(i) A=[212−3153]
Let I=[1001]
Applying IA=A
[1001] [212−3153] =[212−3153]
Applying R1→2R1
[2001][212−3153]=[12−6153]
Applying R2→R2−2R1
[2−201][212−3153]=[10−615183]
Applying R1→R1+18390R2
[183186−2183901][212−3153]=[10015183]
applying R2→18315R2
[183186183−301839018315][212−3153]=[1001]
Sequence of elementary matrix are
[1001] [2001] [2−201][183186−2183901][183186183−301839018315] =A
(ii) Given equation are
x1+x2+x3=3
x1−x2+4x3=4
2x1+3x2−5x3=0
A=⎣⎡1121−1314−5⎦⎤ ,B=⎣⎡340⎦⎤ ,X=⎣⎡x1x2x3⎦⎤
Transforming matrix A to upper triangular matrix(U) using elementary row transformation and Lower triangular matrix(L) simultaneously
Applying R2→R2−R1
U=⎣⎡1021−2313−5⎦⎤ ,L=⎣⎡110010001⎦⎤
Appling R3→R3−2R1
U=⎣⎡1001−2113−7⎦⎤ ,L=⎣⎡112010001⎦⎤
Applying R3→R3+21R2
U=⎣⎡1001−20132−11⎦⎤ ,L=⎣⎡112013−1001⎦⎤
So LY=B
⎣⎡112010001⎦⎤⎣⎡y1y2y3⎦⎤=⎣⎡340⎦⎤
⎣⎡y1y1+y22y1+y3⎦⎤ =⎣⎡340⎦⎤
y1=2,y1+y2=4,y2=2
2y1+y3=0,y3=−2
Now UX=Y
⎣⎡1001−2113−7⎦⎤⎣⎡x1x2x3⎦⎤=⎣⎡22−2⎦⎤
x1+x2+x3=2,−2x2+3x3=2
x2−7x3=−2
Solving these equation we get,
x1=1128,x2=11−8,x3=112
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