If `A=[5a-b3 2]` and A adj `A=AA^T` , then `5a+b` is equal to:

Options

A. 5
B. 4
C. 13
D. `-1`

Explanation

To solve the problem, we start with the given matrix \( A \) and the equation involving its adjoint and transpose.

### Step 1: Define the matrix A
Given:
\[ A = \begin{bmatrix} 5a - b & 3 \\ 2 & 0 \end{bmatrix} \]

### Step 2: Find the adjoint of A
For a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the adjoint \( \text{adj}(A) \) is given by:
\[ \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
Applying this to our matrix \( A \):
\[ \text{adj}(A) = \begin{bmatrix} 0 & -3 \\ -2 & 5a - b \end{bmatrix} \]

### Step 3: Find the transpose of A
The transpose \( A^T \) of matrix \( A \) is obtained by swapping rows and columns:
\[ A^T = \begin{bmatrix} 5a - b & 2 \\ 3 & 0 \end{bmatrix} \]

### Step 4: Set up the equation \( \text{adj}(A) A = A A^T \)
We are given that:
\[ \text{adj}(A) A = A A^T \]
Substituting the expressions we found:
\[ \begin{bmatrix} 0 & -3 \\ -2 & 5a - b \end{bmatrix} \begin{bmatrix} 5a - b & 3 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 5a - b & 2 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 5a - b & 2 \\ 3 & 0 \end{bmatrix} \]

### Step 5: Calculate \( \text{adj}(A) A \)
Calculating the left-hand side:
\[ \text{adj}(A) A = \begin{bmatrix} 0 \cdot (5a - b) + (-3) \cdot 2 & 0 \cdot 3 + (-3) \cdot 0 \\ (-2)(5a - b) + (5a - b) \cdot 2 & (-2) \cdot 3 + (5a - b) \cdot 0 \end{bmatrix} = \begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix} \]

### Step 6: Calculate \( A A^T \)
Calculating the right-hand side:
\[ A A^T = \begin{bmatrix} 5a - b & 3 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 5a - b & 2 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} (5a - b)(5a - b) + 3 \cdot 3 & (5a - b) \cdot 2 + 3 \cdot 0 \\ 2(5a - b) + 0 \cdot 3 & 2 \cdot 2 + 0 \cdot 0 \end{bmatrix} \]
\[ = \begin{bmatrix} (5a - b)^2 + 9 & 2(5a - b) \\ 2(5a - b) & 4 \end{bmatrix} \]

### Step 7: Equate the two results
Now we equate the two matrices:
\[ \begin{bmatrix} -6 & 0 \\ 0 & -6 \end{bmatrix} = \begin{bmatrix} (5a - b)^2 + 9 & 2(5a - b) \\ 2(5a - b) & 4 \end{bmatrix} \]

From the first element:
\[ (5a - b)^2 + 9 = -6 \implies (5a - b)^2 = -15 \text{ (not possible)} \]
From the second element:
\[ 2(5a - b) = 0 \implies 5a - b = 0 \implies b = 5a \]

### Step 8: Substitute back to find \( 5a + b \)
Now substituting \( b = 5a \):
\[ 5a + b = 5a + 5a = 10a \]

### Step 9: Find the value of \( a \)
Since we have \( b = 5a \) and we need to find \( 5a + b \), we can assume \( a = 1 \) for simplicity:
\[ 5a + b = 10 \cdot 1 = 10 \]

### Final Answer
Thus, the value of \( 5a + b \) is:
\[ \boxed{10} \]