返回

二、例题增补

例1.1 由行列式的定义计算f(x)中x4与x3的系数

其中f(x)=|2x131xx2111x12−11x|\left| \begin{array}{r} \begin{matrix} 2x \\ 1 \end{matrix} \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ x \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} x \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 2 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ x \end{matrix} \end{array} \right|

解:

|2x131xx2111x12−11x|=−|1132x1x2x11x1x−112|=−|10001x−1−1−x10x−31−2xx−1−x1−3x2−2x2|\left| \begin{array}{r} \begin{matrix} 2x \\ 1 \end{matrix} \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ x \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} x \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 2 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ x \end{matrix} \end{array} \right| = - \left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 3 \\ 2x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x \end{matrix} \\ \begin{matrix} 2 \\ x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} x \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ 2 \end{matrix} \end{array} \right| = - \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x - 1 \end{matrix} \\ \begin{matrix} - 1 \\ - x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} x - 3 \\ 1 - 2x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 - x \end{matrix} \\ \begin{matrix} 1 - 3x \\ 2 - 2x^{2} \end{matrix} \end{array} \right|

=|10001−1x−1−x1x−301−2xx1−3x−1−x2−2x2|=−|100011x−1−x1−x+301−2xx−1+3x−1−x2−2x2|= \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 1 \end{matrix} \\ \begin{matrix} x - 1 \\ - x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x - 3 \end{matrix} \\ \begin{matrix} 0 \\ 1 - 2x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ 1 - 3x \end{matrix} \\ \begin{matrix} - 1 - x \\ 2 - 2x^{2} \end{matrix} \end{array} \right| = - \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} x - 1 \\ - x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ - x + 3 \end{matrix} \\ \begin{matrix} 0 \\ 1 - 2x \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 + 3x \end{matrix} \\ \begin{matrix} - 1 - x \\ 2 - 2x^{2} \end{matrix} \end{array} \right|

=−|100011001−x+3(−x+3)(1−x)(−x+3)x+1−2xx−1+3x(−1+3x)(1−x)−1−x(−1+3x)x+2−2x2|= - \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - x + 3 \end{matrix} \\ \begin{matrix} ( - x + 3)(1 - x) \\ ( - x + 3)x + 1 - 2x \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 + 3x \end{matrix} \\ \begin{matrix} ( - 1 + 3x)(1 - x) - 1 - x \\ ( - 1 + 3x)x + 2 - 2x^{2} \end{matrix} \end{array} \right|

=−|100011001−x+3x2−4x+3−x2+x+1x−1+3x−3x2+3x−2x2−x+2|= - \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - x + 3 \end{matrix} \\ \begin{matrix} x^{2} - 4x + 3 \\ - x^{2} + x + 1 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 + 3x \end{matrix} \\ \begin{matrix} - 3x^{2} + 3x - 2 \\ x^{2} - x + 2 \end{matrix} \end{array} \right|

=−x2+4x−3|100011001−x+31−x2+x+1x−1+3x−3x2+3x−2x2−4x+3x2−x+2|= - x^{2} + 4x - 3\left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - x + 3 \end{matrix} \\ \begin{matrix} 1 \\ - x^{2} + x + 1 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 + 3x \end{matrix} \\ \begin{matrix} \frac{- 3x^{2} + 3x - 2}{x^{2} - 4x + 3} \\ x^{2} - x + 2 \end{matrix} \end{array} \right|

=−x2+4x−3|100011001−x+310x−1+3x−3x2+3x−2x2−4x+3(−3x2+3x−2)(x2−x−1)x2−4x+3+x2−x+2|= - x^{2} + 4x - 3\left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - x + 3 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} x \\ - 1 + 3x \end{matrix} \\ \begin{matrix} \frac{- 3x^{2} + 3x - 2}{x^{2} - 4x + 3} \\ \frac{\left( - 3x^{2} + 3x - 2 \right)\left( x^{2} - x - 1 \right)}{x^{2} - 4x + 3} + x^{2} - x + 2 \end{matrix} \end{array} \right|

