Determinants Matrices Question 25
Question: If $ A= \begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \\ \end{bmatrix} , $ then det(adj (adj A)) is
Options:
A) $ {{(14)}^{4}} $
B) $ {{(14)}^{3}} $
C) $ {{(14)}^{2}} $
D) $ {{(14)}^{1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] We know that adj (adj A) $ =|A{{|}^{n-2}}A,if|A|\ne 0 $ , provided order of A is n.
$ \therefore $ $ adj( adjA )=| A |A( asn=3 ) $
$ \therefore $ $ det( adj( adjA ) )={{| A |}^{3}}detA={{| A |}^{4}} $
$ But| A |= \begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \\ \end{bmatrix} =14 $
$ \therefore $ det $ (adj(adjA))={{(14)}^{4}} $
BETA
