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(a) Consider an matrix A such that AA=Im. It is necessarily true that? Explain.

(b) Consider an nnmatrix A such that ATA=In. Is it necessarily true that AAT=In? Explain.

Short Answer

Expert verified
  1. Not always true for condition (a).
  2. Always true for condition (b).

Step by step solution

01

Determine AAT≠I3.

To obtain AATI3consider the matrix below.

A=100100

In the matrix A. performing product,

AAT=I2.

Which shows that AATI3

Thus, according to condition (a) it is not necessarily true that AAT=I3.

02

Consider the definition of inverse matrix.

The inverse of matrix is another matrix, which on multiplication with the given matrix gives the multiplicative identity. For a matrix A, its inverse is A-1.

And the formula of inverse matrix is given below.

A-1=1A.AdjA

For example,

A=1-234A=0.40.2-0.30.1

Thus, by consider the definition of inverse matrix above the condition (b) is always true.

Hence, condition (a) is not necessarily true and condition (b) is always true.

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Most popular questions from this chapter

Let Abe annmmatrix. Is the formula(kerA)=im(AT)necessarily true? Explain.

By using paper and pencil, find the least squaresx* of the system Ax=b, whereA=[111011] andb=[333]. Verify that the vectorb-Ax* is perpendicular to the image of A.

Consider the subspaceim(A) of 2 . Where A=[2436] . Find a basis of ker(AT) , and draw a sketch illustrating the formula(imAT)=ker(AT)in this case.

Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14.

8.[5422],[367-2]

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

1.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{*{20}{c}}{{\bf{10}}}\\{ - {\bf{8}}}\\{\bf{2}}\\{\bf{0}}\end{aligned}} \right]\)

Write x as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},{{\bf{u}}_3}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{4}}}} \right\}\).

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