Find the $2\times 2$ matrix $D$ such that $P^{-1}DP=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$

Question

Let $P = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $D$ such that $$P^{-1} DP = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$$

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I think matrix multiplication is not associative, so what else can I do?

Answer 1

Hint:

$$DP=PA\tag{A is that one in RHS}$$

Answer 2

Hint: The matrix $P$ is diagonalizable with distinct eigenvalues $1$ and $2$.

Answer 3

If $$P^{-1}DP=A$$ then $$D=PAP^{-1}$$ so you just need to calculate $P^{-1}$ and multiply.