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

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Filed under algebra, mathematics, multilinear algebra, what is mathematics

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