Let `A=[a_(ij)]_(5xx5)` is a matrix such that `a_(ij)={(3,AA i= j),(0,Aai ne j):}`. If `|(adj(adjA))/(3)|=(sqrt3)^(lambda),` then `lambda` is equal to (where, `adj(M)` represents the adjoint matrix of matrix M)

Answer

To solve the problem, we need to analyze the given matrix \( A \) and find the value of \( \lambda \) such that

\[ \left| \frac{\text{adj}(\text{adj}(A))}{3} \right| = (\sqrt{3})^{\lambda}. \]

### Step 1: Define the matrix \( A \)

The matrix \( A \) is a \( 5 \times 5 \) diagonal matrix where the diagonal entries are all \( 3 \) and all off-diagonal entries are \( 0 \). Thus, we can write:

\[ A = \begin{bmatrix} 3 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \end{bmatrix}. \]

### Step 2: Calculate the determinant of \( A \)

The determinant of a diagonal matrix is the product of its diagonal entries. Therefore,

\[ |A| = 3 \times 3 \times 3 \times 3 \times 3 = 3^5. \]

### Step 3: Use the property of the adjoint

The property of the adjoint states that:

\[ |\text{adj}(A)| = |A|^{n-1}, \]

where \( n \) is the order of the matrix. For our \( 5 \times 5 \) matrix \( A \),

\[ |\text{adj}(A)| = |A|^{5-1} = |A|^4 = (3^5)^4 = 3^{20}. \]

### Step 4: Calculate the adjoint of the adjoint

Now, we need to find \( |\text{adj}(\text{adj}(A))| \). Using the same property again:

\[ |\text{adj}(\text{adj}(A))| = |\text{adj}(A)|^{n-1} = (3^{20})^{5-1} = (3^{20})^4 = 3^{80}. \]

### Step 5: Divide by 3 and equate to \( (\sqrt{3})^{\lambda} \)

Now we need to calculate:

\[ \left| \frac{\text{adj}(\text{adj}(A))}{3} \right| = \frac{|\text{adj}(\text{adj}(A))|}{|3|} = \frac{3^{80}}{3} = 3^{79}. \]

### Step 6: Set up the equation

We need to equate this to \( (\sqrt{3})^{\lambda} \):

\[ 3^{79} = (\sqrt{3})^{\lambda} = 3^{\lambda/2}. \]

### Step 7: Solve for \( \lambda \)

Equating the exponents gives us:

\[ 79 = \frac{\lambda}{2}. \]

Multiplying both sides by \( 2 \):

\[ \lambda = 158. \]

### Final Answer

Thus, the value of \( \lambda \) is

\[ \boxed{158}. \]