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Chapter 2: Q7Q (page 93)

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 \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

Short Answer

Expert verified
  1. \({A^{ - 1}} = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\), and the solutions of the equations are \({A^{ - 1}}{b_1} = \left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right),\)\({A^{ - 1}}{b_2} = \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right),\)\({A^{ - 1}}{b_3} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right),\) and \({A^{ - 1}}{b_4} = \left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\).
  2. The solutions are \(\left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right),\) and \(\left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\), same as in part (a).

Step by step solution

01

Check if A is invertible

\(\begin{aligned}{c}\det A = 1\left( {12} \right) - 2\left( 5 \right)\\ = 12 - 10\\\det A = 2 \ne 0\end{aligned}\)

This implies that A is invertible.

02

Write the inverse of A

(a)

\({A^{ - 1}} = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\)

03

Find the solutions using \({A^{ - {\bf{1}}}}\)

The solution for the system \(Ax = {b_1}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_1}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\3\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{ - 12 - 6}\\{5 + 3}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{ - 18}\\8\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_2}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_2}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\{ - 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12 + 10}\\{ - 5 - 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{22}\\{ - 10}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_3}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_3}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2\\6\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{24 - 12}\\{ - 10 + 6}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}\\{ - 4}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right)\end{aligned}\)

The solution for the system \(Ax = {b_4}\) is obtained as shown below:

\(\begin{aligned}{c}x = {A^{ - 1}}{b_4}\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{12}&{ - 2}\\{ - 5}&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}3\\5\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{36 - 10}\\{ - 15 + 5}\end{aligned}} \right)\\ = \frac{1}{2}\left( {\begin{aligned}{*{20}{c}}{26}\\{ - 10}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

04

Reduce the augmented matrix \(\left(

{\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\)

(b)

\(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{{b_2}}&{{b_2}}&{{b_4}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&2&{ - 1}&1&2&3\\5&{12}&3&{ - 5}&6&5\end{aligned}} \right)\)

At row two, multiply row one by 5 and subtract it from row two, i.e., \({R_2} \to {R_2} - 5{R_1}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&2&{ - 1}&1&2&3\\0&2&8&{ - 10}&{ - 4}&{ - 10}\end{aligned}} \right)\)

Divide row two by 2.

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

At row one, multiply row two by 2 and subtract it from row one, i.e., \({R_1} \to {R_1} - 2{R_2}\).

\( \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 9}&{11}&6&{13}\\0&1&4&{ - 5}&{ - 2}&{ - 5}\end{aligned}} \right)\)

This implies that the solutions are \(\left( {\begin{aligned}{*{20}{c}}{ - 9}\\4\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}{11}\\{ - 5}\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\end{aligned}} \right),\) and \(\left( {\begin{aligned}{*{20}{c}}{13}\\{ - 5}\end{aligned}} \right)\), same as in part (a).

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Most popular questions from this chapter

[M] Suppose memory or size restrictions prevent your matrix program from working with matrices having more than 32 rows and 32 columns, and suppose some project involves \(50 \times 50\) matrices A and B. Describe the commands or operations of your program that accomplish the following tasks.

a. Compute \(A + B\)

b. Compute \(AB\)

c. Solve \(Ax = b\) for some vector b in \({\mathbb{R}^{50}}\), assuming that \(A\) can be partitioned into a \(2 \times 2\) block matrix \(\left[ {{A_{ij}}} \right]\), with \({A_{11}}\) an invertible \(20 \times 20\) matrix, \({A_{22}}\) an invertible \(30 \times 30\) matrix, and \({A_{12}}\) a zero matrix. [Hint: Describe appropriate smaller systems to solve, without using any matrix inverse.]

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 \(AB\).

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

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]
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