Let J_(n,m)=∫01/2x^nx^m-1dx,∀nm and n, m∈N. Consider a matrix A=aij_(3×3) where a_ij= J_(6+i,3)-J_(i+3,3) , i≤j 0 , ij . Then adjA^(-1) is

Question

Let J_(n,m)=∫01/2x^nx^m-1dx,∀n>m and n, m∈N. Consider a matrix A=aij_(3×3) where a_ij= J_(6+i,3)-J_(i+3,3) , i≤j 0 , i>j . Then adjA^(-1) is :

MARKS Content Team · Accepted Answer

Correct answer: (D) (105)^2×2^38. ∵A= a_11 a_12 a_13 0 a_22 a_23 0 0 a_33 ⇒A=a_11a_22a_33 ⇒|A|=J_(7,3)-J_(4,3)·J_(8,3)-J_(5,3)·J_(9,3)-J_(6,3) =∫01/2(x^7-x^4)/(x^3-1)dx·∫01/2(x^8-x^5)/(x^3-1)dx·∫01/2(x^9-x^6)/(x^3-1)dx =∫01/2x^4dx·∫01/2x^5dx·∫01/2x^6dx =1/((210)·(2)^18) Now adjA^(-1)=A-1^2=1/(|A|^2)=(210)·218^2=(105)^2·(2)^38

Let $J_{n, m} = \int_{0}^{1 / 2} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and $n, m \in N .$ Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = \{\begin{matrix} J_{6 + i, 3} - J_{i + 3, 3} & , i \leq j \\ 0 & , i > j \end{matrix}$ . Then $|adj A^{- 1}|$ is :

Solution

Relevant Questions

Let Jn,m=∫01/2xnxm-1dx,∀n>m and n, m∈N. Consider a matrix A=aij3×3 where aij=J6+i,3-Ji+3,3, i≤j0, i>j. Then adjA-1 is :

Solution

Relevant Questions

Let $J_{n, m} = \int_{0}^{1 / 2} \frac{x^{n}}{x^{m} - 1} d x, \forall n > m$ and $n, m \in N .$ Consider a matrix $A = {[a_{i j}]}_{3 \times 3}$ where $a_{i j} = \{\begin{matrix} J_{6 + i, 3} - J_{i + 3, 3} & , i \leq j \\ 0 & , i > j \end{matrix}$ . Then $|adj A^{- 1}|$ is :