Change of basis

Transition matrix.
$$P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$

Inverse. $\det P = -1 - 1 = -2$, so
$$P^{-1} = -\frac12\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$

New matrix.
$$P^{-1}AP = \frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$
Compute:
$$\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 3 & -1 \end{pmatrix},$$
then
$$\frac12\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 3 & -1 \end{pmatrix} = \frac12\begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}.$$

Conclusion. In the basis $\mathcal{B}'$, the matrix is $\operatorname{diag}(3,1)$.