CUET UG 2025 — Mathematics Algebra

Given:

• A and B are 3×3 square matrices
• det(A) = 3
• AB = 3I, where I is the 3×3 identity matrix

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Taking determinant of both sides of AB = 3I:

$$\text{det}(AB) = \text{det}(3I)$$det(AB)=det(3I)

Using the property $\text{det}(AB) = \text{det}(A) \times \text{det}(B)$det(AB)=det(A)×det(B):

$$\text{det}(A) \times \text{det}(B) = \text{det}(3I)$$det(A)×det(B)=det(3I)

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For a 3×3 identity matrix multiplied by scalar 3:

$$3I = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$3I=​300​030​003​​

The determinant of $kI$kI for an $n \times n$n×n matrix is $k^n$kn.

For the 3×3 case:

$$\text{det}(3I) = 3^3 = 27$$det(3I)=33=27

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Substituting into the equation:

$$3 \times \text{det}(B) = 27$$3×det(B)=27

$$\text{det}(B) = \dfrac{27}{3}$$det(B)=327​

$$\text{det}(B) = 9$$det(B)=9

Therefore, the value of det(B) is 9.