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Chapter 2: Q8Q (page 93)

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

Short Answer

Expert verified

The equation \(D = {A^{ - 1}}\) is proved.

Step by step solution

01

Multiply \({A^{ - {\bf{1}}}}\) on both sides of \(AD = I\)

It gives \({A^{ - 1}}AD = {A^{ - 1}}I\).

02

Use the fact of an invertible matrix

Given that A is invertible, then \(A{A^{ - 1}} = {A^{ - 1}}A = I\).

This implies that\(ID = {A^{ - 1}}I\).

03

Use the fact of an identity matrix

It gives \(ID = D\) and \({A^{ - 1}}I = {A^{ - 1}}\). Then, \(D = {A^{ - 1}}\).

Hence, proved.

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Most popular questions from this chapter

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

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, 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.]

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

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

[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.]
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