Question

If $\Delta_{1} = \left| \begin{matrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{matrix} \right|$&$\Delta_{2} = \left| \begin{matrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{matrix} \right|$ then $\frac{\Delta_{1}}{\Delta_{2}}$ =

• A. 1
• B. 2
• C. ½
• D. None

Answer 1

Answer: (A)

1

Answer 2

$\Delta = \left| \begin{matrix} f & 2d & e \\ 2z & 4x & 2y \\ e & 2a & b \end{matrix} \right|$ $\mathrm { C } _ { 1 } \leftrightarrow \mathrm { C } _ { 2 }$

=$- \left| \begin{matrix} 2d & f & e \\ 4x & 2z & 2y \\ 2a & e & b \end{matrix} \right|$ $\mathrm { C } _ { 2 } \leftrightarrow \mathrm { C } _ { 3 }$

=$\left| \begin{matrix} 2d & e & f \\ 4x & 2y & 2z \\ 2a & b & e \end{matrix} \right|$ $\mathrm { R } _ { 1 } \leftrightarrow \mathrm { R } _ { 3 }$ =$\left| \begin{matrix} 2a & b & e \\ 4x & 2y & 2z \\ 2d & e & f \end{matrix} \right|$ $\mathrm { R } _ { 2 } \leftrightarrow \mathrm { R } _ { 3 }$

= $\left| \begin{matrix} 2a & b & e \\ 2d & e & f \\ 4x & 2y & 2z \end{matrix} \right| = \Delta_{2}$

\$\frac{\Delta_{1}}{\Delta_{2}} = 1$