1. Key Concept / Formula

The fundamental property linking the determinant of a matrix $A$ to the determinant of its adjoint, $\text{adj}(A)$, is crucial for solving this problem. For any square matrix $A$ of order $n \times n$, the determinant of its adjoint is given by: $|\text{adj}(A)| = |A|^{n-1}$ In this problem, $A$ is a $3 \times 3$ matrix, so $n=3$. Therefore, the formula simplifies to: $|\text{adj}(A)| = |A|^{3-1} = |A|^2$

2. Step-by-Step Derivation

We are given that $P$ is the adjoint of a $3 \times 3$ matrix $A$, i.e., $P = \text{adj}(A)$. We are also given the matrix $P$ and the determinant of $A$, $|A|=4$. Our goal is to find the value of $\alpha$.

Step 2.1: Calculate the determinant of matrix $P$

First, we need to find the determinant of the given matrix $P$. P = \left[ {\matrix{ 1 & \alpha & 3 \cr 1 & 3 & 3 \cr 2 & 4 & 4 \cr } } \right] We calculate $|P|$ using cofactor expansion along the first row: |P| = 1 \cdot \left| {\matrix{ 3 & 3 \cr 4 & 4 \cr } } \right| - \alpha \cdot \left| {\matrix{ 1 & 3 \cr 2 & 4 \cr } } \right| + 3 \cdot \left| {\matrix{ 1 & 3 \cr 2 & 4 \cr } } \right| Now, we evaluate the $2 \times 2$ determinants:

• \left| {\matrix{ 3 & 3 \cr 4 & 4 \cr } } \right| = (3 \times 4) - (3 \times 4) = 12 - 12 = 0
• \left| {\matrix{ 1 & 3 \cr 2 & 4 \cr } } \right| = (1 \times 4) - (3 \times 2) = 4 - 6 = -2

Substitute these values back into the expression for $|P|$: $|P| = 1(0) - \alpha(-2) + 3(-2)$ $|P| = 0 + 2\alpha - 6$ $|P| = 2\alpha - 6$ This expression gives the determinant of $P$ in terms of $\alpha$.

Step 2.2: Relate $|P|$ to $|A|$ using matrix properties

We are given that $P = \text{adj}(A)$. Therefore, the determinant of $P$ must be equal to the determinant of $\text{adj}(A)$: $|P| = |\text{adj}(A)|$ From the key concept discussed earlier, for a $3 \times 3$ matrix $A$, we know that: $|\text{adj}(A)| = |A|^2$ Combining these two equations, we get: $|P| = |A|^2$ We are given that $|A| = 4$. Substituting this value: $|P| = (4)^2$ $|P| = 16$

Step 2.3: Solve for $\alpha$

Now we have two expressions for $|P|$: one in terms of $\alpha$ ($|P| = 2\alpha - 6$) and one as a numerical value ($|P| = 16$). We can equate these two expressions to solve for $\alpha$: $2\alpha - 6 = 16$ Add 6 to both sides of the equation: $2\alpha = 16 + 6$ $2\alpha = 22$ Divide both sides by 2: $\alpha = \frac{22}{2}$ $\alpha = 11$

Thus, the value of $\alpha$ is 11.

3. Common Mistakes & Tips

• Incorrect Formula: A very common mistake is to use the wrong exponent in the formula $|\text{adj}(A)| = |A|^{n-1}$. Always remember that the exponent is $n-1$, not $n$ or $n+1$. For a $3 \times 3$ matrix, it's $3-1=2$.
• Determinant Calculation Errors: Be meticulous when calculating determinants, especially with signs and multiplications. A small arithmetic error can lead to a completely different answer. For example, $1(12-12)$ correctly evaluates to $0$, but a common slip might be to write $12-12=1$.
• Misinterpreting the Question: Ensure you correctly identify what is given ($P = \text{adj}(A)$, $|A|=4$) and what needs to be found ($\alpha$).
• Verifying the Order: Always note the order ($n$) of the matrix, as it directly impacts the formula for $|\text{adj}(A)|$.

4. Summary / Key Takeaway

This problem effectively tests your understanding of a fundamental property of matrices: the relationship between the determinant of a matrix and the determinant of its adjoint. By correctly applying the formula $|\text{adj}(A)| = |A|^{n-1}$ and performing accurate determinant calculations, we were able to solve for the unknown variable $\alpha$. The key steps involved calculating the determinant of the given matrix $P$, equating it to $|A|^2$ (since $P = \text{adj}(A)$ and $A$ is $3 \times 3$), and then solving the resulting linear equation for $\alpha$.

The final answer is $\boxed{\text{11}}$.