Question

If $D = \left| {\begin{array}{*{20}{c}}  \alpha &\beta \\  \gamma &\delta \end{array}} \right|$, then \[\left| {\begin{array}{*{20}{c}}  {2\alpha }&{2\beta } \\  {2\gamma }&{2\delta } \end{array}} \right|\] is equal to

Answer 1

Hint:  It is given in the question that if $D = \left| {\begin{array}{*{20}{c}}

  \alpha &\beta \\

  \gamma &\delta

\end{array}} \right|$

Then, what is the value of $\left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right|$ .

First, we will assume the $\alpha ,\beta ,\gamma ,\delta $ and put it in the $\left| {\begin{array}{*{20}{c}}

  \alpha &\beta \\

  \gamma &\delta

\end{array}} \right|$ , then after, we will put the value of $\alpha ,\beta ,\gamma ,\delta $ in the $\left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right|$ and finally, after solving further, we will get the answer.

Complete step-by-step answer:

It is given in the question that if $D = \left| {\begin{array}{*{20}{c}}

  \alpha &\beta \\

  \gamma &\delta

\end{array}} \right|$

Then, what is the value of $\left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right|$ .

Let us assume $\alpha = 1,\beta = 2,\gamma = 3,\delta = 4$ .

Now, put the value of $\alpha ,\beta ,\gamma ,\delta $ in the equation $D = \left| {\begin{array}{*{20}{c}}

  \alpha &\beta \\

  \gamma &\delta

\end{array}} \right|$ , we get,

 $\therefore D = \left| {\begin{array}{*{20}{c}}

  1&2 \\

  3&4

\end{array}} \right|$ .

Similarly, Put the value of $\alpha ,\beta ,\gamma ,\delta $ in the $\left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right|$ , we get,

 $ = \left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right| = \left| {\begin{array}{*{20}{c}}

  {2\left( 1 \right)}&{2\left( 2 \right)} \\

  {2\left( 3 \right)}&{2\left( 4 \right)}

\end{array}} \right|$

 $ = \left| {\begin{array}{*{20}{c}}

  2&4 \\

  6&8

\end{array}} \right|$

Now, take out 2 common from the row 1 of the determinant.

 $ = 2\left| {\begin{array}{*{20}{c}}

  1&2 \\

  6&8

\end{array}} \right|$

Now, take out 2 from row 2 of the determinant.

 $ = 2 \times 2\left| {\begin{array}{*{20}{c}}

  1&2 \\

  3&4

\end{array}} \right|$

 $ = 4\left| {\begin{array}{*{20}{c}}

  1&2 \\

  3&4

\end{array}} \right|$

 =4D

Therefore, the value of $\left| {\begin{array}{*{20}{c}}

  {2\alpha }&{2\beta } \\

  {2\gamma }&{2\delta }

\end{array}} \right| = 4D$ .

Note: Students frequently get confused while attempting to distinguish between the properties of a matrix and a determinant. In a matrix we take n common from each element of the matrix. In determinants, we take n common values from each row or column.

For example: in any matrix $A = \left[ {\begin{array}{*{20}{c}}

  1&1 \\

  1&1

\end{array}} \right]$ , $nA = \left[ {\begin{array}{*{20}{c}}

  n&n \\

  n&n

\end{array}} \right]$ , so $\left[ {nA} \right] = n\left[ A \right]$ and

In any determinant $B = \left| {\begin{array}{*{20}{c}}

  1&1 \\

  1&1

\end{array}} \right|$ , $nB = \left| {\begin{array}{*{20}{c}}

  n&n \\

  n&n

\end{array}} \right|$ , so $\left| {nB} \right| = {n^m}\left| B \right|$ , where m is the order of determinant B.