linear algebra @ ohe09nz81r的部落格

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linear algebra

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http://x9e.xanga.com/aa7f5b7042134234698853/w185306877.jpg please provide detail procedure..I really don't have time to modify this....thank you so much

最佳解答:

Q#3 Let [aij] be the inverse of A therefore [k 0 0, 0 1 0, 0 0 1] * [a11 a12 a13, a21 a22 a23, a31 a32 a33] = [ 1 0 0, 0 1 0, 0 0 1] k*a11+0*a21+0*a31 = 1, or a11 = 1/k k*a12+0*a22+0*a32 = 0, or a12 = 0 k*a13+0*a23+0*a33 = 0, or a13 = 0 0*a11+1*a21+0*a31 = 0, or a21 = 0 0*a12+1*a22+0*a32 = 1, or a22 = 1 0*a13+1*a23+0*a33 = 0, or a23 = 0 0*a11+0*a21+1*a31=0, or a31 = 0 0*a12+0*a22+1*a32=0, or a32 = 0 0*a13+0*a23+1*a33=1, or a33 = 1 therefore inverse of A is [ 1/k 0 0, 0 1 0, 0 0 1] 2009-03-03 07:51:21 補充： Let [bij] be the inverse of B Therefore [1 0 0, 0 1 0, k 0 1] * [b11 b12 b13, b21 b22 b23, b31 b32 b33] = [1 0 0, 0 1 0, 0 0 1] 1*b11+0*b21+0*b31 = 1, or b11 = 1 1*b12+0*b22+0*b32 = 0, or b12 = 0 1*b13+0*b23+0*b33 = 0, or b13 = 0 0*b11+1*b21+0*b31 = 0, or b21 = 0 0*b12+1*b22+0*b32 = 1, or b22 = 1 2009-03-03 07:51:33 補充： 0*b13+1*b23+0*b33 = 0, or b23 = 0 k*b11+0*b21+1*b31 = 0, or k+b31 = 0, or b31 = -k k*b12+0*b22+1*b32 = 0, or b32 = 0 k*b13+0*b23+1*b33 = 1, or b33 = 1 Therefore inverse of B is [ 1 0 0, 0 1 0, -k 0 1]

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