Question

Let A = [aij]2x3 and B = [bij]3x2, then |5AB| is equal to

• A. \( 5^2 . |A| \cdot |B| \)
• B. \( 5^3 \cdot |A| \cdot |B| \)
• C. 5^{2} |AB|
• D. 5^{3} |AB|

Hint:

Two Crucial Rules to Remember:

• The order of a product of matrices \( (m \times n) \) * \( (n \times p) \) is \( (m \times p) \).
• The determinant of a scalar multiple \( |kM| \) is \( k^n |M| \), where 'n' is the order of the square matrix \( M \), not necessarily the scalar's exponent in the original expression.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

This question involves properties of matrix multiplication and determinants. We need to determine the size of the product matrix \( AB \) and then apply the property for the determinant of a scalar multiple of a matrix.

Step 2: Key Formula or Approach:

1. Determine the order (dimensions) of the product matrix \( AB \).
2. Use the property of determinants: For a square matrix \( M \) of order \( n \) and a scalar \( k \), \( |kM| = k^n |M| \).

Step 3: Detailed Explanation:

First, let's find the order of the product matrix \( AB \).

The order of matrix \( A \) is \( 2 \times 3 \).
The order of matrix \( B \) is \( 3 \times 2 \).
For the product \( AB \) to be defined, the number of columns in \( A \) must equal the number of rows in \( B \). Here, it is 3, so the product is defined.
The order of the resulting matrix \( AB \) is (number of rows of \( A \)) \( \times \) (number of columns of \( B \)), which is \( 2 \times 2 \).
So, \( AB \) is a square matrix of order \( n = 2 \).

Now we need to find the determinant of \( 5AB \). We use the property \( |kM| = k^n |M| \).

Here, \( M = AB \), \( k = 5 \), and the order \( n = 2 \).

\[ |5AB| = 5^2 |AB| = 25 |AB| \]

Note that options (1) and (2) are incorrect because the determinant is only defined for square matrices. Since \( A \) (2x3) and \( B \) (3x2) are not square, \( |A| \) and \( |B| \) are not defined.

Step 4: Final Answer:

The value of \( |5AB| \) is \( 5^2 |AB| \).