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  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?

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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.

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

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