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Solution

✅ Correct Option: 2

Given a matrix AA of order 3×33 \times 3 with ∣A∣=3|A| = 3.

The adjoint of matrix AA satisfies the property:

A(adj A)=∣A∣×IA(\text{adj } A) = |A| \times I

where II is the identity matrix of the same order.


Substituting ∣A∣=3|A| = 3:

A(adj A)=3IA(\text{adj } A) = 3I


Taking the determinant of both sides:

∣A(adj A)∣=∣3I∣|A(\text{adj } A)| = |3I|

For a 3×33 \times 3 identity matrix, when multiplied by scalar kk:

∣kI∣=kn×∣I∣|kI| = k^n \times |I|

where nn is the order of the matrix.


Since the matrix is 3×33 \times 3:

∣3I∣=33×∣I∣|3I| = 3^3 \times |I|

∣3I∣=33×1|3I| = 3^3 \times 1

∣3I∣=27|3I| = 27


Therefore, ∣A(adj A)∣=27|A(\text{adj } A)| = 27

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