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

Options

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

Answer

To solve the problem step-by-step, we start with the given matrix \( A \) and the condition involving its adjoint.

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

### Step 2: Use the Property of Adjoint
We know that:
\[ A \text{ adj } A = \det(A) I_2 \]
where \( I_2 \) is the identity matrix of order 2.

### Step 3: Calculate the Determinant of A
The determinant of matrix \( A \) is calculated as follows:
\[ \det(A) = (5a)(2) - (-b)(3) = 10a + 3b \]

### Step 4: Write the Equation for A adj A
Using the property mentioned:
\[ A \text{ adj } A = \det(A) I_2 = (10a + 3b) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{pmatrix} \]

### Step 5: Calculate A A^T
Next, we need to calculate \( A A^T \):
\[ A^T = \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix} \]
Now, calculate \( A A^T \):
\[ A A^T = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix} \]
Calculating the elements:
- First element: \( 5a \cdot 5a + (-b) \cdot (-b) = 25a^2 + b^2 \)
- Second element: \( 5a \cdot 3 + (-b) \cdot 2 = 15a - 2b \)
- Third element: \( 3 \cdot 5a + 2 \cdot (-b) = 15a - 2b \)
- Fourth element: \( 3 \cdot 3 + 2 \cdot 2 = 9 + 4 = 13 \)

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

### Step 6: Set the Two Equations Equal
From the properties, we have:
\[ \begin{pmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{pmatrix} = \begin{pmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{pmatrix} \]

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

### Step 7: Solve the Second Equation
From the second equation:
\[ 10a + 3b = 13 \quad \text{(Equation 1)} \]

### Step 8: Substitute into the First Equation
Substituting \( 10a + 3b = 13 \) into the first equation:
\[ 13 = 25a^2 + b^2 \quad \text{(Equation 2)} \]

### Step 9: Express b in terms of a
From Equation 1, we can express \( b \):
\[ b = \frac{13 - 10a}{3} \]

### Step 10: Substitute b into Equation 2
Substituting \( b \) into Equation 2:
\[ 13 = 25a^2 + \left(\frac{13 - 10a}{3}\right)^2 \]
Expanding and simplifying:
\[ 13 = 25a^2 + \frac{(13 - 10a)^2}{9} \]
Multiply through by 9 to eliminate the fraction:
\[ 117 = 225a^2 + (13 - 10a)^2 \]
Expanding \( (13 - 10a)^2 \):
\[ 117 = 225a^2 + 169 - 260a + 100a^2 \]
Combine like terms:
\[ 117 = 325a^2 - 260a + 169 \]
Rearranging gives:
\[ 0 = 325a^2 - 260a + 52 \]

### Step 11: Solve the Quadratic Equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{260 \pm \sqrt{(-260)^2 - 4 \cdot 325 \cdot 52}}{2 \cdot 325} \]
Calculating the discriminant and simplifying will yield values for \( a \).

### Step 12: Find b and Calculate 5a + b
Once \( a \) is found, substitute back to find \( b \) and compute \( 5a + b \).

### Final Result
After calculating, we find that:
\[ 5a + b = 5 \]