Question

Let \( A \) be a square matrix of order 3 and \( |A| \) denote the determinant of \( A \). If \( A = kB \), where \( k \) is a scalar, then \( |A| \) is equal to

• A. \( |B| \)
• B. \( 3|B| \)
• C. \( k|B| \)
• D. \( k^3|B| \)

Hint:

For a square matrix of order \( n \), if the matrix is multiplied by a scalar \( k \), the determinant is multiplied by \( k^n \).

Answer 1

The Correct Option is D

Step 1: Multiplying a matrix by a scalar \( k \) multiplies EVERY row by \( k \), not just one row or one entry.
Step 2: Determinants scale multiplicatively per row: scaling one row by \( k \) scales the determinant by \( k \) once. Since \( A = kB \) is a \( 3 \times 3 \) matrix, all 3 rows are scaled, contributing \( k \times k \times k = k^3 \).
Step 3: So \( |A| = |kB| = k^3|B| \), matching the general rule \( |kB| = k^n|B| \) with \( n = 3 \).
\[ \boxed{|A| = k^3|B|} \]

Answer 2

To see how the determinant behaves under scalar multiplication of a \( 3 \times 3 \) matrix, it helps to test the relationship on a simple concrete example before checking the options.

1. \( |B| \): Take \( B \) as the \( 3 \times 3 \) identity matrix, so \( |B| = 1 \). If \( k = 2 \), then \( A = 2B \) is diagonal with entries \( 2,2,2 \), and \( |A| = 2 \times 2 \times 2 = 8 \). Since \( |B| = 1 \), this option's value of \( 1 \) does not equal the actual \( 8 \), so it is false.
2. \( 3|B| \): Here \( 3|B| = 3 \times 1 = 3 \), which again does not equal the actual determinant \( 8 \).
3. \( k|B| \): Here \( k|B| = 2 \times 1 = 2 \), still not equal to \( 8 \), showing a single power of \( k \) cannot account for scaling all three rows.
4. \( k^3|B| \): Here \( k^3|B| = 2^3 \times 1 = 8 \), which matches the actual computed determinant exactly. Repeating this test with any other value of \( k \) or any other invertible \( B \) always confirms the same relationship, because each of the 3 rows contributes one independent factor of \( k \).

The worked numerical example confirms that only the \( k^3 \) scaling reproduces the true determinant, consistent with \( |kB| = k^n|B| \).

Therefore, the correct answer is \( k^3|B| \).