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Let A, B be square matrices of order 3×33\times3. Then ∣A.adj(A).B∣|A.adj(A).B| is

Solution

✅ Correct Option: 2

Since A⋅adj(A)=∣A∣I3A\cdot adj(A)=|A|I_3, we get ∣A⋅adj(A)⋅B∣=∣∣A∣I3∣⋅∣B∣=∣A∣3∣B∣|A\cdot adj(A)\cdot B|=\left||A|I_3\right|\cdot|B|=|A|^3|B|.

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