Question

Let a third order determinant $\Delta_1=\left\{a_{i j}\right\} ; i , j \in\{1,2,3\}$ and the determinant $\Delta_2$ is constructed by multiplying all the elements of $\Delta_1$ by $2^{i-j}$, i.e. $\Delta_2=\left\{2^{i-j} a_{i j}\right\}$ and $\Delta_2=\lambda \Delta_1$ then $\lambda$ is equal to

• A. -1
• B. 1
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\Delta_1=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}$ and $\Delta_2=\begin{vmatrix}a_{11} & \frac{1}{2} a_{12} & \frac{1}{2^2} a_{13} \\ 2 a_{21} & a_{22} & \frac{1}{2} a_{23} \\ 2^2 a_{31} & 2 a_{32} & a_{33}\end{vmatrix}$
$=\frac{1}{2^2} \cdot \frac{1}{2}\begin{vmatrix}2^2 a_{11} & 2 a_{12} & a_{13} \\ 2^2 a_{21} & 2 a_{22} & a_{23} \\ 2^2 a_{31} & 2 a_{32} & a_{33}\end{vmatrix}=\begin{vmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}=\Delta_1$