例 2.37

求下列 $n$ 阶矩阵的逆矩阵 $(a_{i} \neq = 0)$ :

A = 1 + a_{1} 11 ⋮ 1 1 1 + a_{2} 1 ⋮ 1 11 1 + a_{3} ⋮ 1 \dots \dots \dots \dots 111 ⋮ 1 + a_{n}

解对 $(A; I_{n})$ 用初等变换法,将第 $i$ 行乘以 $a_{i}^{- 1} (i = 1, 2, \dots, n)$ ,有

1 + a_{1} 11 ⋮ 1 1 1 + a_{2} 1 ⋮ 1 11 1 + a_{3} ⋮ 1 \dots \dots \dots \dots 111 ⋮ 1 + a_{n} ⋮ 1 ⋮ 0 ⋮ 0 ⋮ 0 010 ⋮ 0 001 ⋮ 0 \dots \dots \dots \dots 000 ⋮ 1 \to

1 + \frac{1}{a _{1}} \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} \frac{1}{a _{1}} 1 + \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} \frac{1}{a _{1}} \frac{1}{a _{2}} 1 + \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} \dots \dots \dots \dots \frac{1}{a _{1}} \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ 1 + \frac{1}{a _{n}} \frac{1}{a _{1}} 00 ⋮ 0 0 \frac{1}{a _{2}} 0 ⋮ 0 00 \frac{1}{a _{3}} ⋮ 0 \dots \dots \dots ⋮ \dots 000 \frac{1}{a _{n}} ⋮ .

将下面的行都加到第一行上,并令 $s = 1 + \frac{1}{a _{1}} + \frac{1}{a _{2}} + \dots + \frac{1}{a _{n}}$ ,则上面的矩阵变为

s \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} s 1 + \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} s \frac{1}{a _{2}} 1 + \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} \dots \dots \dots \dots s \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ 1 + \frac{1}{a _{n}} \frac{1}{a _{1}} \frac{1}{1} \frac{1}{1} ⋮ \frac{1}{1} \frac{1}{a _{2}} 00 ⋮ 0 \frac{1}{a _{3}} \frac{1}{a _{2}} 0 ⋮ 0 \dots 0 \frac{1}{a _{3}} ⋮ 0 \frac{1}{a _{n}} \dots \dots \dots 00 ⋮ \frac{1}{a _{n}} \to

1 \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} 1 1 + \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} 1 \frac{1}{a _{2}} 1 + \frac{1}{a _{3}} ⋮ \frac{1}{a _{n}} \dots \dots \dots \dots 1 \frac{1}{a _{2}} \frac{1}{a _{3}} ⋮ 1 + \frac{1}{a _{n}} \frac{1}{s a _{1}} \frac{1}{0} \frac{1}{0} ⋮ 0 \frac{1}{s a _{2}} \frac{1}{a _{2}} 0 ⋮ 0 \frac{1}{s a _{3}} 00 ⋮ 0 \dots \dots \frac{1}{a _{3}} ⋮ \dots \frac{1}{s a _{n}} 0 \dots \frac{1}{a _{n}} 0 ⋮ \to

100 ⋮ 0 110 ⋮ 0 101 ⋮ 0 \dots \dots \dots \dots 100 ⋮ 1 \frac{1}{s a _{1}} - \frac{1}{s a _{2} a _{1}} - \frac{1}{s a _{3} a _{1}} ⋮ - \frac{1}{s a _{n} a _{1}} \frac{1}{s a _{2}} \frac{s a _{2} - 1}{s a _{2}^{2}} - \frac{1}{s a _{3} a _{2}} ⋮ - \frac{1}{s a _{n} a _{2}} \frac{1}{s a _{3}} - \frac{1}{s a _{2} a _{3}} \frac{s a _{3} - 1}{s a _{3}^{2}} ⋮ - \frac{1}{s a _{n} a _{3}} \dots \dots \dots ⋮ \dots \frac{1}{s a _{n}} - \frac{1}{s a _{2} a _{n}} - \frac{1}{s a _{3} a _{n}} \frac{s a _{n} - 1}{s a _{n}^{2}}

再消去第一行的 1 就得到

A^{- 1} = - \frac{1}{s} \frac{1 - s a _{1}}{a _{1}^{2}} \frac{1}{a _{2} a _{1}} \frac{1}{a _{3} a _{1}} ⋮ \frac{1}{a _{n} a _{1}} \frac{1}{a _{1} a _{2}} \frac{1 - s a _{2}}{a _{2}^{2}} \frac{1}{a _{3} a _{2}} ⋮ \frac{1}{a _{n} a _{2}} \frac{1}{a _{1} a _{3}} \frac{1}{a _{2} a _{3}} \frac{1 - s a _{3}}{a _{3}^{2}} ⋮ \frac{1}{a _{n} a _{3}} \dots \dots \dots \dots \frac{1}{a _{1} a _{n}} \frac{1}{a _{2} a _{n}} \frac{1}{a _{3} a _{n}} ⋮ \frac{1 - s a _{n}}{a _{n}^{2}} . □

Youliang Zhong