Let A=a_ij be a square matrix of order 3 and B=b_ij be a matrix such that b_ij=2^(i-j)a_ij for 1≤i,j≤3, ∀i,j∈N. If the determinant of A is same as its…

Question

Let A=a_ij be a square matrix of order 3 and B=b_ij be a matrix such that b_ij=2^(i-j)a_ij for 1≤i,j≤3, ∀i,j∈N. If the determinant of A is same as its order, then the value of B^T-1 is

MARKS Content Team · Accepted Answer

Correct answer: (A) 1/3. ∵A=3 B= a_11 1/2a_12 1/4a_13 2a_21 a_22 1/2a_23 4a_31 2a_32 a_33 B=8 a_11 1/2a_12 1/4a_13 a_21 1/2a_22 1/4a_23 a_31 1/2a_32 1/4a_33 = a_11 a_12 a_13 a_21 a_22 a_23 a_31 a_32 a_33 =A=3 B^T-1=1/(B^T)=1/B=1/3

Let $A = [a_{i j}]$ be a square matrix of order $3$ and $B = [b_{i j}]$ be a matrix such that $b_{i j} = 2^{i - j} a_{i j}$ for $1 \leq i, j \leq 3, \forall i, j \in N .$ If the determinant of $A$ is same as its order, then the value of $|{(B^{T})}^{- 1}|$ is

Solution

More Determinants Questions

Let A=aij be a square matrix of order 3 and B=bij be a matrix such that bij=2i-jaij for 1≤i,j≤3, ∀i,j∈N. If the determinant of A is same as its order, then the value of BT-1 is

Solution

More Determinants Questions

Let $A = [a_{i j}]$ be a square matrix of order $3$ and $B = [b_{i j}]$ be a matrix such that $b_{i j} = 2^{i - j} a_{i j}$ for $1 \leq i, j \leq 3, \forall i, j \in N .$ If the determinant of $A$ is same as its order, then the value of $|{(B^{T})}^{- 1}|$ is