CUET UG 2025 — Mathematics Algebra

For any scalar $k$k and $n \times n$n×n matrix $M$M:

$$\det(kM) = k^n \times \det(M)$$det(kM)=kn×det(M)

For any $n \times n$n×n matrix $M$M:

$$\det(\text{adj}(M)) = [\det(M)]^{n-1}$$det(adj(M))=[det(M)]n−1

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$$A = \begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$$A=​203​−12−5​−2−10​​

Expanding along Row 1:

$$\det(A) = 2 \times \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - (-1) \times \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2) \times \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}$$det(A)=2×​2−5​−10​​−(−1)×​03​−10​​+(−2)×​03​2−5​​

$$\det(A) = 2[(2)(0) - (-1)(-5)] + 1[(0)(0) - (-1)(3)] - 2[(0)(-5) - (2)(3)]$$det(A)=2[(2)(0)−(−1)(−5)]+1[(0)(0)−(−1)(3)]−2[(0)(−5)−(2)(3)]

$$\det(A) = 2(0 - 5) + 1(0 + 3) - 2(0 - 6)$$det(A)=2(0−5)+1(0+3)−2(0−6)

$$\det(A) = 2(-5) + 1(3) - 2(-6)$$det(A)=2(−5)+1(3)−2(−6)

$$\det(A) = -10 + 3 + 12$$det(A)=−10+3+12

$$\det(A) = 5$$det(A)=5

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

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

$$\det(2A) = 8 \times 5$$det(2A)=8×5

$$\det(2A) = 40$$det(2A)=40

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Since $2A$2A is a $3 \times 3$3×3 matrix:

$$\det(\text{adj}(2A)) = [\det(2A)]^{3-1}$$det(adj(2A))=[det(2A)]3−1

$$\det(\text{adj}(2A)) = [\det(2A)]^2$$det(adj(2A))=[det(2A)]2

$$\det(\text{adj}(2A)) = (40)^2$$det(adj(2A))=(40)2

$$\det(\text{adj}(2A)) = 1600$$det(adj(2A))=1600