Krdytkyu

In the official solution of an university exam I see the following question:

Given the base vectors

$$ B = { begin{pmatrix}
-1 \
1
end{pmatrix} , begin{pmatrix}
1 \
1
end{pmatrix} } $$

and a second set of base vectors:
$$ B^* = { begin{pmatrix}
1 \
2
end{pmatrix} , begin{pmatrix}
2 \
3
end{pmatrix} } $$

give the equations of the change from B to B* basis and apply it to vector of components (3,-1) in B basis.

The official solution is:

$$ C(B,B^*) = begin{pmatrix}
-5 & 1 \
-3 & -1
end{pmatrix} $$

and vector in B* basis is (-16,-8)

The method has been to place side by side the two matrices and apply some row multiplication and additions till make diagonal first one.

This result surprises me. If I call T the matrix formed by the base vectors in B and $T^*$ the one for $B^*$, I think the result should be ${(T^*)}^{-1}T$, by the simple logic of first T to pass to cartesian coordinates and back to B^* with the inverse of its matrix. That gives me a result of:

$$ C(B,B^*) = begin{pmatrix}
5 & -1 \
-3 & 1
end{pmatrix} $$

and vector in B* equal to (16,-10).

Please, could you help me to known which one is the correct solution?

Your solution is correct. You can tell because the columns of the change of basis matrix should give the coordinates of the basis elements in $B$ in terms of the basis $B^{*}$.

Indeed, we have
$$begin{pmatrix}-1\ 1end{pmatrix}=5begin{pmatrix}1\2end{pmatrix}-3begin{pmatrix}2\3end{pmatrix}$$
and
$$begin{pmatrix}1\ 1end{pmatrix}=-begin{pmatrix}1\2end{pmatrix}+begin{pmatrix}2\3end{pmatrix}$$

You're right and the official solution is wrong.

The vector with components $3$ and $-1$ with respect to $B$ is
$$3begin{pmatrix}-1\1end{pmatrix} - 1begin{pmatrix}1\1end{pmatrix}= begin{pmatrix}-4\2end{pmatrix}$$
in standard coordinates.

The vector with components $-16$ and $-8$ with respect to $B^*$ is
$$-16begin{pmatrix}1\2end{pmatrix} - 8begin{pmatrix}2\3end{pmatrix}= begin{pmatrix}-32\-56end{pmatrix}$$
in standard coordinates - definitely not a match.

The vector with components $16$ and $-10$ with respect to $B^*$ is
$$16begin{pmatrix}1\2end{pmatrix} - 10begin{pmatrix}2\3end{pmatrix}= begin{pmatrix}-4\2end{pmatrix}$$
in standard coordinates. Confirmed.

The method used in the solution is basically sound. As you say, the change-of-basis matrix should be $(T^*)^{-1}T$, which you can obtain by concatenating the two matrices and row-reducing: $$left[begin{array}{c|c} T^* & T end{array}right] to left[begin{array}{c|c} I & (T^*)^{-1}T end{array}right].$$ Comparing the correct and incorrect results, I would guess that a simple sign error was made somewhere along the way in the given solution.