Let $A=\left[a_{\mathrm{ij}}\right]_{2\times 3}$ and $B=\left[b_{\mathrm{ij}}\right]_{3\times 2}$ , then

$—\; 5AB—$ is equal to:

The correct option is 5² —AB—.

Here is the step-by-step explanation and derivation:

Step 1: Determine the order of the product matrix AB
We are given that matrix $A$ is of dimension $2\times 3$ and matrix $B$ is of dimension $3\times 2$ .
When multiplying a matrix of dimension $m\times n$ with a matrix of dimension $n\times p$ , the resulting matrix product $AB$ will have a dimension of $m\times p$ .
Consequently, $AB$ is a square matrix of order $2\times 2$ .

Step 2: Understand the scaling property of determinants
For any square matrix $M$ of order $k\times k$ and a scalar $c$ , the determinant of $cM$ scales by the constant raised to the power of the matrix dimension:
$$\left|cM\right|=c^k\left|M\right|$$
Here, the vertical bars represent the determinant of the matrix.

Step 3: Evaluate the determinant of -5AB
Substituting $M=AB$ (which is of order $k=2$ ) and the scalar $c=-5$ into the determinant scaling formula:
$$|-5AB|={(-5)}^2|AB$$

Evaluating the exponent gives:
$${(-5)}^2=5^2$$
Thus, the expression simplifies to:
$$5^2\left|AB\right|$$
Representing the determinant using the question's notation (—AB—), we obtain:
5² —AB—