find the canonical form of the quadratic form 2x12 +x22 + x32 +2x1x2 – 2x1x3 – 4x2x3
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Expert's answer
2022-01-20T08:39:47-0500
2x12+2x1x2−2x1x3+x22−4x2x3+x32=0a11=1,a12=−2,a13=−1,a14=0,a22=1,a23=1,a24=0,a33=2,a34=0,a44=0The variant of the equation when converting coordinates are determinants I1=a11+a22+a33I2=∣∣a11a12a12a22∣∣+∣∣a22a23a23a33∣∣+∣∣a11a13a13a33∣∣I3=∣∣a11a12a13a12a22a23a13a23a33∣∣I(λ)=∣∣a11−λa12a13a12a22−λa23a13a23a33−λ∣∣k2=∣∣a11a14a14a44∣∣+∣∣a12a24a24a44∣∣+∣∣a33a34a34a44∣∣k3=∣∣a11a12a14a12a22a24a14a24a44∣∣+∣∣a22a13a14a23a33a34a24a34a44∣∣+∣∣a11a13a14a13a33a34a14a34a44∣∣WehaveI1=4I2=∣∣1−2−21∣∣+∣∣1112∣∣+∣∣1−1−12∣∣=−1I3=∣∣1−2−1−211−112∣∣=−4I(λ)=∣∣1−λ−2−1−21−λ1−112−λ∣∣I(λ)=−λ3+4λ2+λ−4k2=0,k3=0λ3−4λ2−λ+4=0bysolvingthepolynomial,wegetλ1=4,λ2=1,λ3=−1thecanonicalformisx12−x22−4x32=0
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