Key Concepts and Formulas

Before we dive into the solution, let's review some fundamental properties of matrices that are essential for solving this problem:

1. Transpose Properties: For matrices $A$ and $B$ of the same order, and a scalar $k$:

   • $(A+B)^T = A^T + B^T$ (Transpose of a sum is the sum of transposes)
   • $(kA)^T = k A^T$ (Transpose of a scalar multiple is the scalar multiple of the transpose)
   • $(A^T)^T = A$ (Transpose of a transpose is the original matrix)
   • $I^T = I$ (The identity matrix is symmetric)

2. Determinant Properties: For an $n \times n$ matrix $A$ and a scalar $k$:

   • $|kA| = k^n |A|$ (Determinant of a scalar multiple)
   • $|Adj(A)| = |A|^{n-1}$ (Determinant of the Adjoint matrix)

These properties will be instrumental in simplifying the given matrix equation and ultimately finding the required determinant.

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Step-by-Step Solution

We are given a $3 \times 3$ real matrix $P$ that satisfies the equation: $P^T = aP + (a-1)I \quad \text{(Equation 1)}$ where $a>1$ and $I$ is the $3 \times 3$ identity matrix. Our goal is to determine the value of $|Adj(P)|$. To achieve this, we first need to find the matrix $P$ itself, or at least its determinant $|P|$.

Step 1: Apply the Transpose Operation to the Given Equation To establish another relationship between $P$ and $P^T$, a common strategy in matrix problems is to take the transpose of the given equation. This often helps in creating a system of equations that can be solved for $P$.

Taking the transpose of both sides of Equation 1: $(P^T)^T = (aP + (a-1)I)^T$ Now, we apply the transpose properties listed above:

• $(P^T)^T = P$
• $(aP + (a-1)I)^T = (aP)^T + ((a-1)I)^T$ (using $(A+B)^T = A^T + B^T$)
• $(aP)^T = aP^T$ (using $(kA)^T = k A^T$)
• $((a-1)I)^T = (a-1)I^T = (a-1)I$ (using $(kA)^T = k A^T$ and $I^T = I$)

Substituting these back, we get: $P = aP^T + (a-1)I \quad \text{(Equation 2)}$

Step 2: Eliminate $P^T$ to Solve for $P$ We now have two equations involving $P$ and $P^T$. We can eliminate $P^T$ by substituting Equation 1 into Equation 2. This will give us an equation solely in terms of $P$, which we can then solve.

Substitute $P^T = aP + (a-1)I$ (from Equation 1) into Equation 2: $P = a(aP + (a-1)I) + (a-1)I$

Step 3: Simplify the Equation and Determine $P$ Now, we perform algebraic simplification to isolate $P$: $P = a^2 P + a(a-1)I + (a-1)I$ Factor out $(a-1)I$ from the last two terms: $P = a^2 P + (a(a-1) + (a-1))I$ $P = a^2 P + (a-1)(a+1)I$ $P = a^2 P + (a^2-1)I$ Gather all terms involving $P$ on one side: $P - a^2 P = (a^2-1)I$ Factor out $P$: $(1-a^2)P = (a^2-1)I$ We can rewrite $(1-a^2)$ as $-(a^2-1)$: $-(a^2-1)P = (a^2-1)I$ Since we are given that $a>1$, it implies $a^2 > 1$, and therefore $a^2-1 \neq 0$. This allows us to safely divide both sides by $(a^2-1)$: $-P = I$ Multiplying by $-1$, we find: $P = -I$ So, $P$ is the negative of the $3 \times 3$ identity matrix.