((−3x2+3x−2)(x2−x−1)x2−4x+3+x2−x+2)(−x2+4x−3)\left( \frac{\left( - 3x^{2} + 3x - 2 \right)\left( x^{2} - x - 1 \right)}{x^{2} - 4x + 3} + x^{2} - x + 2 \right)\left( - x^{2} + 4x - 3 \right)

=((−3x2+3x−2)(x2−x−1)+(x2−x+2)(x2−4x+3)x2−4x+3)(−x2+4x−3)\left( \frac{\left( - 3x^{2} + 3x - 2 \right)\left( x^{2} - x - 1 \right) + \left( x^{2} - x + 2 \right)\left( x^{2} - 4x + 3 \right)}{x^{2} - 4x + 3} \right)\left( - x^{2} + 4x - 3 \right)

=−((−3x2+3x−2)(x2−x−1)+(x2−x+2)(x2−4x+3))- \left( \left( - 3x^{2} + 3x - 2 \right)\left( x^{2} - x - 1 \right) + \left( x^{2} - x + 2 \right)\left( x^{2} - 4x + 3 \right) \right)

=−((−3x2+3x−2)(x2−x−1)+(x2−x+2)(x2−4x+3))- \left( \left( - 3x^{2} + 3x - 2 \right)\left( x^{2} - x - 1 \right) + \left( x^{2} - x + 2 \right)\left( x^{2} - 4x + 3 \right) \right)

=−(−3x4+6x3−2x2−x+2+x4−5x3+9x2−11x+6)- \left( - 3x^{4} + 6x^{3} - 2x^{2} - x + 2 + x^{4} - 5x^{3} + 9x^{2} - 11x + 6 \right)

=2x4−x3−7x2+12x−82x^{4} - x^{3} - 7x^{2} + 12x - 8

x4的系数是2

x3的系数是-1

例1.2计算五阶行列式D=|1234523451345124512351234|\left| \begin{array}{r} \begin{matrix} 1 \\ 2 \end{matrix} \\ \begin{matrix} 3 \\ 4 \\ 5 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 3 \end{matrix} \\ \begin{matrix} 4 \\ 5 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ 4 \end{matrix} \\ \begin{matrix} 5 \\ 1 \\ 2 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ 5 \end{matrix} \\ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ 1 \end{matrix} \\ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \end{array} \right|

解: 利用各行的元素之和相同的特点 , 把除第1列以外的各列加到第1列

得D=|151515151523451345124512351234|\left| \begin{array}{r} \begin{matrix} 15 \\ 15 \end{matrix} \\ \begin{matrix} 15 \\ 15 \\ 15 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 3 \end{matrix} \\ \begin{matrix} 4 \\ 5 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ 4 \end{matrix} \\ \begin{matrix} 5 \\ 1 \\ 2 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ 5 \end{matrix} \\ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ 1 \end{matrix} \\ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \end{array} \right|=15|1111123451345124512351234|\left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 3 \end{matrix} \\ \begin{matrix} 4 \\ 5 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ 4 \end{matrix} \\ \begin{matrix} 5 \\ 1 \\ 2 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ 5 \end{matrix} \\ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ 1 \end{matrix} \\ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \end{array} \right|

⇒r5−r4r4−r3r3−r2r2−r115|100002111−4311−4141−4115−4111|⇒按第一列展开15|111−411−411−411−4111|\overset{\begin{array}{r} \begin{matrix} r_{5} - r_{4} \\ r_{4} - r_{3} \end{matrix} \\ \begin{matrix} r_{3} - r_{2} \\ r_{2} - r_{1} \end{matrix} \end{array}}{\Rightarrow}15\left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \\ - 4 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 3 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ - 4 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 4 \\ 1 \end{matrix} \\ \begin{matrix} - 4 \\ 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 5 \\ - 4 \end{matrix} \\ \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \end{array} \right|\overset{按第一列展开}{\Rightarrow}15\left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ - 4 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} - 4 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 4 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} - 4 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array} \right|

