If A,B,C are three square matrices of third order such that A=⎡⎢⎣x020y000z⎤⎥⎦ & |B|=2232,|C|=2,x,y,z∈I+ & |adj(adjABC)|=2163874 then number of distinct possible matrices A is
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Solution
|A|4|B|4|C|4=2163874 |A|4=2474
|A| =xyz xyz=2×7 (x,y,z)→(2,1,7)↓6ways,(14,1,1)↓3ways
Total →9 ways