91影视

Two matrices A and B are similar if there exists a nonsingular matrix P such that \(B = P^{-1}AP\). Since A is similar to B, we have \(B = P^{-1}AP\), where P is a nonsingular matrix.

We are given that A is nonsingular, meaning that A has an inverse, denoted as A鈦宦�. Now let's find the inverse of B. We can multiply both sides of the equation \(B = P^{-1}AP\) by their inverse matrices, i.e., first multiply both sides by B鈦宦�, and then multiply by P. Beginning with the left side, B鈦宦笲: \(B^{-1}B = B^{-1}(P^{-1}AP)\) Next, multiply both sides by P: \((B^{-1}B)P = (B^{-1}(P^{-1}AP))P\) Now, simplify: \(B^{-1}BP = B^{-1}(P^{-1}APP)\) Since BP = BP, the equation becomes: \(B^{-1} = B^{-1}(P^{-1}A)\) Now, multiply both sides of the equation by A鈦宦�: \(B^{-1}A^{-1} = B^{-1}(P^{-1}A)A^{-1}\) Simplifying, we get: \(B^{-1}A^{-1} = B^{-1}(P^{-1})\) Finally, multiply both sides by P鈦宦� to get: \(B^{-1}(A^{-1}P^{-1}) = B^{-1}\) This equation shows that B鈦宦� exists, and therefore B is nonsingular.

Now, we need to show that A鈦宦� and B鈦宦� are similar. We can do this by finding a relationship between A鈦宦� and B鈦宦� using the matrix P. We know already that \(B = P^{-1}AP\), and we know that A and B are nonsingular. Therefore, we can multiply both sides of the equation by A鈦宦� and P鈦宦�. First, multiply both sides by A鈦宦�: \(A^{-1}B = A^{-1}(P^{-1}AP)\) Next, multiply both sides by P鈦宦�: \(A^{-1}BP^{-1} = A^{-1}(P^{-1}AP)P^{-1}\) Now simplify: \(A^{-1}(BP^{-1}) = A^{-1}(P^{-1}A)\) We can rewrite this equation: \(BP^{-1} = P^{-1}A\) Now, multiply both sides of the equation by B鈦宦�: \(B^{-1}BP^{-1} = B^{-1} (P^{-1}A)\) Simplifying, we get: \(B^{-1} = P^{-1}AA^{-1}\) Finally, multiply both sides by P: \(PB^{-1} = P(P^{-1}AA^{-1})\) Simplifying, we get: \(PA^{-1} = B^{-1}P\), which implies that \(B^{-1} =PAPA^{-1}\), by substituting A鈦宦�. That is, the inverse matrices A鈦宦� and B鈦宦� are similar as required.