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Question
Let A=⎧⎪⎨⎪⎩−1111−1111−1⎫⎪⎬⎪⎭ then |Adj(AdjA)|
Let A=⎧⎪⎨⎪⎩−1111−1111−1⎫⎪⎬⎪⎭ then |Adj(AdjA)|
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Solution
The correct option is B 256
From the above matrix, we get
det(A)=−(−1−1)+(1+1)
=4
det(AdjA)=[det(A)]n−1.
Where n is order of square matrix.
Hence,
det(adjA)=43−1=42
=16
det(adj(adjA))=(16)3−1
=162
=256
From the above matrix, we get
det(A)=−(−1−1)+(1+1)
=4
det(AdjA)=[det(A)]n−1.
Where n is order of square matrix.
Hence,
det(adjA)=43−1=42
=16
det(adj(adjA))=(16)3−1
=162
=256
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