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Chapter 2: Q2.4-21Q (page 93)

a. Verify that \({A^2} = I\) when \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\).

b. Use partitioned matrices to show that \({M^2} = I\) when\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).

Short Answer

Expert verified
  1. It is verified that \({A^2} = I\).
  2. It is proved that \({M^2} = I\).

Step by step solution

01

Verify that \({A^2} = I\)

(a)

If \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\),

\(\begin{array}{c}{A^2} = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + 0}&{0 + 0}\\{3 - 3}&{0 + {{\left( { - 1} \right)}^2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = I.\end{array}\)

Thus, it is verified that \({A^2} = I\).

02

Determine the partitioned matrix of M

(b)

Matrix \(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\)can be written as a \(2 \times 2\) partitioned matrix.

\(\begin{array}{c}M = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\end{array}\)

03

Use the partitioned matrix to show that \({M^2} = I\)

If \(M = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\),

\[\begin{array}{c}{M^2} = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{A^2} + 0}&{0 + 0}\\{A - A}&{0 + {{\left( { - A} \right)}^2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\,\,\,\,\,\,{\mathop{\rm since}\nolimits} \,\,\,{A^2} = I\\ = I.\end{array}\]

Thus, it is proved that \({M^2} = I\).

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Most popular questions from this chapter

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 exercises 11 and 12, mark each statement True or False. Justify each answer.

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

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.)

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)
See all solutions

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