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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

Consider the following geometric 2D transformations: D, a dilation (in which x-coordinates and y-coordinates are scaled by the same factor); R, a rotation; and T a translation. Does D commute with R? That is, is \(D\left( {R\left( {\bf{x}} \right)} \right) = R\left( {D\left( {\bf{x}} \right)} \right)\)for all \({\bf{x}}\) in \({\mathbb{R}^{\bf{2}}}\)? Does D commute with T? Does R commute with T?

Verify the boxed statement preceding Example 1 i.e., Let A and B be square matrices. If \[AB = I\], then Aand B are both invertible, with \[B = {A^{ - {\bf{1}}}}\] and \[A = {B^{ - {\bf{1}}}}\].

Exercises 23-26 display a matrix A and echelon form of A. Find a basis for Col A and a basis for Nul A.

\[A = \left[ {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{5}}&{\bf{9}}&{ - {\bf{2}}}\\{\bf{6}}&{\bf{5}}&{\bf{1}}&{{\bf{12}}}\\{\bf{3}}&{\bf{4}}&{\bf{8}}&{ - {\bf{3}}}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{6}}&{ - {\bf{5}}}\\{\bf{0}}&{\bf{1}}&{\bf{5}}&{ - {\bf{6}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right]\]

Use matrix multiplication to find the image of the triangle with data matrix \(D = \left[ {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{2}}&{\bf{4}}\\{\bf{0}}&{\bf{2}}&{\bf{3}}\end{array}} \right]\) under the transformation that reflects points through the y-axis. Sketch both the original triangle and its image.

Suppose the first two columns, \({{\bf{b}}_1}\) and \({{\bf{b}}_2}\), of Bare equal. What can you say about the columns of AB(if ABis defined)? Why?
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