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Let P=1004101641 and I be the identity matrix of order 3. If Q = [qij] is a matrix such that P50Q=I, then q31+q32q21=

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detailed solution

103

P2=10041016411004101641=1008104881P3=10081048811004101641=100121096121Pn=1004n108n2+n4n1P50=200108×50(52)2001p50Q=I Equating we get q21=200q31=400×51q32=200q31+q32q21=400×51+200200=2(51)+1=103


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