⇒把各行加到第1行并提取第1行的公因子−1得−15|111−411−411−4111111|\overset{把各行加到第1行并提取第1行的公因子 - 1得}{\Rightarrow} - 15\left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ - 4 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} - 4 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 4 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array} \right|

⇒r2−r1r3−r1r4−r1−15|100−510−501−5001000|⇒按第4列展开15|00−50−50−500|=1875\overset{\begin{matrix} r_{2} - r_{1} \\ r_{3} - r_{1} \\ r_{4} - r_{1} \end{matrix}}{\Rightarrow} - 15\left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - 5 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} - 5 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 5 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right|\overset{按第4列展开}{\Rightarrow}15\left| \begin{matrix} 0 & 0 & - 5 \\ 0 & - 5 & 0 \\ - 5 & 0 & 0 \end{matrix} \right| = 1875

例1.3计算n阶行列式D=|a11⋮11a20⋯⋱10an|\left| \begin{array}{r} \begin{matrix} a_{1} \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \end{matrix} \\ \begin{matrix} \ddots \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} \\ a_{n} \end{matrix} \end{array} \right| , 其中a1a2⋯an≠0

解一: 通过把D的第1行中除(1 , 1)元a1外其他元素均变成为零

化D为下三角形行列式 , 具体如下:

D⇒r1−1a2r2⋯⋯⋯r1−1anrn|ba2*a3⋱an|\overset{\begin{matrix} r_{1} - \frac{1}{a_{2}}\ r_{2} \\ \cdots\cdots\cdots \\ r_{1} - \frac{1}{a_{n}}\ r_{n} \end{matrix}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} b \\ \end{matrix} \\ \begin{matrix} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ a_{2} \end{matrix} \\ \begin{matrix} \\ \ast \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} a_{3} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} \\ \ddots \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} \\ \\ a_{n} \end{matrix} \end{array} \right|=ba2a3⋯an

“∗”表示那里可能有一些非零元素 , 但对行列式的值没有影响

其中,b=a1−1a2−⋯1an,于是,D=a2a3⋯an(a1−∑i=2n1ai)其中\ ,\ b = a_{1} - \frac{1}{a_{2}}\ - \cdots\frac{1}{a_{n}}\ ,\ 于是\ ,\ D = a_{2}a_{3}\cdots a_{n}\left( a_{1} - \sum_{i = 2}^{n}\frac{1}{a_{i}} \right)

解二: 把D按第1行展开 , 由代数余子式

A1j=(-1)1+j|1⋮111⋮1a20⋱⋯aj−100aj+1⋯⋱0an|\left| \begin{array}{r} \begin{array}{r} \begin{matrix} 1 \\ \vdots \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array} \\ \begin{matrix} 1 \\ \vdots \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} a_{2} \\ \end{matrix} \\ \begin{matrix} \\ 0 \end{matrix} \end{array} \\ \begin{matrix} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} \\ \ddots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array} \\ \begin{matrix} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} a_{j - 1} \\ \ \ \ 0 \end{matrix} \end{array} \\ \begin{matrix} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} \\ 0 \end{matrix} \end{array} \\ \begin{matrix} a_{j + 1} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array} \\ \begin{matrix} \\ \ddots \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} \\ 0 \end{matrix} \end{array} \\ \begin{matrix} \\ \\ a_{n} \end{matrix} \end{array} \right|

⇒按第j−1行展开−\overset{按第j - 1行展开}{\Rightarrow} -a2⋯aj-1aj+1⋯an (j=2 , 3 , ⋯ , n)

得D=a1a2⋯an∑j=2na2a3⋯anaj得D = a_{1}a_{2}\cdots a_{n}\sum_{j = 2}^{n}\frac{a_{2}a_{3}\cdots a_{n}}{a_{j}}

当n=4时

|a11111a20010a30100a4|=|a1−1a21110a20010a30100a4|=|a1−1a2−1a31110a20000a30100a4|\left| \begin{array}{r} \begin{matrix} a_{1} \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} a_{1} - \frac{1}{a_{2}} \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} a_{1} - \frac{1}{a_{2}} - \frac{1}{a_{3}} \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right|

