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)

$det(A) ≥ 0$

Answer 2

Given that:

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

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 this is the answer