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: Determinant properties.
When a matrix \( A \) is equal to a scalar \( k \) multiplied by matrix \( B \), i.e., \( A = kB \), the determinant of \( A \) is related to the determinant of \( B \) by the property: \[ |A| = |kB| = k^n |B| \] where \( n \) is the order of the square matrix. For a 3x3 matrix, \( n = 3 \).
Step 2: Conclusion.
Thus, \( |A| = k^3 |B| \). Therefore, the correct answer is (4) \( k^3|B| \).

Answer 2

The question asks how the determinant of a 3×3 matrix changes when the matrix is multiplied by a scalar \( k \), i.e. when \( A = kB \). Since determinant scaling depends on how the scalar affects each row of the matrix, let's examine every option using the row-factoring property of determinants.

1. \( |B| \): This would be true only if multiplying the matrix by \( k \) left the determinant unchanged, as if \( k \) had no effect on any row. But \( A = kB \) means every entry of every row of \( B \) is scaled by \( k \), so the determinant cannot stay at \( |B| \) unless \( k = 1 \). This option ignores the scalar entirely.
2. \( 3|B| \): This treats the effect of the scalar as if it contributed a fixed factor of \( 3 \) regardless of what \( k \) actually is. Since \( k \) is a variable scalar and determinants scale multiplicatively with each row, a fixed factor of \( 3 \) cannot be correct in general.
3. \( k|B| \): This would be correct if only ONE row of the matrix were multiplied by \( k \), since scaling a single row of a determinant scales the whole determinant by that factor once. But \( A = kB \) scales ALL THREE rows of \( B \) by \( k \), so the factor of \( k \) must appear three times, not once.
4. \( k^3|B| \): Using the property that scaling a single row of a determinant by a constant scales the determinant by that same constant, we can factor \( k \) out of row 1, then row 2, then row 3, one at a time: \[ |kB| = k \times |\text{row 1 scaled}| = k \times k \times |\text{rows 1,2 scaled}| = k \times k \times k \times |B| = k^3|B| \] Each of the three rows contributes one factor of \( k \), giving \( k^3 \) overall, matching the general rule \( |kB| = k^n|B| \) for an \( n\times n \) matrix, here \( n=3 \).

Factoring row by row confirms all three rows each contribute a factor of \( k \), so the determinant scales by \( k^3 \), not by \( 1 \), \( 3 \), or a single factor of \( k \).

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