Animated Solution for Mathematics - Matrices and Determinants: If $P$ is a $3\times 3$ real matrix such that $P^T=aP+(a-1)I$, where $a>1$, then

Welcome, fellow traveler of the mathematical landscape. Today, we confront a matrix equation that seems to hide its true nature behind a veil of transposes and identity matrices.

We are given a $3\times 3$ real matrix $P$ satisfying the equation $P^T=aP+(a-1)I$, with the condition $a>1$. At first glance, this looks like a tangled mess, but in the world of linear algebra, every equation is a story waiting to be told.

When you see a matrix equation involving its transpose, think of it as a mirror. The equation $P^T=aP+(a-1)I$ gives us one perspective. By taking the transpose of the entire equation, we obtain:

Using the fundamental properties of transposes—specifically that $(P^T{)}^T=P$ and that the transpose distributes over addition—this simplifies beautifully to:

Now, we have two equations: the original and this new, mirrored version. This is the key to our breakthrough.

We want to isolate $P$. We have an expression for $P^T$ from our first equation, so let us substitute it into our second equation:

Expanding this carefully, we multiply the scalar $a$ through the brackets:

Grouping the terms with the identity matrix $I$, we see a common factor of $(a-1)$:

We bring all the $P$ terms to one side to solve for the matrix:

Here is where the condition $a>1$ saves us. Because $a>1$, we know that $a^2>1$, which means $1-a^2$ is non-zero. We can safely divide by $(1-a^2)$:

How elegant! The entire complex equation collapses into the simple identity matrix, scaled by $-1$.

Now that we know $P=-I$, finding the determinant is straightforward. For a $3\times 3$ matrix, the determinant of $-I$ is $(-1{)}^3|I|=-1$.

Finally, we need the determinant of the adjoint of $P$. Recall the property:

With $n=3$, this becomes $|P{|}^2$. Substituting our value of $|P|=-1$, we get:

The complexity vanishes, leaving behind a clean, solid result. The final answer is 1.