asked 39.8k views
5 votes
Suppose A is a 3 x 3 matrix such that det (A) = 9. Prove det (3 (A-!') is equal to 3

asked
User Ersin
by
7.6k points

1 Answer

4 votes

Answer: The proof is done below.

Step-by-step explanation: Given that A is a 3 x 3 matrix such that det (A) = 9.

We are to prove the following :


det(3A^(-1))=3.

For a non-singular matrix B of order n, we have two two properties of its determinant :


(i)~det(B^(-1))=(1)/(det(B)),\\\\\\(ii)~det(kB)=k^ndet(B),~\textup{k is a scalar.}

Therefore, we get


det(A^(-1))=(1)/(det(A))=(1)/(9),

and so,


det(3A^(-1))~~~~~~~[\textup{since A is of order 3}]\\\\=3^3det(A^(-1))\\\\=27*(1)/(9)\\\\=3.

Hence proved.

answered
User Shreyansh Sharma
by
7.4k points
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