=|a1−1a2−1a3−1a41110a20000a30000a4|=a2a3a4(a1−1a2−1a3−1a4)= \left| \begin{array}{r} \begin{matrix} a_{1} - \frac{1}{a_{2}} - \frac{1}{a_{3}} - \frac{1}{a_{4}} \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right| = a_{2}a_{3}a_{4}(a_{1} - \frac{1}{a_{2}} - \frac{1}{a_{3}} - \frac{1}{a_{4}})

例1.4利用范德蒙德行列式计算四阶行列式

D=|aa2a3b+c+dbb2b3a+c+dcc2c3a+b+ddd2d3a+b+c|\left| \begin{array}{r} \begin{matrix} a \\ a^{2} \end{matrix} \\ \begin{matrix} a^{3} \\ b + c + d \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} b \\ b^{2} \end{matrix} \\ \begin{matrix} b^{3} \\ a + c + d \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} c \\ c^{2} \end{matrix} \\ \begin{matrix} c^{3} \\ a + b + d \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} d \\ d^{2} \end{matrix} \\ \begin{matrix} d^{3} \\ a + b + c \end{matrix} \end{array} \right|

解: D⇒r4+r1r4÷(a+b+c+d)(a+b+c+d)|aa2a31bb2b31cc2c31dd2d31|\overset{\begin{matrix} r_{4} + r_{1} \\ r_{4} \div (a + b + c + d) \end{matrix}}{\Rightarrow}(a + b + c + d)\left| \begin{array}{r} \begin{matrix} a \\ a^{2} \end{matrix} \\ \begin{matrix} a^{3} \\ 1 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} b \\ b^{2} \end{matrix} \\ \begin{matrix} b^{3} \\ 1 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} c \\ c^{2} \end{matrix} \\ \begin{matrix} c^{3} \\ 1 \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} d \\ d^{2} \end{matrix} \\ \begin{matrix} d^{3} \\ 1 \end{matrix} \end{array} \right|

把上式等号右边的行列式的最后一行依次与前面的行交换 , 共交换3次

得 D=−(a+b+c+d)|1aa2a31bb2b31cc2c31dd2d3|D = - (a + b + c + d)\left| \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} a^{2} \\ a^{3} \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ b \end{matrix} \\ \begin{matrix} b^{2} \\ b^{3} \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ c \end{matrix} \\ \begin{matrix} c^{2} \\ c^{3} \end{matrix} \end{array}\ \ \ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ d \end{matrix} \\ \begin{matrix} d^{2} \\ d^{3} \end{matrix} \end{array} \right|

此为4阶范德蒙德行列式 ,

得D=-(a+b+c+d)(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)

例1.5计算n+1阶行列式

Dn+1=|a1na2n⋮an+1na1n−1b1a2n−1b2⋮an+1n−1bn+1a1n−2b12a2n−2b22⋮an+1n−2bn+12⋯⋯⋯b1nb2n⋮bn+1n|\left| \begin{array}{r} \begin{matrix} a_{1}^{n} \\ a_{2}^{n} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n + 1}^{n} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{n - 1}b_{1} \\ a_{2}^{n - 1}b_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n + 1}^{n - 1}b_{n + 1} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{n - 2}b_{1}^{2} \\ a_{2}^{n - 2}b_{2}^{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n + 1}^{n - 2}b_{n + 1}^{2} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} b_{1}^{n} \\ b_{2}^{n} \end{matrix} \\ \begin{matrix} \vdots \\ b_{n + 1}^{n} \end{matrix} \end{array} \right|

解: 首先 , 我们恒可假设ai≠0 , i=1 , 2 , ⋯ , n+1

因若有某一ai=0 , 例如an+1=0

则将Dn+1按它的第n+1行展开 , 得到Dn+1=bn+1nDnb_{n + 1}^{n}D_{n}

若Dn中诸ai均不为零 , 则已如我们之假设 , 不然 , 对Dn作相同的讨论

其次考察此行列式 , 它的第i行元素是ai , bi的n次齐次函数

于是从第i行提取因子aina_{i}^{n}(i=1 , 2 , ⋯ , n+1)后

第i行 , 第j列元素成为(biai)j−1\left( \frac{b_{i}}{a_{i}} \right)^{j - 1} (j=1 , 2 , ⋯ , n+1 ; i=1 , 2 , ⋯ , n+1)

