Research Space for Linear Algebra & Discrete Mathematics
這種猜反矩陣的, 題庫時講了不少, 把它看成純量1/(x+2)去猜, ...因為A^2 - A + I = O=> A^2 = A - I先猜(A + 2I)^(-1) = aA + bII = (A + 2I)(aA + bI)= aA^2 + (2a+b)A + 2bI= a(A - I) + (2a+b)A + 2bI= (3a + b)A + (2b - a)I得3a + b = 0, 2b - a = 1解得a = -1/7, b = 3/7因此(A + 2I)^(-1) = (-1/7)A+(3/7)I
張貼留言
1 則留言:
這種猜反矩陣的, 題庫時講了不少, 把它看成純量1/(x+2)去猜, ...
因為A^2 - A + I = O
=> A^2 = A - I
先猜(A + 2I)^(-1) = aA + bI
I = (A + 2I)(aA + bI)
= aA^2 + (2a+b)A + 2bI
= a(A - I) + (2a+b)A + 2bI
= (3a + b)A + (2b - a)I
得3a + b = 0, 2b - a = 1
解得a = -1/7, b = 3/7
因此(A + 2I)^(-1) = (-1/7)A+(3/7)I
張貼留言