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$If A = [\begin{array}{ccc} 0 & 1 & - 1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{array}] then [\begin{array}{lll} A (adj A) A^{- 1} \end{array}] A =$

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600060006

−6000−6000−6

01/6−1/61/31/61/21/21/31/6

none

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detailed solution

Correct option is A

det A=−12−9−14−3=7−1=6AAdj AA−1A=AAdj AA−1A=Aadj A=det(A)I=6I

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If A=01−1213321 then A(adj⁡A)A−1A=

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$If A = [\begin{array}{ccc} 0 & 1 & - 1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{array}] then [\begin{array}{lll} A (adj A) A^{- 1} \end{array}] A =$