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Question-219696




Question Number 219696 by Spillover last updated on 01/May/25
Answered by SdC355 last updated on 01/May/25
E= ((1,0,0),(0,1,0),(0,0,1) )  cus A= ((3,1,1),(5,1,2),(4,2,7) )   swap row 1 and row 2   ((5,1,2),(3,1,1),(4,2,7) ) and Subtract (3/5)× row 1 from row 2   ((5,(  1),(     2)),(0,(2/5),(−(1/5))),(4,(  2),(     7)) )    and Subtract (4/5)×row 1 from row 3   ((5,(  1),(      2)),(0,(2/5),(−(1/5))),(0,(6/5),(    ((27)/5))) )  swap row 2 and row 3   ((5,1,(    2)),(0,(6/5),(    ((27)/5))),(0,(2/5),(−(1/5))) )  subtract (1/3)×row 2 from row 3   ((5,1,2),(0,(6/5),((27)/5)),(0,0,(−2)) )  Divide row 3 by −2   ((5,1,2),(0,(6/5),((27)/5)),(0,0,1) )  substract ((27)/5)× row3 from row2   ((5,1,2),(0,(6/5),0),(0,0,1) )  substract 2×row 3 from row 1   ((5,1,0),(0,(6/5),0),(0,0,1) )  multifly row 2 by (5/6)   ((5,1,0),(0,1,0),(0,0,1) )  divide row 1 by 5   ((1,0,0),(0,1,0),(0,0,1) )
$$\boldsymbol{\mathrm{E}}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{cus}\:\mathrm{A}=\begin{pmatrix}{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\: \\ $$$$\mathrm{swap}\:\mathrm{row}\:\mathrm{1}\:\mathrm{and}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{7}}\end{pmatrix}\:\mathrm{and}\:\mathrm{Subtract}\:\frac{\mathrm{3}}{\mathrm{5}}×\:\mathrm{row}\:\mathrm{1}\:\mathrm{from}\:\mathrm{row}\:\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\:\:\mathrm{1}}&{\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{4}}&{\:\:\mathrm{2}}&{\:\:\:\:\:\mathrm{7}}\end{pmatrix}\:\: \\ $$$$\mathrm{and}\:\mathrm{Subtract}\:\frac{\mathrm{4}}{\mathrm{5}}×\mathrm{row}\:\mathrm{1}\:\mathrm{from}\:\mathrm{row}\:\mathrm{3} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\:\:\mathrm{1}}&{\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\:\:\:\:\frac{\mathrm{27}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\mathrm{swap}\:\mathrm{row}\:\mathrm{2}\:\mathrm{and}\:\mathrm{row}\:\mathrm{3} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\:\:\:\:\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\:\:\:\:\frac{\mathrm{27}}{\mathrm{5}}}\\{\mathrm{0}}&{\frac{\mathrm{2}}{\mathrm{5}}}&{−\frac{\mathrm{1}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\mathrm{subtract}\:\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{row}\:\mathrm{2}\:\mathrm{from}\:\mathrm{row}\:\mathrm{3} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\frac{\mathrm{27}}{\mathrm{5}}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{Divide}\:\mathrm{row}\:\mathrm{3}\:\mathrm{by}\:−\mathrm{2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\frac{\mathrm{27}}{\mathrm{5}}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{substract}\:\frac{\mathrm{27}}{\mathrm{5}}×\:\mathrm{row3}\:\mathrm{from}\:\mathrm{row2} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{2}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{substract}\:\mathrm{2}×\mathrm{row}\:\mathrm{3}\:\mathrm{from}\:\mathrm{row}\:\mathrm{1} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{\mathrm{6}}{\mathrm{5}}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{multifly}\:\mathrm{row}\:\mathrm{2}\:\mathrm{by}\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\begin{pmatrix}{\mathrm{5}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{divide}\:\mathrm{row}\:\mathrm{1}\:\mathrm{by}\:\mathrm{5} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$
Answered by Spillover last updated on 01/May/25

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