If A is a square matrix of order 3 such that |2(adj A)| = 288, then the value of A is

The correct option is ±6.

Let us understand the step-by-step mathematical derivation to find the value of $\left|A\right|$ (the determinant of matrix $A$).

We are given that $A$ is a square matrix of order $n=3$.
We are also given the relation:
$\left|2\right(\text{adj}A\left)\right|=288$

To solve this, we will use two standard properties of determinants:

Property 1: For any scalar $k$ and a square matrix $M$ of order $n$,
$\left|kM\right|=k^n\left|M\right|$

Property 2: For any square matrix $A$ of order $n$, the determinant of its adjoint matrix is given by:
$\left|\text{adj}A\right|={\left|A\right|}^{n-1}$

Let us apply Property 1 to the given equation $\left|2\right(\text{adj}A\left)\right|=288$. Here, the scalar $k=2$ and the matrix $M=\text{adj}A$ which is of order $n=3$:
$2^3\left|\text{adj}A\right|=288$

Simplifying this, we get:
$8\left|\text{adj}A\right|=288$

Dividing both sides by 8:
$\left|\text{adj}A\right|=\frac{288}{8}$
$\left|\text{adj}A\right|=36$

Now, let us apply Property 2 with $n=3$:
$\left|\text{adj}A\right|={\left|A\right|}^{3-1}$
$\left|\text{adj}A\right|={\left|A\right|}^2$

Substitute this back into our simplified equation:
${\left|A\right|}^2=36$

Taking the square root of both sides gives:
$\left|A\right|=\pm \sqrt{36}$
$\left|A\right|=\pm 6$

Thus, the value of the determinant of $A$ is $\pm 6$.