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Chapter 2: Q31Q (page 93)

Find the inverse of matrices in Exercise 29-32, if they exist. Use the algorithm introduced in this section.

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

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\)

Step by step solution

01

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

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

02

Apply the row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row three, multiply row one by 2 and subtract it from row three, i.e., \({R_3} \to {R_3} - 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\{ - 3}&1&4\\0&{ - 3}&8\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\{ - 2}&0&1\end{aligned}} \right)\)

03

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row two, multiply row one by 3 and add it to row two, i.e., \({R_2} \to {R_2} + 3{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\0&1&{ - 2}\\0&{ - 3}&8\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\3&1&0\\{ - 2}&0&1\end{aligned}} \right)\)

04

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row three, multiple row two by 3 and add it to row three, i.e., \({R_3} \to {R_3} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\0&1&{ - 2}\\0&0&2\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\3&1&0\\7&3&1\end{aligned}} \right)\)

05

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row two, add rows two and three, i.e., \({R_2} \to {R_2} + {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&{ - 2}\\0&1&0\\0&0&2\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{10}&4&1\\7&3&1\end{aligned}} \right)\)

06

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)

At row one, add rows three and one, i.e., \({R_1} \to {R_1} + {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&2\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\7&3&1\end{aligned}} \right)\)

Divide row three by 2, i.e., \({R_3} \to \frac{{{R_3}}}{2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\,\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\)

So, the inverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}8&3&1\\{10}&4&1\\{\frac{7}{2}}&{\frac{3}{2}}&{\frac{1}{2}}\end{aligned}} \right)\).

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Most popular questions from this chapter

Suppose the second column of Bis all zeros. What can you

say about the second column of AB?

Let Ube the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of Ulists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

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 \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).
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