Given that matrix `A[(x,3,2),(1,y,4),(2,2,z)]`. If `xyz=60` and `8x+4y+3z=20`, then A(adj A) is equal to

Options

A. `[(64,0,0),(0,64,0),(0,0,64)]`
B. `[(88,0,0),(0,88,0),(0,0,88)]`
C. `[(68,0,0),(0,68,0),(0,0,68)]`
D. `[(34,0,0),(0,34,0),(0,0,34)]`

Explanation

To solve the problem, we need to find \( A \cdot \text{adj}(A) \) for the given matrix \( A \) and the conditions provided.

### Step 1: Define the matrix \( A \)
Given the matrix:
\[ A = \begin{pmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)
The determinant of a 3x3 matrix is calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \( A \):
\[ \text{det}(A) = x(yz - 8) - 3(1z - 8) + 2(2y - 2) \]
Expanding this gives:
\[ \text{det}(A) = xyz - 8x - 3z + 24 + 4y - 4 \]
Simplifying further:
\[ \text{det}(A) = xyz - 8x + 4y + 20 - 3z \]

### Step 3: Substitute the given values
We know from the problem:
1. \( xyz = 60 \)
2. \( 8x + 4y + 3z = 20 \)

Substituting \( xyz = 60 \) into the determinant expression:
\[ \text{det}(A) = 60 - 8x + 4y - 3z + 20 \]

### Step 4: Rearranging the determinant equation
From the second equation, we can express \( 8x + 4y + 3z = 20 \) as:
\[ -8x + 4y - 3z = -20 \]
Thus, we can substitute this into the determinant:
\[ \text{det}(A) = 60 - (-20) = 60 + 20 = 80 \]

### Step 5: Find \( A \cdot \text{adj}(A) \)
Using the property:
\[ A \cdot \text{adj}(A) = \text{det}(A) \cdot I \]
where \( I \) is the identity matrix. Therefore:
\[ A \cdot \text{adj}(A) = 80 \cdot I \]
This results in:
\[ A \cdot \text{adj}(A) = \begin{pmatrix} 80 & 0 & 0 \\ 0 & 80 & 0 \\ 0 & 0 & 80 \end{pmatrix} \]

### Final Answer
Thus, \( A \cdot \text{adj}(A) \) is equal to:
\[ \begin{pmatrix} 80 & 0 & 0 \\ 0 & 80 & 0 \\ 0 & 0 & 80 \end{pmatrix} \]