CUET UG 2025 — Mathematics Algebra

For any $n \times n$n×n matrix: $\det(kA) = k^n \times \det(A)$det(kA)=kn×det(A)

Since $A$A is a $3 \times 3$3×3 matrix:

$$\det(2A) = 2^3 \times \det(A)$$det(2A)=23×det(A)

$$\det(2A) = 8 \times \det(A)$$det(2A)=8×det(A)

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Given: $A = \begin{bmatrix} 2 & 0 & 3 \\ -1 & 1 & 3 \\ 0 & -4 & 0 \end{bmatrix}$A=​2−10​01−4​330​​

Expanding along Row 3 (which contains the most zeros):

$$\det(A) = 0 \cdot C_{31} - (-4) \cdot C_{32} + 0 \cdot C_{33}$$det(A)=0⋅C31​−(−4)⋅C32​+0⋅C33​

$$\det(A) = 4 \cdot M_{32}$$det(A)=4⋅M32​

where $M_{32}$M32​ is the minor obtained by removing Row 3 and Column 2.

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$$M_{32} = \begin{vmatrix} 2 & 3 \\ -1 & 3 \end{vmatrix}$$M32​=​2−1​33​​

$$M_{32} = (2)(3) - (3)(-1)$$M32​=(2)(3)−(3)(−1)

$$M_{32} = 6 - (-3)$$M32​=6−(−3)

$$M_{32} = 6 + 3$$M32​=6+3

$$M_{32} = 9$$M32​=9

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$$\det(A) = 4 \times 9$$det(A)=4×9

$$\det(A) = 36$$det(A)=36

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$$\det(2A) = 8 \times \det(A)$$det(2A)=8×det(A)

$$\det(2A) = 8 \times 36$$det(2A)=8×36

$$\det(2A) = 288$$det(2A)=288

Therefore, the value of $\det(2A) = 288$det(2A)=288.