3
$\begingroup$
  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?

  • 0
    @noname Why do you think $B$ is necessarily diagonal?2012-11-04

1 Answers 1

2

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.

  • 0
    now I find the answer. thanks for your help and kind explanation.2012-11-04