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標題:
發問:
A is 3 x 3 matrix. B is a 3 x 1 matrix. If AB = 0, is it true that det A = 0 ? Please prove. 更新: To : YA HOO ! 知識+管理員 If B is not a zero matrix, will it be true that det A = 0 ? Please give example, thanks.
最佳解答:
No. It may not be true. For example, A= [ 1 0 0 ] [ 0 1 0 ] [ 0 0 1 ] B= [ 0 ] [ 0 ] [ 0 ] AB = [ 1 0 0 ][ 0 ] [ 0 1 0 ][ 0 ] [ 0 0 1 ][ 0 ] = [ 0 ] [ 0 ] [ 0 ] ∴AB = 0 However, det A=1≠0 ∴No, it isn't true. 2014-06-09 20:21:19 補充： 如果改了題目，最簡單就係用 老虎貓 知識長在意見區回答過嘅方法（Contradiction）作答。 Assume that det A ≠ 0. ∴A^(-1) exists. AB = 0 A^(-1) AB = A^(-1) 0 B = 0 Since B is not a zero matrix, but in the above calculation, B = 0. ∴It has a contradiction. ∴If AB = 0 and B is not a zero matrix, "det A = 0" is true. 2014-06-09 20:27:20 補充： 老虎貓 知識長在意見區也用了不同的方法作答， 你也可在意見區看看。
其他解答:
［ａｂｃ］［ｘ］ Ａ＝［ｄｅｆ］Ｂ＝［ｙ］ ［ｇｈｉ］［ｚ］ ＡＢ＝０ ［ａ］［ｂ］［ｃ］［０］ ｘ［ｄ］＋ｙ［ｅ］＋ｚ［ｆ］＝［０］ ［ｇ］［ｈ］［ｉ］［０］ 2014-06-09 11:39:58 補充： WLOG, let z ≠ 0, since B is a non-zero vector. ［ｃ］［ａ］［ｂ］ ［ｆ］＝－ｘ／ｚ［ｄ］－ｙ／ｚ［ｅ］ ［ｉ］［ｇ］［ｈ］ Thus, for matrix A, one of its columns is a linear combination of the other two columns. That means, det(A) = 0. Therefore, the argument is proved, and the answer is YES. 2014-06-09 11:43:16 補充： Or if you do not like the above direct proof. I give you a proof by contradiction. To prove that for B being a non-zero vector, AB = 0 ? det(A) = 0. Assume that det(A) ≠ 0, then A?1 exists. AB = 0 A?1(AB) = A?1(0) B = 0 Contradiction!!! Thus, AB = 0 (with B ≠ 0) ? det(A) = 0.
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Matrix發問:
A is 3 x 3 matrix. B is a 3 x 1 matrix. If AB = 0, is it true that det A = 0 ? Please prove. 更新: To : YA HOO ! 知識+管理員 If B is not a zero matrix, will it be true that det A = 0 ? Please give example, thanks.
最佳解答:
No. It may not be true. For example, A= [ 1 0 0 ] [ 0 1 0 ] [ 0 0 1 ] B= [ 0 ] [ 0 ] [ 0 ] AB = [ 1 0 0 ][ 0 ] [ 0 1 0 ][ 0 ] [ 0 0 1 ][ 0 ] = [ 0 ] [ 0 ] [ 0 ] ∴AB = 0 However, det A=1≠0 ∴No, it isn't true. 2014-06-09 20:21:19 補充： 如果改了題目，最簡單就係用 老虎貓 知識長在意見區回答過嘅方法（Contradiction）作答。 Assume that det A ≠ 0. ∴A^(-1) exists. AB = 0 A^(-1) AB = A^(-1) 0 B = 0 Since B is not a zero matrix, but in the above calculation, B = 0. ∴It has a contradiction. ∴If AB = 0 and B is not a zero matrix, "det A = 0" is true. 2014-06-09 20:27:20 補充： 老虎貓 知識長在意見區也用了不同的方法作答， 你也可在意見區看看。
其他解答:
［ａｂｃ］［ｘ］ Ａ＝［ｄｅｆ］Ｂ＝［ｙ］ ［ｇｈｉ］［ｚ］ ＡＢ＝０ ［ａ］［ｂ］［ｃ］［０］ ｘ［ｄ］＋ｙ［ｅ］＋ｚ［ｆ］＝［０］ ［ｇ］［ｈ］［ｉ］［０］ 2014-06-09 11:39:58 補充： WLOG, let z ≠ 0, since B is a non-zero vector. ［ｃ］［ａ］［ｂ］ ［ｆ］＝－ｘ／ｚ［ｄ］－ｙ／ｚ［ｅ］ ［ｉ］［ｇ］［ｈ］ Thus, for matrix A, one of its columns is a linear combination of the other two columns. That means, det(A) = 0. Therefore, the argument is proved, and the answer is YES. 2014-06-09 11:43:16 補充： Or if you do not like the above direct proof. I give you a proof by contradiction. To prove that for B being a non-zero vector, AB = 0 ? det(A) = 0. Assume that det(A) ≠ 0, then A?1 exists. AB = 0 A?1(AB) = A?1(0) B = 0 Contradiction!!! Thus, AB = 0 (with B ≠ 0) ? det(A) = 0.
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