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Question-211902




Question Number 211902 by Gangadhar last updated on 23/Sep/24
Commented by BHOOPENDRA last updated on 24/Sep/24
3
$$\mathrm{3} \\ $$
Answered by BHOOPENDRA last updated on 24/Sep/24
 [((1        2      ),3),((3        4),4),((7        10  ),(12)) ]   [(3,6,9),(3,4,4),(7,(10),(12)) ] multiply first row by 3   [(1,2,3),(0,(−2),(−5)),(7,(10),(12)) ]row1−row2   [(7,(14),(21)),(0,(−2),(−5)),(7,(10),(12)) ]row1×7   [(1,2,3),(0,(−2),(−5)),(0,(−4),(−9)) ]subract row1 from 3 row   [(1,2,3),(0,(−4),(−10)),(0,(−4),(−9)) ]2nd row×2   [(1,2,3),(0,(−2),(−5)),(0,0,1) ]  sub 2nd row from 3rd row &restore 2nd row   [(1,2,3),(0,(−2),(−5)),(0,0,1) ]  matrix rank=3
$$\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:}&{\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{4}}&{\mathrm{4}}\\{\mathrm{7}\:\:\:\:\:\:\:\:\mathrm{10}\:\:}&{\mathrm{12}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{9}}\\{\mathrm{3}}&{\mathrm{4}}&{\mathrm{4}}\\{\mathrm{7}}&{\mathrm{10}}&{\mathrm{12}}\end{bmatrix}\:{multiply}\:{first}\:{row}\:{by}\:\mathrm{3} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{−\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{7}}&{\mathrm{10}}&{\mathrm{12}}\end{bmatrix}{row}\mathrm{1}−{row}\mathrm{2} \\ $$$$\begin{bmatrix}{\mathrm{7}}&{\mathrm{14}}&{\mathrm{21}}\\{\mathrm{0}}&{−\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{7}}&{\mathrm{10}}&{\mathrm{12}}\end{bmatrix}{row}\mathrm{1}×\mathrm{7} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{−\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{0}}&{−\mathrm{4}}&{−\mathrm{9}}\end{bmatrix}{subract}\:{row}\mathrm{1}\:{from}\:\mathrm{3}\:{row} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{−\mathrm{4}}&{−\mathrm{10}}\\{\mathrm{0}}&{−\mathrm{4}}&{−\mathrm{9}}\end{bmatrix}\mathrm{2}{nd}\:{row}×\mathrm{2} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{−\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$${sub}\:\mathrm{2}{nd}\:{row}\:{from}\:\mathrm{3}{rd}\:{row}\:\&{restore}\:\mathrm{2}{nd}\:{row} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{−\mathrm{2}}&{−\mathrm{5}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$${matrix}\:{rank}=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Commented by Spillover last updated on 24/Sep/24
correct
$${correct} \\ $$

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