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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

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

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.]

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

Suppose AB = AC, where Band Care \(n \times p\) matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible.

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).
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