If `A=|{:(,5a,-b),(,3,2):}|` and A adj `A=A A^(T)`, then 5a+b is equal to

Options

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

Answer

To solve the problem, we need to find the value of \(5a + b\) given the matrix \(A\) and the condition \(A \cdot \text{adj}(A) = A \cdot A^T\).

### Step-by-Step Solution:

1. **Define the Matrix \(A\)**:
\[ A = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} \]

2. **Find the Adjoint of Matrix \(A\)**:
For a \(2 \times 2\) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the adjoint is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Thus, for our matrix \(A\):
\[ \text{adj}(A) = \begin{pmatrix} 2 & b \\ -3 & 5a \end{pmatrix} \]

3. **Calculate \(A \cdot \text{adj}(A)\)**:
\[ A \cdot \text{adj}(A) = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & b \\ -3 & 5a \end{pmatrix} \]
Performing the multiplication:
- First row, first column: \(5a \cdot 2 + (-b) \cdot (-3) = 10a + 3b\)
- First row, second column: \(5a \cdot b + (-b) \cdot 5a = 5ab - 5ab = 0\)
- Second row, first column: \(3 \cdot 2 + 2 \cdot (-3) = 6 - 6 = 0\)
- Second row, second column: \(3 \cdot b + 2 \cdot 5a = 3b + 10a\)

Thus,
\[ A \cdot \text{adj}(A) = \begin{pmatrix} 10a + 3b & 0 \\ 0 & 3b + 10a \end{pmatrix} \]

4. **Calculate \(A \cdot A^T\)**:
First, find \(A^T\):
\[ A^T = \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix} \]
Now calculate \(A \cdot A^T\):
\[ A \cdot A^T = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix} \]
Performing the multiplication:
- First row, first column: \(5a \cdot 5a + (-b) \cdot (-b) = 25a^2 + b^2\)
- First row, second column: \(5a \cdot 3 + (-b) \cdot 2 = 15a - 2b\)
- Second row, first column: \(3 \cdot 5a + 2 \cdot (-b) = 15a - 2b\)
- Second row, second column: \(3 \cdot 3 + 2 \cdot 2 = 9 + 4 = 13\)

Thus,
\[ A \cdot A^T = \begin{pmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{pmatrix} \]

5. **Set the Two Results Equal**:
From the condition \(A \cdot \text{adj}(A) = A \cdot A^T\), we equate:
\[ \begin{pmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{pmatrix} = \begin{pmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{pmatrix} \]

This gives us two equations:
- \(10a + 3b = 25a^2 + b^2\) (1)
- \(10a + 3b = 13\) (2)

6. **Solve the Equations**:
From equation (2):
\[ 10a + 3b = 13 \]
Rearranging gives:
\[ 3b = 13 - 10a \quad \Rightarrow \quad b = \frac{13 - 10a}{3} \]

Substitute \(b\) into equation (1):
\[ 10a + 3\left(\frac{13 - 10a}{3}\right) = 25a^2 + \left(\frac{13 - 10a}{3}\right)^2 \]
Simplifying:
\[ 10a + 13 - 10a = 25a^2 + \frac{(13 - 10a)^2}{9} \]
This reduces to:
\[ 13 = 25a^2 + \frac{169 - 260a + 100a^2}{9} \]

Multiply through by 9 to eliminate the fraction:
\[ 117 = 225a^2 + 169 - 260a + 100a^2 \]
Combine like terms:
\[ 0 = 325a^2 - 260a + 52 \]

Solving this quadratic equation using the quadratic formula:
\[ a = \frac{-(-260) \pm \sqrt{(-260)^2 - 4 \cdot 325 \cdot 52}}{2 \cdot 325} \]

After solving for \(a\), substitute back to find \(b\).

7. **Find \(5a + b\)**:
Finally, substitute the values of \(a\) and \(b\) into \(5a + b\) to find the answer.