## thomas5267 one year ago Two proofs needed: 1. Prove or provide a counterexample that matrices of the following form must be singular. $\begin{pmatrix} n&n+1&n+2\\ n+3&n+4&n+5\\ n+6&n+7&n+8 \end{pmatrix}$ 2. Prove the triple product identity. $\underline{a}\times (\underline{b}\times \underline{c}) = \underline{b}(\underline{a}\cdot\underline{c}) - \underline{c}(\underline{a}\cdot\underline{b})$

1. ganeshie8

for first question try below : R2 - R1 R3 - R1

2. thomas5267

It is very unfortunate that the lecutrer try to use the vectors $\underline{a}=\begin{pmatrix}1&2&3\end{pmatrix}^T\\ \underline{b}=\begin{pmatrix}4&5&6\end{pmatrix}^T\\ \underline{c}=\begin{pmatrix}7&8&9\end{pmatrix}^T$ to demonstrate the scalar triple product. He wanted to demonstrate that the scalar triple product corresponded to the volume of the parallelepiped of formed by that three vectors. As $\underline{a}\cdot(\underline{b}\times\underline{c})= \begin{vmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{vmatrix}$ and matrix of the form $\begin{pmatrix} n&n+1&n+2\\ n+3&n+4&n+5\\ n+6&n+7&n+8 \end{pmatrix}$ are always singular, you can see how this goes!

3. thomas5267

Proven by standard Gaussian elimination. I was expecting some ingenious prove without using it.

4. thomas5267

*proofs