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If AA and BB square matrices of order 3 such that ∣A∣=−1|A| = -1, ∣B∣=5|B| = 5, then the value of ∣2AB∣|2AB| is:

Solution

✅ Correct Option: 1

Given AA and BB are square matrices of order 3 with ∣A∣=−1|A| = -1 and ∣B∣=5|B| = 5.


To find ∣2AB∣|2AB|, use the property that for scalar kk and matrix AA of order nn:

∣kA∣=kn×∣A∣|kA| = k^n \times |A|

For matrix product:

∣AB∣=∣A∣×∣B∣|AB| = |A| \times |B|


Applying these properties:

∣2AB∣=∣2A∣×∣B∣|2AB| = |2A| \times |B|

For ∣2A∣|2A| with order n=3n = 3:

∣2A∣=23×∣A∣|2A| = 2^3 \times |A|

∣2A∣=8×(−1)|2A| = 8 \times (-1)

∣2A∣=−8|2A| = -8


Therefore:

∣2AB∣=∣2A∣×∣B∣|2AB| = |2A| \times |B|

∣2AB∣=(−8)×5|2AB| = (-8) \times 5

∣2AB∣=−40|2AB| = -40

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