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

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.