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Let A=[aij]2×3A = [a_{ij}]_{2 \times 3} and B=[bij]3×2B = [b_{ij}]_{3 \times 2}, then ∣5AB∣|5AB| is equal to

Solution

✅ Correct Option: 3

Given:

  • Matrix AA is a 2×32 \times 3 matrix (2 rows, 3 columns)
  • Matrix BB is a 3×23 \times 2 matrix (3 rows, 2 columns)

When multiplying A×BA \times B:

Result =(2×3)×(3×2)=2×2= (2 \times 3) \times (3 \times 2) = 2 \times 2 matrix

So ABAB is a square matrix of order 2×22 \times 2.


Determinants (the ∣∣| | symbol) only exist for square matrices.

AA is 2×32 \times 3 → not square → no determinant exists

BB is 3×23 \times 2 → not square → no determinant exists

ABAB is 2×22 \times 2 → square → determinant exists


When a scalar kk multiplies an n×nn \times n matrix MM:

∣kM∣=kn∣M∣|kM| = k^n |M|

where nn is the order of the square matrix.


For this problem:

ABAB is a 2×22 \times 2 matrix, so n=2n = 2

Scalar multiplier k=5k = 5

Applying the formula:

∣5AB∣=52∣AB∣|5AB| = 5^2 |AB|

∣5AB∣=25∣AB∣|5AB| = 25|AB|

Therefore, ∣5AB∣=52∣AB∣|5AB| = 5^2|AB|.

option (A) does not satisfy the result; it is not given as correct in the NTA key, it is due to the fact that determinants ∣A∣|A| and ∣B∣|B| are not defined for non-square matrices.

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