Question

If \(A =[5a-b 32]\) and A.adjA=AA^T,then which of the following statements is true

• A. \(5a - b = -5\)
• B. \(det(A) < 0 \)
• C. \( 5a + b = 10\)
• D. A is symmetric
• E. \(det(A) ≥ 0\)

Answer 1

Answer: (E)

Given matrix \( A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \) and the condition \( A \cdot \text{adj}(A) = AA^T \).

First, compute \( \text{adj}(A) \) and \( A^T \): \[ \text{adj}(A) = \begin{bmatrix} 2 & b \\ -3 & 5a \end{bmatrix}, \quad A^T = \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix} \]

Compute \( A \cdot \text{adj}(A) \): \[ A \cdot \text{adj}(A) = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & b \\ -3 & 5a \end{bmatrix} = \begin{bmatrix} 10a + 3b & 0 \\ 0 & 3b + 10a \end{bmatrix} \]

Compute \( AA^T \): \[ AA^T = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 5a & 3 \\ -b & 2 \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{bmatrix} \]

Set \( A \cdot \text{adj}(A) = AA^T \): \[ \begin{bmatrix} 10a + 3b & 0 \\ 0 & 10a + 3b \end{bmatrix} = \begin{bmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{bmatrix} \]

This gives the system of equations: 1. \( 10a + 3b = 25a^2 + b^2 \) 2. \( 15a - 2b = 0 \) 3. \( 10a + 3b = 13 \)

From equation (2): \( b = \frac{15}{2}a \) Substitute into equation (3): \[ 10a + 3\left(\frac{15}{2}a\right) = 13 \] \[ 10a + \frac{45}{2}a = 13 \] \[ \frac{65}{2}a = 13 \] \[ a = \frac{2}{5} \] Then \( b = \frac{15}{2} \cdot \frac{2}{5} = 3 \)

Now check the options: (A) \( 5a - b = 2 - 3 = -1 \neq -5 \) → False (B) \( 5a + b = 2 + 3 = 5 \neq 10 \) → False (C) \( \det(A) = 10a + 3b = 13 > 0 \) → False (D) \( A \) is not symmetric → False (E) \( \det(A) = 13 \geq 0 \) → True

The correct statement is (E).

Answer 2

Given

\(adj(A) = | 27 -(5a - b) | | -32 5a |\)

and  \(A × adj(A) = A × A^T\)

Now, as pe r the question let us calculate the following:

\(A × adj(A) = | 5a - b 32 | × | 27 -(5a - b) | = | 135a - 27b - (5a - b)(32) 0 | | 32 5a | | -(32) 0 |\)

\(A × A^T = | 5a - b 32 | × | 5a - b 32 | = | (5a - b)^2 + 32^2 (5a - b)(32) | | 32 5a | | (5a - b)(32) 32^2 + 5a^2 |\)

 As per the given data \(A × adj(A) = A × A^T\)

For the (1,1) element: \(135a - 27b - (5a - b)(32) = (5a - b)^2 + 32^2\)

                                     \(=  135a - 27b - 160a + 32b = 25a^2 - 10ab + b^2 + 1024\)

                                     \(=25a^2 - 10ab + b^2 - 27a + 5b - 1024 = 0\)------(1)

For the (1,2) element: 0 = (5a - b)(32)

Since the determinant of a matrix is the product of its diagonal elements, we can use the (1,1) and (1,2) elements to find the determinant of A:

Therefore, \(det(A) = (5a - b) * 0 = 0\)

Now let's go through each option:

a)For, \(5a - b = -5\) We cannot conclude this from the equations we derived. So, it is not true.

b)For,  \(det(A) < 0\)We calculated det(A) as 0, which is not less than 0. So, it is not true.

c)For, \(5a + b = 10\) We cannot conclude this from the equations we derived. So, it is not true.

d) For, A is symmetric A matrix is symmetric if it is equal to its transpose. However, from the derived equations, we can see that A is not symmetric. So, it is not true.

e) For, \(det(A) ≥ 0\) We found that det(A) = 0. Since 0 is greater than or equal to 0, this statement is true.------ 

Hence the correct option is (E)