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If A is a matrix of order 3×33 \times 3 such that ∣A∣=4|A| = 4 and ∣k⋅adj A∣=32∣A∣,k∈R|k \cdot adj\, A| = 32|A|, k \in \mathbb{R}; then the value of k is:

Solution

✅ Correct Option: 2

∣adj A∣=∣A∣2=16|adj\,A| = |A|^2 = 16 for a 3×33\times3 matrix.

So ∣k adj A∣=k3(16)=32(4)=128|k\,adj\,A| = k^3(16) = 32(4) = 128, giving k3=8k^3 = 8 and k=2k = 2.

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