Given A and B are square matrices of order 3 with ∣A∣=−1 and ∣B∣=5.
To find ∣2AB∣, use the property that for scalar k and matrix A of order n:
∣kA∣=kn×∣A∣
For matrix product:
∣AB∣=∣A∣×∣B∣
Applying these properties:
∣2AB∣=∣2A∣×∣B∣
For ∣2A∣ with order n=3:
∣2A∣=23×∣A∣
∣2A∣=8×(−1)
∣2A∣=−8
Therefore:
∣2AB∣=∣2A∣×∣B∣
∣2AB∣=(−8)×5
∣2AB∣=−40