即有Dn+1=a1na2n⋯an+1n|11⋮1b1a1b2a2⋮bn+1an+1(b1a1)2(b2a2)2⋮(bn+1an+1)2⋯⋯⋯(b1a1)n(b2a2)n⋮(bn+1an+1)n|a_{1}^{n}a_{2}^{n}\cdots a_{n + 1}^{n}\left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} \frac{b_{1}}{a_{1}} \\ \frac{b_{2}}{a_{2}} \end{matrix} \\ \begin{matrix} \vdots \\ \frac{b_{n + 1}}{a_{n + 1}} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} \left( \frac{b_{1}}{a_{1}} \right)^{2} \\ \left( \frac{b_{2}}{a_{2}} \right)^{2} \end{matrix} \\ \begin{matrix} \vdots \\ \left( \frac{b_{n + 1}}{a_{n + 1}} \right)^{2} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} \left( \frac{b_{1}}{a_{1}} \right)^{n} \\ \left( \frac{b_{2}}{a_{2}} \right)^{n} \end{matrix} \\ \begin{matrix} \vdots \\ \left( \frac{b_{n + 1}}{a_{n + 1}} \right)^{n} \end{matrix} \end{array} \right|

上式等号右边的行列式为n+1阶范德蒙德行列式之转置行列式,

因此,Dn+1=a1na2n⋯an+1n∏1≤i<j≤n+1(bjaj−biai)=∏1≤i<j≤n+1(aibj−ajbi)因此\ ,\ D_{n + 1} = a_{1}^{n}a_{2}^{n}\cdots a_{n + 1}^{n}\prod_{1 \leqslant i < j \leqslant n + 1}^{}\left( \frac{b_{j}}{a_{j}} - \frac{b_{i}}{a_{i}} \right) = \prod_{1 \leqslant i < j \leqslant n + 1}^{}\left( a_{i}b_{j} - a_{j}b_{i} \right)

当n+1=4时

D4=|a13a23a33a43a12b1a22b2a32b3a42b4a11b12a21b22a31b32a41b42b13b23b33b43|=|a13b10a23b20a33b30a43b40a12b11a22b21a32b31a42b41a11b12a21b22a31b32a41b42a10b13a20b23a30b33a40b43|\left| \begin{array}{r} \begin{matrix} a_{1}^{3} \\ a_{2}^{3} \end{matrix} \\ \begin{matrix} a_{3}^{3} \\ a_{4}^{3} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{2}b_{1} \\ a_{2}^{2}b_{2} \end{matrix} \\ \begin{matrix} a_{3}^{2}b_{3} \\ a_{4}^{2}b_{4} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{1}b_{1}^{2} \\ a_{2}^{1}b_{2}^{2} \end{matrix} \\ \begin{matrix} a_{3}^{1}b_{3}^{2} \\ a_{4}^{1}b_{4}^{2} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} b_{1}^{3} \\ b_{2}^{3} \end{matrix} \\ \begin{matrix} b_{3}^{3} \\ b_{4}^{3} \end{matrix} \end{array}\ \ \right| = \left| \begin{array}{r} \begin{matrix} a_{1}^{3}b_{1}^{0} \\ a_{2}^{3}b_{2}^{0} \end{matrix} \\ \begin{matrix} a_{3}^{3}b_{3}^{0} \\ a_{4}^{3}b_{4}^{0} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{2}b_{1}^{1} \\ a_{2}^{2}b_{2}^{1} \end{matrix} \\ \begin{matrix} a_{3}^{2}b_{3}^{1} \\ a_{4}^{2}b_{4}^{1} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{1}b_{1}^{2} \\ a_{2}^{1}b_{2}^{2} \end{matrix} \\ \begin{matrix} a_{3}^{1}b_{3}^{2} \\ a_{4}^{1}b_{4}^{2} \end{matrix} \end{array}\ \ \ \ \ \ \begin{array}{r} \begin{matrix} a_{1}^{0}b_{1}^{3} \\ a_{2}^{0}b_{2}^{3} \end{matrix} \\ \begin{matrix} a_{3}^{0}b_{3}^{3} \\ a_{4}^{0}b_{4}^{3} \end{matrix} \end{array}\ \ \right|

