Δ=−2⋅(6−1)−1⋅(−3−1)+1⋅(1+2)=−10+4+3=−3=0,
Δ1=a⋅(6−5)−1⋅(−3b−c)+1⋅(b+2c)=a+4b+3c,
Δ2=−2⋅(−3b−c)−a⋅(−3−1)+1⋅(c−b)=4a+5b+3c,
Δ3=−2⋅(−2c−b)−1⋅(c−b)+a⋅(1+2)=3a+3b+3c,
x=ΔΔ1=−3a−34b−c,
y=ΔΔ2=−34a−35b−c,
z=ΔΔ3=−a−b−c,
there is a solution for any a, b, c.
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