Key Concept: Properties of Adjoint and Determinant

This problem leverages a fundamental property relating the determinant of a matrix to the determinant of its adjoint. For an $n \times n$ square matrix $M$, the determinant of its adjoint, $Adj(M)$, is given by: $|Adj(M)| = |M|^{n-1}$ When dealing with nested adjoints, such as $Adj(Adj(A))$, we apply this property iteratively. Let $M_1 = Adj(A)$. Then, $|Adj(Adj(A))| = |M_1|^{n-1}$ Now, substitute the formula for $|M_1| = |Adj(A)| = |A|^{n-1}$ into the expression: $|Adj(Adj(A))| = (|A|^{n-1})^{n-1} = |A|^{(n-1)^2}$ In this problem, the given matrix $Adj(Adj(A))$ is a $3 \times 3$ matrix. This implies that the original matrix $A$ must also be a $3 \times 3$ matrix. Thus, $n=3$. Therefore, for this specific problem, the key relation becomes: $|Adj(Adj(A))| = |A|^{(3-1)^2} = |A|^{2^2} = |A|^4$

Step-by-Step Solution

Step 1: Establish the relationship between $|Adj(Adj(A))|$ and $|A|$ Based on the key concept for an $n \times n$ matrix $A$, we know that $|Adj(Adj(A))| = |A|^{(n-1)^2}$. Since the given matrix $Adj(Adj(A))$ is a $3 \times 3$ matrix, the original matrix $A$ must also be a $3 \times 3$ matrix, meaning $n=3$. Substituting $n=3$ into the formula: $|Adj(Adj(A))| = |A|^{(3-1)^2} = |A|^{2^2} = |A|^4$ This equation is crucial as it connects the determinant of the given matrix directly to the determinant of $A$, which we need to find.

Step 2: Calculate the determinant of the given matrix $Adj(Adj(A))$ We are given the matrix: Adj(Adj(A)) = \left( {\matrix{ {14} & {28} & { - 14} \cr { - 14} & {14} & {28} \cr {28} & { - 14} & {14} \cr } } \right) Let's find its determinant. To simplify the calculation, we observe that 14 is a common factor in every element of each row. We can factor out 14 from the first row ($R_1$), 14 from the second row ($R_2$), and 14 from the third row ($R_3$). Explanation: When a common factor $k$ is taken out from each of the $n$ rows of an $n \times n$ matrix, the determinant of the matrix is multiplied by $k^n$. |Adj(Adj(A))| = (14)(14)(14) \left| {\matrix{ {14/14} & {28/14} & { - 14/14} \cr { - 14/14} & {14/14} & {28/14} \cr {28/14} & { - 14/14} & {14/14} \cr } } \right| |Adj(Adj(A))| = (14)^3 \left| {\matrix{ 1 & 2 & { - 1} \cr { - 1} & 1 & 2 \cr 2 & { - 1} & 1 \cr } } \right| Now, we calculate the determinant of the simplified $3 \times 3$ matrix: \left| {\matrix{ 1 & 2 & { - 1} \cr { - 1} & 1 & 2 \cr 2 & { - 1} & 1 \cr } } \right| Expanding along the first row: = 1 \cdot \text{det}\left( {\matrix{ 1 & 2 \cr { - 1} & 1 \cr } } \right) - 2 \cdot \text{det}\left( {\matrix{ { - 1} & 2 \cr 2 & 1 \cr } } \right) + (-1) \cdot \text{det}\left( {\matrix{ { - 1} & 1 \cr 2 & { - 1} \cr } } \right) $= 1((1)(1) - (2)(-1)) - 2((-1)(1) - (2)(2)) - 1((-1)(-1) - (1)(2))$ $= 1(1 + 2) - 2(-1 - 4) - 1(1 - 2)$ $= 1(3) - 2(-5) - 1(-1)$ $= 3 + 10 + 1$ $= 14$ Substituting this value back into our expression for $|Adj(Adj(A))|$: $|Adj(Adj(A))| = (14)^3 \times 14 = (14)^4$

Step 3: Equate the expressions and solve for $|A|$ From Step 1, we have the relationship: $|Adj(Adj(A))| = |A|^4$ From Step 2, we calculated the determinant of the given matrix: $|Adj(Adj(A))| = (14)^4$ Equating these two expressions, we get: $|A|^4 = (14)^4$ To solve for $|A|$, we take the fourth root of both sides: $|A| = \pm \sqrt[4]{(14)^4}$ $|A| = \pm 14$ The question specifically asks for the positive value of the determinant of matrix $A$. Therefore, we choose the positive root: $|A| = 14$

Tips and Common Mistakes:

• Formula Recall: Make sure to correctly recall the formula for the determinant of an adjoint: $|Adj(M)| = |M|^{n-1}$. A common error is to use $n$ instead of $n-1$ as the exponent.
• Nested Adjoints: For nested adjoints like $Adj(Adj(A))$, apply the formula step-by-step or remember the generalized form $|Adj(Adj(A))| = |A|^{(n-1)^2}$. This simplifies to $|A|^4$ for a $3 \times 3$ matrix.
• Determinant Calculation: Be meticulous with signs and arithmetic when expanding determinants. Factoring out common terms (like 14 in this case) is an excellent strategy to simplify calculations, but remember to multiply by $k^n$ (where $k$ is the factor and $n$ is the matrix order) when you factor $k$ from each of the $n$ rows/columns.
• Question Specifics: Always pay close attention to details in the question, such as