Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. If $B=M^{-1} AM$, why is $\det B=\det A$? Show also that $\det A^{-1}B=1$.

  2. If the points $(x,y,z)$, $(2,1,0)$ and $(1,1,1)$ lie on a plane through the origin, what determinant is zero? Are the vectors $(1,0,-1)$, $(2,1,0)$, $(1,1,1)$ independent?

share|cite|improve this question
Are you aware of the fact that $\det(C_1 C_2) = \det(C_1) \det(C_2)$? – user17762 Nov 4 '12 at 3:32
For the second part of problem 1, you'll need to assume that $A$ is invertible. – Cameron Buie Nov 4 '12 at 3:33
I know B is a diagonal matrix, but I'm not sure why detB=detA. detB=det(M−1)det(A)det(M), the lhs is "product of the pivots" but the rhs?? I'm not sure about that. – noname Nov 4 '12 at 3:37
@noname Why do you think $B$ is necessarily diagonal? – EuYu Nov 4 '12 at 3:48

Here's a hint for part one:

$$\det ABC = \det A \det B \det C.$$

The determinant is a real number, and so satisfies the commutative property.

Finally, $AA^{-1} = I$. $\det I = 1$. Use the first hint to show what $\det A^{-1}$ is.

share|cite|improve this answer
now I find the answer. thanks for your help and kind explanation. – noname Nov 4 '12 at 3:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.