f=x₁²+2x₂²+5x₃²+2x₁x₂+2x₁x₃+8x₂x₃
=[x₁²+2x₁(x₂+x₃)]+2x₂²+5x₃²+8x₂x₃
=(x₁+x₂+x₃)²-(x₂+x₃)²+2x₂²+5x₃²+8x₂x₃
=(x₁+x₂+x₃)²+x₂²+6x₂x₃+4x₃²
=(x₁+x₂+x₃)²+(x₂+3x₃)²-5x₃²,
令 y₁=x₁+x₂+x₃,y₂=x₂+3x₃,y₃=x₃,
则 x₁=y₁-y₂+2y₃,x₂=y₂-3y₃,x₃=y₃,
故在此线性变换下
f=y₁²+y₂²-5y₃².
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