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Chapter 2: Q36Q (page 93)

(M) Let \(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{25}}}&{ - {\bf{9}}}&{ - {\bf{27}}}\\{{\bf{546}}}&{{\bf{180}}}&{{\bf{537}}}\\{{\bf{154}}}&{{\bf{50}}}&{{\bf{149}}}\end{aligned}} \right)\). Find the second and third columns of \({A^{ - {\bf{1}}}}\) without computing the first column.

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\)

Step by step solution

01

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{2}}}}&{{e_{\bf{3}}}}\end{aligned}} \right)\)

The matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_2}}&{{e_3}}\end{aligned}} \right)\) can be expressed as

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_2}}&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\\{546}&{180}&{537}&1&0\\{154}&{50}&{149}&0&1\end{aligned}} \right)\).

02

Convert the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{2}}}}&{{e_{\bf{3}}}}\end{aligned}} \right)\) into the row-reduced echelon form

Consider the matrix\(\left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\\{546}&{180}&{537}&1&0\\{154}&{50}&{149}&0&1\end{aligned}} \right)\).

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\end{aligned};{\rm{ }}\begin{aligned}{*{20}{c}}{546}&{180}&{537}&1&0\end{aligned};{\rm{ }}\begin{aligned}{*{20}{c}}{154}&{50}&{149}&0&1\end{aligned}{\rm{ }}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{aligned}\)

\(\left( {\begin{aligned}{*{20}{c}}1&0&0&{\frac{3}{2}}&{ - \frac{9}{2}}\\0&1&0&{ -

\frac{{433}}{6}}&{\frac{{439}}{2}}\\0&0&1&{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\)

03

Find the column of the inverse matrix

In the matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&0&{\frac{3}{2}}&{ - \frac{9}{2}}\\0&1&0&{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\0&0&1&{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\), the second and third columns are \(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\).

So, the third column of the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\).

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Most popular questions from this chapter

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.]

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

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.

15. a. If A and B are \({\bf{2}} \times {\bf{2}}\) with columns \({{\bf{a}}_1},{{\bf{a}}_2}\) and \({{\bf{b}}_1},{{\bf{b}}_2}\) respectively, then \(AB = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}{{\bf{b}}_1}}&{{{\bf{a}}_2}{{\bf{b}}_2}}\end{aligned}} \right)\).

b. Each column of ABis a linear combination of the columns of Busing weights from the corresponding column of A.

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

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

e. The transpose of a product of matrices equals the product of their transposes in the same order.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

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

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{4}}}\\{\bf{7}}&{ - {\bf{8}}}\end{aligned}} \right)\).
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