Q.
Let α,β,γ be three real numbers satisfying [αβγ]⎣⎡2−1−1−1−121−21⎦⎤=[000] . If the point A(α,β,γ) lies on the plane 2x+y+3z=2 , then 3α+2β−6γ is equal to
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Solution:
[αβγ]⎣⎡2−1−1−1−121−21⎦⎤=[000] [2α−β−γ−α−β+2γα−2β+γ]=[000] ⇒2α−β−γ=0 −α−β+2γ=0 α−2β+γ=0
Using cramer's rule
Δ=∣∣2−11−1−1−2−121∣∣=2(−1+4)+1(−1−2)−1(2+1)=6−3−3=0
Using cramer's rule Δ=∣∣2−11−1−1−2−121∣∣=2(−1+4)+1(−1−2)−1(2+1) =6−3−3=0
So, the system of equations has infinite solutions Put γ=k 2α−β=kα+β=2k}⇒γ=k,β=k ⇒(α,β,γ)≡(k,k,k) ∵(α,β,γ) satisfies 2x+y+3z=2 ⇒6k=2⇒k=31 (α,β,γ)≡(31,31,31) ⇒3α+2β−6γ=−31