Prove that
$$\begin{vmatrix}a+b & b+c & c+a\\l+m & m+n & n+l\\p+q & q+r & r+p\end{vmatrix}\div \begin{vmatrix}a & b & c\\l & m & n\\p & q & r\end{vmatrix} = 2$$
I have tried solving this after expanding it fully but i cant seem to find an answer.
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the question is to prove that the given can be resolved into two – Joseph_Kandathil Jul 04 '17 at 17:34
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I withdrew my comment because I think I misread the question. The link is somewhat hard to read. I had read the arithmetic symbol as "$+$" but I expect it really is "$\div$". You still have to assume that at least the second determinant is non-zero. – lulu Jul 04 '17 at 17:41
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Learn some basic Mathjax for posting here. – StubbornAtom Jul 04 '17 at 18:03
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IT asks to do addition – Joseph_Kandathil Jul 04 '17 at 23:46
1 Answers
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Let $A = \begin{pmatrix} a & b & c \\ l & m & n \\ p & q & r \end{pmatrix} = \begin{pmatrix} C_1 & C_2 & C_3 \end{pmatrix}$ where $C_i$ is the $i$-th column.
Let $B = \begin{pmatrix} a+b & b+c & c+a \\ l+m & m+n & n+l \\ p+q & q+r & r+p \end{pmatrix}$
Then $B = \begin{pmatrix} C_1 + C_2 & C_2 + C_3 & C_3 + C_1 \end{pmatrix}$
And you're asked to show that $|B| = 2|A|$.
It's quite easy to see that $\begin{pmatrix} C_1+C_2 & C_2+C_3 & C_3 \end{pmatrix}$ and $A$ share the same determinant, now how can you turn $C_3$ into $C_3+C_1$ using the columns $C_1+C_2$ and $C_2+C_3$ ?
krirkrirk
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@Joseph_Kandathil No ; $|A | + |B| = 2$ is generally false, as shown by the example $A=I$ – krirkrirk Jul 05 '17 at 00:13