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 step by step, we need to find the value of \( \text{adj}(A) A \) for the given matrix \( A \) and the conditions provided.

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

### Step 2: Calculate the Determinant of A
To find \( \text{adj}(A) A \), we first need to find the determinant of \( A \) (denoted as \( \det(A) \)). We can calculate the determinant using the formula for a 3x3 matrix:
\[ \det(A) = x \begin{vmatrix} y & 4 \\ 2 & z \end{vmatrix} - 3 \begin{vmatrix} 1 & 4 \\ 2 & z \end{vmatrix} + 2 \begin{vmatrix} 1 & y \\ 2 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} y & 4 \\ 2 & z \end{vmatrix} = yz - 8 \)
2. \( \begin{vmatrix} 1 & 4 \\ 2 & z \end{vmatrix} = z - 8 \)
3. \( \begin{vmatrix} 1 & y \\ 2 & 2 \end{vmatrix} = 2 - 2y \)

Substituting these back into the determinant formula:
\[ \det(A) = x(yz - 8) - 3(z - 8) + 2(2 - 2y) \]
\[ = xyz - 8x - 3z + 24 + 4 - 4y \]
\[ = xyz - 8x - 3z + 28 - 4y \]

### Step 3: Use the Given Conditions
We have two conditions:
1. \( xyz = 2\lambda \)
2. \( 8x + 4y + 3x = \lambda + 28 \)

From the second condition, we simplify:
\[ 11x + 4y = \lambda + 28 \]
Thus, we can express \( \lambda \) as:
\[ \lambda = 11x + 4y - 28 \]

### Step 4: Substitute into the Determinant
Now substituting \( xyz = 2\lambda \) into the determinant:
\[ \det(A) = 2\lambda - 8x - 3z + 28 - 4y \]
Substituting \( \lambda \):
\[ = 2(11x + 4y - 28) - 8x - 3z + 28 - 4y \]
\[ = 22x + 8y - 56 - 8x - 3z + 28 - 4y \]
\[ = (22x - 8x) + (8y - 4y) - 3z - 56 + 28 \]
\[ = 14x + 4y - 3z - 28 \]

### Step 5: Find adjoint(A) A
Using the property:
\[ \text{adj}(A) A = \det(A) I \]
where \( I \) is the identity matrix. Thus:
\[ \text{adj}(A) A = \det(A) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Substituting \( \det(A) \):
\[ \text{adj}(A) A = (14x + 4y - 3z - 28) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ = \begin{pmatrix} 14x + 4y - 3z - 28 & 0 & 0 \\ 0 & 14x + 4y - 3z - 28 & 0 \\ 0 & 0 & 14x + 4y - 3z - 28 \end{pmatrix} \]

### Final Result
Thus, the final result for \( \text{adj}(A) A \) is:
\[ \text{adj}(A) A = \begin{pmatrix} 14x + 4y - 3z - 28 & 0 & 0 \\ 0 & 14x + 4y - 3z - 28 & 0 \\ 0 & 0 & 14x + 4y - 3z - 28 \end{pmatrix} \]