the determinants of the elementary matrices given in Exercise 25-30.

27. \(\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\k&0&1\end{array}} \right]\).

Question: Use Cramer鈥檚 rule to compute the solutions of the systems in Exercises1-6.

4. \(\begin{array}{c} - 5{x_1} + 2{x_2} = 9\\3{x_1} - {x_2} = - 4\end{array}\)

Compute the determinant in Exercise 7 using a cofactor expansion across the first row.

7. \[\left| {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{3}}&{\bf{0}}\\{\bf{6}}&{\bf{5}}&{\bf{2}}\\{\bf{9}}&{\bf{7}}&{\bf{3}}\end{array}} \right|\]
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