A = [2 1 1 4 −6 0 −2 7 2]andb = [6 8 3]only answere of part d(a) Basis for Null Space and Column space of B.(b) Dimension of Null Space and column space of B.(c) State the rank of system through rank theorem.(d) When the column Space of B become R3Space?Space?i asked this question but your tutor asked we solve on 3 subparts and it is the part d so the answer is give below short course i upload only 2 images From the row reduced echolon formot B.-from the origihal matnix :entryiharow.a leadinghe columns ĕantaihing leadihg ones are thePivot columms, To obtain a basis for theSlumn space, we inet Use the pivot calumnsone i2 the first nonzerro etryinarow.This is the vequired basis for coumn & paceof B.(b) The dimension of null Space of B i8 1,Since the bauis contains only oneelement.The dimengion of column space Of B ů 3,Shce the basis of columm space ofBContains 3 vectors,the Rank theorem,-If A is a mabrix with n columns g thendimension of column SpaceAt dimension of null8paceof A =n.i.e, pank (A) +hullity ofA)-nHere n- 4., nullity (eB)-1there forerankCB)-4-I= 3 R=Rp+)R3RI=R,+(-)R2This 18 the row veducedechelen form of BNow, Solving the matin equationd21If we take 24=t, thent, x, = 3t8.4Thorefore,3t4-5t-5This is a null space.The basis for the nul space is-5- O. O

Given A=2114-60-272b=683 Define…

Answered: A = [2 1 1 4 −6 0 −2 7 2]…

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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Question

A = [2 1 1
4 −6 0
−2 7 2]
and

b = [6

only answere of part d

(a) Basis for Null Space and Column space of B.
(b) Dimension of Null Space and column space of B.
(c) State the rank of system through rank theorem.
(d) When the column Space of B become R3
Space?

Space?

i asked this question but your tutor asked we solve on 3 subparts and it is the part d so the answer is give below short course i upload only 2 images

From the row reduced echolon formot B.
-from the origihal matnix :
entryiharow.
a leading
he columns ĕantaihing leadihg ones are the
Pivot columms, To obtain a basis for the
Slumn space, we inet Use the pivot calumns
one i2 the first nonzerro etryinarow.
This is the vequired basis for coumn & pace
of B.
(b) The dimension of null Space of B i8 1,
Since the bauis contains only one
element.
The dimengion of column space Of B ů 3,
Shce the basis of columm space ofB
Contains 3 vectors,
the Rank theorem,-
If A is a mabrix with n columns g then
dimension of column SpaceAt dimension of null
8paceof A =n.
i.e, pank (A) +hullity ofA)-n
Here n- 4., nullity (eB)-1
there fore
rankCB)-4-I
= 3

R=Rp+)R3
RI=R,+(-)R2
This 18 the row veduced
echelen form of B
Now, Solving the matin equation
d2
1
If we take 24=t, then
t, x, = 3t
8.
4
Thorefore,
3t
4
-5t
-5
This is a null space.
The basis for the nul space is
-5
- O. O

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