
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
486.6k+ views
Hint: It is given in the question that if $D = \left| {\begin{array}{*{20}{c}}
Complete step-by-step answer:
Note: Students frequently get confused while attempting to distinguish between the properties of a matrix and a determinant. In a matrix we take n common from each element of the matrix. In determinants, we take n common values from each row or column.
\alpha &\beta \\
\gamma &\delta
\end{array}} \right|$
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.
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 $\left| {\begin{array}{*{20}{c}}
{2\alpha }&{2\beta } \\
{2\gamma }&{2\delta }
\end{array}} \right| = 4D$ .
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.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
