Let matrix `A=[(x,3,2),(1,y,4),(2, 2,z)], " if " xyz=2lambda and 8x+4y+3x=lambda+28`, then (adj A) A equals :

Options

A. `[(lambda+1,0,0),(0,lambda+1,0),(0,0,lambda+1)]`
B. `[(lambda,0,0),(0,lambda,0),(0,0,lambda)]`
C. `[(lambda^(2),0,0),(0,lambda^(2),0),(0,0,lambda^(2))]`
D. `[(lambda+2,0,0),(0,lambda+2,0),(0,0,lambda+2)]`

Answer

To solve the problem, we need to find the value of \((\text{adj } A) A\) given the matrix \(A\) and certain conditions involving \(\lambda\).

### Step-by-Step Solution:

1. **Define the Matrix A**:
The matrix \(A\) is given as:
\[ A = \begin{pmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{pmatrix} \]

2. **Calculate the Determinant of A**:
The determinant of a 3x3 matrix can be 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:
\[ = xyz - 8x - 3z + 24 + 4y - 4 \]
Thus, we have:
\[ \text{det } A = xyz - 8x + 4y + 20 - 3z \]

3. **Substitute the Given Conditions**:
We know from the problem statement:
\[ xyz = 2\lambda \quad \text{and} \quad 8x + 4y + 3z = \lambda + 28 \]
Substitute \(xyz\) into the determinant:
\[ \text{det } A = 2\lambda - 8x + 4y + 20 - 3z \]

4. **Rearranging the Second Condition**:
From \(8x + 4y + 3z = \lambda + 28\), we can express \(8x + 4y + 3z\) as:
\[ 8x + 4y + 3z - \lambda - 28 = 0 \]
Rearranging gives:
\[ \lambda = 8x + 4y + 3z - 28 \]

5. **Substituting Back into the Determinant**:
Now, substitute \(\lambda\) back into the determinant:
\[ \text{det } A = 2(8x + 4y + 3z - 28) - 8x + 4y + 20 - 3z \]
Simplifying this:
\[ = 16x + 8y + 6z - 56 - 8x + 4y + 20 - 3z \]
\[ = (16x - 8x) + (8y + 4y) + (6z - 3z) - 56 + 20 \]
\[ = 8x + 12y + 3z - 36 \]

6. **Setting the Determinant Equal to Zero**:
For \((\text{adj } A) A = \text{det } A \cdot I\), we need to find the value of \(\text{det } A\). Since we have already simplified it:
\[ \text{det } A = \lambda \]
Thus, we can conclude:
\[ (\text{adj } A) A = \text{det } A \cdot I = \lambda I \]

7. **Final Result**:
The adjoint of \(A\) multiplied by \(A\) gives:
\[ (\text{adj } A) A = \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} \]

### Conclusion:
The final answer is:
\[ (\text{adj } A) A = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} \]