想知道4*4方陣的秩的計算過程

Question

[....1......5......3.......2..]

[....1/5....1.....1/4....7..]

[....1/3....4......1......9..]

[....1/2....1/7....1/9...1..]

想了解這個陣列的秩的計算過程

謝謝

Answer 1

最佳解答

想了解這個陣列的秩的計算過程謝謝|1....5....3....2|...a

|1/5..1....1/4..7|...b*20

|1/3..4....1....9|...c*3

|1/2..1/7..1/9..1|...d*126=|1...5...3...2..|...a-c

.|4...20..5...140|...b

.|1...12..3...27.|...c

.|63..18..14..126|...d=|1...5...3...2..|...a

.|4...20..5...140|...b

.|0...7...0...25.|...c

.|63..18..14..126|...d

..p...q...r...s......-25q+7s=|1...5...3..-111|...a

.|4...20..5...480|...b

.|0...7...0...0..|...c

.|63..18..14..432|...d=|1...3..-111|

.|4...5...480|*(-7)

.|63..14..432|=-7(5*432+189*480-56*111+315*111-12*432-14*480)=-7(-7*432+175*480+259*111)=-7(-1728+84000+28749)=-777147=\=0Rank=4........ans

Answer 2

數字蠻大的,版主題目沒抄錯吧,

如果麻辣大計算沒錯,行列式值不為0,那rank就等於4,

但萬一行列式值為0, rank<4,

還是要回到列運算的基本做法.