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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Let $u,v,w$ be the three columns in a 3 x 3 - matrix $A$. Determinant to matrix $A$ can then be considered as a function of $u,v,w$. Assume that $\det(u,v,w)= \det(A)=4$ then find $\det(v+9w, w+3u, u+3v)$.

My progress:

I start with:

$\det(A+B)= \det(A)+\det(B)$

$\det(v+9w, w+3u, u+3v)=\det(v, w+3u,u+3v)+\det(9w, w+3u, u+3v) =$

$\det( v, w, u+3v) + \det(v, 3u, u+3v) + \det(9w, w, u+3v) + \det(9w, 3u, u+3v) = $

$\det(v, w, u) + \det(v, w , 3v) + \det(v, 3u, u) + \det(v, 3u, 3v) + \det(9w, w, u) + $

$\det(9w, w, 3v) + \det(9w, 3u, u) + \det(9w, u, 3v)$

My progress:

I start with:

$\det(A+B)= \det(A)+\det(B)$

$\det(v+9w, w+3u, u+3v)=\det(v, w+3u,u+3v)+\det(9w, w+3u, u+3v) =$

$\det( v, w, u+3v) + \det(v, 3u, u+3v) + \det(9w, w, u+3v) + \det(9w, 3u, u+3v) = $

$\det(v, w, u) + \det(v, w , 3v) + \det(v, 3u, u) + \det(v, 3u, 3v) + \det(9w, w, u) + $

$\det(9w, w, 3v) + \det(9w, 3u, u) + \det(9w, u, 3v)$

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