证明:将题目中的n阶行列式记做An A1=a+B==,144=(+82-08=.命题成立 (2)假设n=k时命题成立,则n=k+1时 Ak+1l=(a+B))Ak|-aB Ak-lI (a+B) 命题成立 试用克莱姆法则解线性方程组 ∫x +x2+2x3+3x4=1 +3x2-x x1+2x2+3r3 解:方程组的系数行列式为 1123 A 153 23-1-1 141≠0,由克莱姆法则,方程组有如下唯一解: A2=|A A 6y²: ÚK8•n1™Pâ|An|. (1) n = 1, 2 |A1| = α + β = α 2−β 2 α−β , |A2| = (α + β) 2 − αβ = α 3−β 3 α−β . ·K§·. (2) bn = kû·K§·, Kn = k + 1û |Ak+1| = (α + β)|Ak| − αβ|Ak−1| = (α + β) α k+1 − β k+1 α − β − αβ α k − β k α − β = α k+2 − β k+2 α − β . ·K§·. 9 £^é40{K)Ç5êß|    x1 + x2 + 2x3 + 3x4 = 1 3x1 − x2 − x3 − 2x4 = −4 2x1 + 3x2 − x3 − x4 = −6 x1 + 2x2 + 3x3 − x4 = −4 . ): êß|XÍ1™è |A| =         1 1 2 3 3 −1 −1 −2 2 3 −1 −1 1 2 3 −1         = −153. |A| 6= 0, dé40{K, êß|kXeçò): x1 = |A1| |A| , x2 = |A2| |A| , x3 = |A3| |A| , x4 = |A4| |A| , 6