If A is a square matrix of order 3 and |A| = -3, then √78 value of |2AA^T | is

The correct option is 72.

Let us solve this step-by-step using the properties of determinants.

We are given that $A$ is a square matrix of order $n=3$ and its determinant is:

$\left|A\right|=-3$

We need to find the value of $\left|2A{A}^T\right|$.

First, we use the property of determinants for scalar multiplication. For any square matrix $M$ of order $n$ and a scalar $k$:
$\left|kM\right|=k^n\left|M\right|$

Applying this property with $k=2$, $n=3$, and $M=A{A}^T$, we get:
$\left|2A{A}^T\right|=2^3\left|A{A}^T\right|=8\left|A{A}^T\right|$

Next, we use the multiplicative property of determinants, which states that for any two square matrices $A$ and $B$ of the same order:
$\left|AB\right|=\left|A\right|·\left|B\right|$

Applying this to our expression gives:
$\left|A{A}^T\right|=\left|A\right|·\left|A^T\right|$

We also know that the determinant of a matrix is equal to the determinant of its transpose:
$\left|A^T\right|=\left|A\right|$

Substituting $\left|A^T\right|=\left|A\right|$ into the equation, we get:
$\left|A{A}^T\right|=\left|A\right|·\left|A\right|={\left|A\right|}^2$

Now, substituting this back into our original expression:
$\left|2A{A}^T\right|=8{\left|A\right|}^2$

Given that $\left|A\right|=-3$, we substitute this value:
$\left|2A{A}^T\right|=8·{(-3)}^2$

Calculating the final value:
$\left|2A{A}^T\right|=8·9=72$