Question

Let a third order determinant ${\Delta }_1=\left\{a_{ij}\right\};i,j\in \{1,2,3\}$ and the determinant ${\Delta }_2$ is constructed by multiplying all the elements of ${\Delta }_1$ by $2^{i-j}$, i.e. ${\Delta }_2=\left\{2^{i-j}{a}_{ij}\right\}$ 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)

${\Delta }_1=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}{a}_{11}& \frac{1}{2}{a}_{12}& \frac{1}{2^2}{a}_{13}\\ 2a_{21}& a_{22}& \frac{1}{2}{a}_{23}\\ 2^2{a}_{31}& 2a_{32}& a_{33}\end{array}\right|$
$=\frac{1}{2^2}\cdot \frac{1}{2}\left|\begin{array}{ccc}{2}^2{a}_{11}& 2a_{12}& a_{13}\\ 2^2{a}_{21}& 2a_{22}& a_{23}\\ 2^2{a}_{31}& 2a_{32}& a_{33}\end{array}\right|=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|={\Delta }_1$