# simu exam comment…

The inversion of a matrix.

For a change of basis in $\mathbb{R}^3$:

$\eta_1=\varepsilon_1$

$\eta_2=\varepsilon_1+2\varepsilon_2$

$\eta_3=\varepsilon_2-\varepsilon_3$

we have as a change-of-basis-matrix:

$\left(\begin{array}{ccc}1&1&0\\ 0&2&1\\ 0&0&-1\end{array}\right)$

and by solving for:

$\varepsilon_1=\eta_1$

$\varepsilon_2=-\frac{1}{2}\eta_1+\frac{1}{2}\eta_2$

$\varepsilon_2=-\frac{1}{2}\eta_1+\frac{1}{2}\eta_2-\eta_3$

then we get as an inverse of that matrix:

$\left(\begin{array}{ccc}1&-1/2&-1/2\\ 0&1/2&1/2\\ 0&0&-1\end{array}\right)$

All that inside a recent exam did at the dept of maths