Complete parts a and b below for the matrix A.- 496328 - 35 - 21214928 - 42 - 49144235 - 35A=- 354942 - 3542 - 14 - 5621 - 35 - 5674928 - 56- 425635 - 28 - 28 - 63 -21a. Construct matrices C and N whose columns are bases for Col A and Nul A, respectively, and construct a matrix R whose rows form a basis for Row A.c=0C =N=OR=b. Construct a matrix M whose columns form a basis for Nul A', form the matrices S= RT N and T= C M, and explain why S and T should be square. Verify that both S and T are invertible.M=The matrix S= RT N isx because the columns of RT and N are in RU and dim Row A + dim Nul A =.[RT N]#The matrix T=[C M ]is xO because the columns of C and M are in R and dim Col A + dim Nul Aby the Rank Theorem, since Col A = Row AT.The columns of the matrix S are linearlyThe columns of S form a basis of RThus, by the Invertible Matrix Theorem, S is invertible.The columns of the matrix T are linearlyThe columns of T form a basis of RU. Thus, by the Invertible Matrix Theorem, T is invertible.

find the row reduced echelon form of A. To get the row space, consider the non zero rows of RREF(A).…

Complete parts a and b below for the matrix A. -49 63 28 - 35 - 21 21 49 28 - 42 - 49 14 42 35 - 35 A= - 35 49 42 - 35 42 - 14 - 56 21 - 35 - 56 7 49 28 - 56 -42 56 35 - 28 - 28- 63 -21 a. Construct matrices C and N whose columns are bases for Col A and NulA, respectively, and construct a matrix R whose rows form a basis for Row A. c-O N=O R= b. Construct a matrix M whose columns form a basis for Nul A", form the matrices S=[RT N]and T=[c m], and explain why S and T should be square. Verify that both S and T are invertible. M=[ The matrix S= [RT N]is -O because the columns of R"and N are in Rand dim Row A+ dim Nul A=D. The matrix T=[CM]isI×O because the columns of C and M are in R and dim Col A + dim Nul AT =O by the Rank Theorem, since Col A = Row A".

Complete parts a and b below for the matrix A. -49 63 28 - 35 - 21 21 49 28 - 42 - 49 14 42 35 - 35 A= - 35 49 42 - 35 42 - 14 - 56 21 - 35 - 56 7 49 28 - 56 -42 56 35 - 28 - 28- 63 -21 a. Construct matrices C and N whose columns are bases for Col A and NulA, respectively, and construct a matrix R whose rows form a basis for Row A. c-O N=O R= b. Construct a matrix M whose columns form a basis for Nul A", form the matrices S=[RT N]and T=[c m], and explain why S and T should be square. Verify that both S and T are invertible. M=[ The matrix S= [RT N]is -O because the columns of R"and N are in Rand dim Row A+ dim Nul A=D. The matrix T=[CM]isI×O because the columns of C and M are in R and dim Col A + dim Nul AT =O by the Rank Theorem, since Col A = Row A".

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied under linear algebra. The first-degree equations involving algebraic expressions are called linear equations. That is, if you represent it by drawing a graph you will get a straight line.

Question

Complete parts a and b below for the matrix A.
- 49
63
28 - 35 - 21
21
49
28 - 42 - 49
14
42
35 - 35
A=
- 35
49
42 - 35
42 - 14 - 56
21 - 35 - 56
7
49
28 - 56
- 42
56
35 - 28 - 28 - 63 -21
a. Construct matrices C and N whose columns are bases for Col A and Nul A, respectively, and construct a matrix R whose rows form a basis for Row A.
c=0
C =
N=O
R=
b. Construct a matrix M whose columns form a basis for Nul A', form the matrices S= RT N and T= C M, and explain why S and T should be square. Verify that both S and T are invertible.
M=
The matrix S= RT N isx because the columns of RT and N are in RU and dim Row A + dim Nul A =.
[RT N]#
The matrix T=[C M ]is xO because the columns of C and M are in R and dim Col A + dim Nul A
by the Rank Theorem, since Col A = Row AT.
The columns of the matrix S are linearly
The columns of S form a basis of R
Thus, by the Invertible Matrix Theorem, S is invertible.
The columns of the matrix T are linearly
The columns of T form a basis of RU. Thus, by the Invertible Matrix Theorem, T is invertible.

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