求一道线性代数题目的解答已知二次型f(x1,x2,x3)=x1^2+(a+4)x2^2+3x3^2+4x1x2-4x2x3经正交变换x=Qy化为标准型by1^2+5y2^2-y3^2求a,b及所用的正交变换.
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![求一道线性代数题目的解答已知二次型f(x1,x2,x3)=x1^2+(a+4)x2^2+3x3^2+4x1x2-4x2x3经正交变换x=Qy化为标准型by1^2+5y2^2-y3^2求a,b及所用的正交变换.](/uploads/image/z/10038890-2-0.jpg?t=%E6%B1%82%E4%B8%80%E9%81%93%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E9%A2%98%E7%9B%AE%E7%9A%84%E8%A7%A3%E7%AD%94%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%9E%8Bf%28x1%2Cx2%2Cx3%29%3Dx1%5E2%2B%28a%2B4%29x2%5E2%2B3x3%5E2%2B4x1x2-4x2x3%E7%BB%8F%E6%AD%A3%E4%BA%A4%E5%8F%98%E6%8D%A2x%3DQy%E5%8C%96%E4%B8%BA%E6%A0%87%E5%87%86%E5%9E%8Bby1%5E2%2B5y2%5E2-y3%5E2%E6%B1%82a%2Cb%E5%8F%8A%E6%89%80%E7%94%A8%E7%9A%84%E6%AD%A3%E4%BA%A4%E5%8F%98%E6%8D%A2.)
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