Determinant of a $$2\times2$$ Matrix

To find the least value of $$\det(A)$$, we use the properties of skew-symmetric matrices and substitute them into the updated matrix equation.

Step 1: Understand the property of matrix P

The given condition is:

$$P + P^T = 0 \implies P^T = -P$$

This means $$P$$ is a skew-symmetric matrix of order 2. Any real skew-symmetric matrix of order 2 can be expressed as:

$$P = \begin{pmatrix} 0 & k \\ -k & 0 \end{pmatrix}$$

where $$k$$ is a real number.

Step 2: Express matrix A in terms of P

From the second equation, isolate matrix $$A$$:

$$A + 2P = 3I$$

$$A = 3I - 2P$$

Step 3: Construct matrix A

Substitute the identity matrix $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ and the matrix $$P$$ into the expression for $$A$$:

$$A = 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - 2\begin{pmatrix} 0 & k \\ -k & 0 \end{pmatrix}$$

$$A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} - \begin{pmatrix} 0 & 2k \\ -2k & 0 \end{pmatrix}$$

$$A = \begin{pmatrix} 3 & -2k \\ 2k & 3 \end{pmatrix}$$

Step 4: Calculate the determinant of A

Evaluate $$\det(A)$$:

$$\det(A) = (3)(3) - (-2k)(2k)$$

$$\det(A) = 9 + 4k^2$$

Step 5: Find the minimum value

Since $$k$$ is a real number, its square is always non-negative ($$k^2 \ge 0$$), which means:

$$4k^2 \ge 0$$

Therefore:

$$\det(A) \ge 9$$

The least value of $$\det(A)$$ is $$9$$, which occurs when $$k = 0$$

Given

$$X=\begin{pmatrix}a&b\\ c&-a\end{pmatrix}$$

Then,

$$X^2=\begin{pmatrix}a^2+bc&ab-ab\\ ac-ac&a^2+bc\end{pmatrix}$$

$$=\begin{pmatrix}a^2+bc&0\\0&a^2+bc\end{pmatrix}$$

Since

$$X^2=I,$$

$$a^2+bc=1\quad\cdots(1)$$

Also,

$$\det(X)=-a^2-bc$$

Using

$$(1),$$

$$\det(X)=-1$$

But the question gives $$\det(X)=1,$$ which is impossible.

Hence no such matrix exists.

Therefore, the number of such matrices is

$$\boxed{0}$$