Solution:
Let E be the identity matrix:
E=⎝⎛10⋮001⋮0……⋱…00⋮1⎠⎞=⎝⎛…Ei…Ej…⎠⎞.
Let Pij be the permutation matrix:
Pij=⎝⎛…Ej…Ei…⎠⎞.
Then the result of left multiplication of the matrix Pij by the matrix
A=⎝⎛…Ai…Aj…⎠⎞
is the matrix obtained from the original matrix A by permuting its i-th and j-th rows:
PijA=⎝⎛…Aj…Ai…⎠⎞.
For example,
P23A=⎝⎛100001010⎠⎞⎝⎛a11a21a31a12a22a32a13a23a33⎠⎞=⎝⎛a11a31a21a12a32a22a13a33a23⎠⎞.
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5.1) E1A=B
E1⎝⎛−2 387 4 1 115⎠⎞=⎝⎛83−2 1 4 7 511⎠⎞.
Pij=P13⇒E1=P13=⎝⎛001010100⎠⎞.
Check:
⎝⎛001010100⎠⎞⎝⎛−2 387 4 1 115⎠⎞=⎝⎛83−2 1 4 7 511⎠⎞.
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5.2) E1B=A
E1⎝⎛83−2 1 4 7 511⎠⎞=⎝⎛−2 387 4 1 115⎠⎞.
Pij=P13⇒E1=P13=⎝⎛001010100⎠⎞.
Check:
⎝⎛001010100⎠⎞⎝⎛83−2 1 4 7 511⎠⎞=⎝⎛−2 387 4 1 115⎠⎞.
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Let A=⎝⎛…Ai…Aj…⎠⎞,A=⎝⎛…Ai+λAj…Aj…⎠⎞, then
A=Nij(λ)⎝⎛…Ai…Aj…⎠⎞=⎝⎛10⋮0⋮001⋮0⋮0……⋱…⋱…00⋮1⋮0……⋱…⋱…00⋮λ⋮0……⋱…⋱…00⋮0⋮1⎠⎞⎝⎛…Ai…Aj…⎠⎞=⎝⎛…Ai+λAj…Aj…⎠⎞. --------------------------------------------------------------------------------------------------------------------
5.3) E2A=C
E2⎝⎛−2 387 4 1 115⎠⎞=⎝⎛−2 3274−7 113⎠⎞.
1-st occasion:
(2−73)=(815)+λ(−271).
⎩⎨⎧2=8−2λ,−7=1+7λ,3=5+λ.
λ∈∅.
2-nd occasion:
(2−73)=(815)+λ(341).
⎩⎨⎧2=8+3λ,−7=1+4λ,3=5+λ.
λ=−2.
i=3,j=2,λ=−2⇒Nij(λ)=N32(−2).
E2=N32(−2)=⎝⎛10001−2001⎠⎞.
Check:
⎝⎛10001−2001⎠⎞⎝⎛−2 387 4 1 115⎠⎞=⎝⎛−2 3274−7 113⎠⎞.
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5.4) E3C=A
E3⎝⎛−23274−7113⎠⎞=⎝⎛−238741115⎠⎞.
1-st occasion:
(815)=(2−73)+λ(−271).
⎩⎨⎧8=2−2λ,1=−7+7λ,5=3+λ.
λ∈∅.
2-nd occasion:
(815)=(2−73)+λ(341).
⎩⎨⎧8=2+3λ,1=−7+4λ,5=3+λ.
λ=2.
i=3,j=2,λ=2⇒Nij(λ)=N32(2).
E3=N32(2)=⎝⎛100012001⎠⎞.
Check:
⎝⎛100012001⎠⎞⎝⎛−23274−7113⎠⎞=⎝⎛−238741115⎠⎞.
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Answer:
E1=⎝⎛001010100⎠⎞,E2=⎝⎛10001−2001⎠⎞,E3=⎝⎛100012001⎠⎞
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