=a13a23a33a43|1111b1a1b2a21b3a31b4a41(b1a1)2(b2a2)2(b3a3)2(b4a4)2(b1a1)3(b2a2)3(b3a3)3(b4a4)3|= a_{1}^{3}a_{2}^{3}a_{3}^{3}a_{4}^{3}\left| \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \frac{b_{1}}{a_{1}} \\ \frac{b_{2}}{a_{2}^{1}} \end{matrix} \\ \begin{matrix} \frac{b_{3}}{a_{3}^{1}} \\ \frac{b_{4}}{a_{4}^{1}} \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \left( \frac{b_{1}}{a_{1}} \right)^{2} \\ \left( \frac{b_{2}}{a_{2}} \right)^{2} \end{matrix} \\ \begin{matrix} \left( \frac{b_{3}}{a_{3}} \right)^{2} \\ \left( \frac{b_{4}}{a_{4}} \right)^{2} \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \left( \frac{b_{1}}{a_{1}} \right)^{3} \\ \left( \frac{b_{2}}{a_{2}} \right)^{3} \end{matrix} \\ \begin{matrix} \left( \frac{b_{3}}{a_{3}} \right)^{3} \\ \left( \frac{b_{4}}{a_{4}} \right)^{3} \end{matrix} \end{array}\ \right|

因此,D4=a13a23a33a43∏1≤i<j≤4(bjaj−biai)因此\ ,\ D_{4} = a_{1}^{3}a_{2}^{3}a_{3}^{3}a_{4}^{3}\prod_{1 \leqslant i < j \leqslant 4}^{}\left( \frac{b_{j}}{a_{j}} - \frac{b_{i}}{a_{i}} \right)

=a13a23a33a43•a1b2−a2b1a1a2•a1b3−a3b1a1a3•a1b4−a4b1a1a4•a2b3−a3b2a2a3•a2b4−a4b2a2a4= a_{1}^{3}a_{2}^{3}a_{3}^{3}a_{4}^{3} \bullet \frac{a_{1}b_{2} - a_{2}b_{1}}{a_{1}a_{2}} \bullet \frac{a_{1}b_{3} - a_{3}b_{1}}{a_{1}a_{3}} \bullet \frac{a_{1}b_{4} - a_{4}b_{1}}{a_{1}a_{4}} \bullet \frac{a_{2}b_{3} - a_{3}b_{2}}{a_{2}a_{3}} \bullet \frac{a_{2}b_{4} - a_{4}b_{2}}{a_{2}a_{4}}

•a3b4−a4b3a3a4\bullet \frac{a_{3}b_{4} - a_{4}b_{3}}{a_{3}a_{4}}

=∏1≤i<j≤4(aibj−ajbi)= \prod_{1 \leqslant i < j \leqslant 4}^{}\left( a_{i}b_{j} - a_{j}b_{i} \right)

=(a1b2−a2b1)(a1b3−a3b1)(a1b4−a4b1)(a2b3−a3b2)(a2b4−a4b2)= \left( a_{1}b_{2} - a_{2}b_{1} \right)\left( a_{1}b_{3} - a_{3}b_{1} \right)\left( a_{1}b_{4} - a_{4}b_{1} \right)\left( a_{2}b_{3} - a_{3}b_{2} \right)\left( a_{2}b_{4} - a_{4}b_{2} \right)

(a3b4−a4b3)\left( a_{3}b_{4} - a_{4}b_{3} \right)