Recipe

Compute (or confirm) a similarity B = P^{-1}AP

Use when: you have A and an invertible P, and either need B=P^{-1}AP as a specific matrix, or need to confirm/deny a claimed B

Goal: either the actual matrix B, or a settled yes/no on a claimed similarity — whichever the problem needs

Taught in Similar matrices

1. 1

   Compute P^{-1} with the 2\times2 swap-negate-divide formula, or the general method for a larger matrix.

   Example

   For P=\begin{pmatrix}1&1\\1&2\end{pmatrix}: \det P = 1(2)-1(1)=1, so P^{-1}=\dfrac{1}{1}\begin{pmatrix}2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}.

   ⚡ Shortcut

   If the problem only asks you to CONFIRM a claimed similarity — a specific B is already given and you just need yes or no — skip inverting P altogether. Multiply B=P^{-1}AP on the left by P to get the equivalent, inverse-free condition AP=PB: compute both ordinary products and check they match. This is exactly the shortcut the second worked example below uses.

   Why: producing B as an actual matrix means undoing P's change of basis, and P^{-1} is the only thing that does that; there's no way around it when B itself is unknown.

2. 2

   Multiply A by P first — B=P^{-1}(AP) groups the product this way, so AP has to exist before P^{-1} can be applied.

   Example

   A=\begin{pmatrix}3&1\\0&2\end{pmatrix}, P=\begin{pmatrix}1&1\\1&2\end{pmatrix}: AP=\begin{pmatrix}3(1)+1(1)&3(1)+1(2)\\0(1)+2(1)&0(1)+2(2)\end{pmatrix}=\begin{pmatrix}4&5\\2&4\end{pmatrix}.

   Why: matrix multiplication isn't commutative, so computing AP before left-multiplying matches the definition's grouping exactly.

3. 3

   Multiply the P^{-1} from Step 1 by the AP from Step 2, entry by entry as dot products of rows with columns.

   Example

   P^{-1}(AP)=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}\begin{pmatrix}4&5\\2&4\end{pmatrix}=\begin{pmatrix}2(4)-1(2)&2(5)-1(4)\\-1(4)+1(2)&-1(5)+1(4)\end{pmatrix}=\begin{pmatrix}6&6\\-2&-1\end{pmatrix}

   Why: this executes the definition B=P^{-1}AP exactly, left to right — it's the only step that actually produces B's entries.

4. 4

   Multiply the B you just found by P and confirm you land back on the same AP computed in Step 2.

   Example

   PB=\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}6&6\\-2&-1\end{pmatrix}=\begin{pmatrix}1(6)+1(-2)&1(6)+1(-1)\\1(6)+2(-2)&1(6)+2(-1)\end{pmatrix}=\begin{pmatrix}4&5\\2&4\end{pmatrix}=AP\ \checkmark

   Why: this is the identity Step 1's shortcut already exploits — running it as a final check costs two products and catches an arithmetic slip in either P^{-1} or the multiplications before it reaches your final answer.