Question

Let A be an invertible matrix of size 4x4 with complex entries. If the determinant of adj(A) is 5,then the number of possible value of determinant of A is

• A. $1$
• B. $4$
• C. $6$
• D. $3$
• E. $2$

Answer 1

Answer: (A)

$1$

Answer 2

Given data:
Let, A be an invertible $4×4$ matrix with complex entries.

The determinant of A is denoted as $\text{det}(A),$ and the adjoint of $A$ is denoted as $\text{adj}(A).$

Given that $\text{det}(\text{adj}(A)) = 5$,

Using the property of determinants:

$det(A × adj(A)) = det(A)^{(n-1)}$

where $n$ is the size of the square matrix (in this case, $n = 4$).

We have $A × \text{adj}(A) = \text{det}(A) × I$, (where $I$ is the identity matrix.)

So, det(A) × det(adj(A)) = d

Substituting det(adj(A)) = 5 and n = 4:

$det(A) × 5 = det(A)^3$

Now, solve for $det(A)$:

$det(A)^3 - 5×det(A) = 0$

$⇒det(A) × (det(A)^2 - 5) = 0$

$det(A) = 0$ or $det(A)^2 - 5 = 0$

If $det(A) = 0$ : If $det(A) = 0$, then A is a singular matrix, and it won't be invertible. However, we are given that A is an invertible matrix, so this case is not possible.

If $det(A)^2 - 5 = 0: det(A)^2 = 5.$

$|det(A)| = √(5)$

So, considering the positive values for $det(A)$ are $+√(5)$.

Therefore, the number of possible values of the determinant of $A$ is $1$.(_Ans)