Step 4: Calculate the Determinant of $P$ To find $|Adj(P)|$, we first need the determinant of $P$. $|P| = |-I|$ For an $n \times n$ matrix $A$ and a scalar $k$, we know $|kA| = k^n|A|$. Here, $k=-1$ and $P$ is a $3 \times 3$ matrix, so $n=3$. $|P| = (-1)^3 |I|$ Since the determinant of the identity matrix $|I|$ is $1$: $|P| = (-1) \times 1$ $|P| = -1$

Step 5: Calculate the Determinant of the Adjoint of $P$ Finally, we use the property that for an $n \times n$ matrix $A$, the determinant of its adjoint is given by $|Adj(A)| = |A|^{n-1}$. For our matrix $P$, which is $3 \times 3$ (so $n=3$): $|Adj(P)| = |P|^{3-1}$ $|Adj(P)| = |P|^2$ Substitute the value of $|P| = -1$ that we just calculated: $|Adj(P)| = (-1)^2$ $|Adj(P)| = 1$

Step 6: Compare with the Options The calculated value $|Adj(P)| = 1$ matches option (A).

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Alternative Method (Element-wise Comparison)

While matrix algebra is generally more efficient, one could also solve this by equating individual elements. Let $P = [p_{ij}]$. Then $P^T = [p_{ji}]$. The given equation $P^T = aP + (a-1)I$ can be written in terms of elements as: $p_{ji} = a p_{ij} + (a-1)\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta (1 if $i=j$, 0 if $i \neq j$).

1. For diagonal elements ($i=j$): $p_{ii} = a p_{ii} + (a-1)\delta_{ii}$ Since $\delta_{ii} = 1$: $p_{ii} = a p_{ii} + (a-1)$ $p_{ii} - a p_{ii} = a-1$ $p_{ii}(1-a) = a-1$ Since $a>1$, $1-a \neq 0$, so we can divide by $(1-a)$: $p_{ii} = \frac{a-1}{1-a} = -1$ Thus, all diagonal elements of $P$ are $-1$.

2. For off-diagonal elements ($i \neq j$): $p_{ji} = a p_{ij} + (a-1)\delta_{ij}$ Since $\delta_{ij} = 0$ for $i \neq j$: $p_{ji} = a p_{ij}$

   Now, consider the transposed equation $P = aP^T + (a-1)I$. In element form for $i \neq j$: $p_{ij} = a p_{ji}$

   Substitute $p_{ji} = a p_{ij}$ into $p_{ij} = a p_{ji}$: $p_{ij} = a(a p_{ij})$ $p_{ij} = a^2 p_{ij}$ $p_{ij} - a^2 p_{ij} = 0$ $p_{ij}(1-a^2) = 0$ Since $a>1$, $a^2 > 1$, so $1-a^2 \neq 0$. Therefore, $p_{ij} = 0$ for all $i \neq j$.

Combining these results, $P = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} = -I$. The subsequent steps to calculate $|P|$ and $|Adj(P)|$ remain the same as in the main solution.

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Tips and Common Mistakes

• Prioritize Matrix Algebra: For problems involving matrix equations, it is almost always more efficient and less prone to errors to use matrix algebra properties (like transposing an entire equation) rather than comparing elements individually. The element-wise method can be tedious and increases the chance of minor calculation errors, especially with larger matrices.
• Master Transpose Properties: Ensure you correctly apply the properties of transpose, especially $(A+B)^T = A^T+B^T$, $(kA)^T=kA^T$, and $(A^T)^T = A$. Remember that $I^T=I$.
• Recall Determinant Properties: The formulas $|kA| = k^n|A|$ and $|Adj(A)| = |A|^{n-1}$ are fundamental for matrices and frequently appear in competitive exams. Memorizing and understanding their application can save significant time.
• Leverage Conditions: The condition $a>1$ is crucial. It ensures that expressions like $(1-a)$ and $(a^2-1)$ are non-zero, allowing you to divide by them without worrying about undefined operations. Always pay attention to such conditions provided in the problem.
• Don't Rush Simplification: Take your time with algebraic simplification steps. A small error in distributing or grouping terms can lead to a completely wrong result.

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Summary and Key Takeaway

This problem demonstrates a classic approach to solving matrix equations involving transposes. By applying the transpose operation to the given equation, we created a system of two equations. Eliminating $P^T$ allowed us to solve for $P$ in terms of the identity matrix. Once $P$ was found, its determinant $|P|$ was easily calculated, which then enabled us to find $|Adj(P)|$ using the standard formula. The key takeaway is the power of matrix algebra and the importance of knowing fundamental matrix and determinant properties.