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# If $\Delta = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&5 \\ 3&6&{12} \end{array}} \right|$and $\Delta ' = \left| {\begin{array}{*{20}{c}} 4&8&{15} \\ 3&6&{12} \\ 2&3&5 \end{array}} \right|$, then$A.\Delta ' = 2\Delta \\ B.\Delta ' = - 2\Delta \\ C.\Delta = - \Delta ' \\ D.\Delta ' = \Delta \\ E.\Delta ' = 3\Delta \\$

Last updated date: 24th Jun 2024
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Hint: In order to solve this problem you need to simply go through by applying row transformation. That is solving algebraically the elements of row with the elements of another row. Here two determinants are given. We need to get the relation between them by applying the row transformation. We will get two values of both the determinants. Then we can find a relation between these two values.

We can solve this problem by making one of the determinants equal to another so that an equation can be made and we can get the relation between the determinants by applying the row transformation.

As given in question
$\Delta = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&5 \\ 3&6&{12} \end{array}} \right|$and $\Delta ' = \left| {\begin{array}{*{20}{c}} 4&8&{15} \\ 3&6&{12} \\ 2&3&5 \end{array}} \right|$
We have $\Delta ' = \left| {\begin{array}{*{20}{c}} 4&8&{15} \\ 3&6&{12} \\ 2&3&5 \end{array}} \right|$
Now, on exchanging ${R_2}$with ${R_3}$
We get, $\Delta ' = - \left| {\begin{array}{*{20}{c}} 4&8&{15} \\ 2&3&5 \\ 3&6&{12} \end{array}} \right|$………(1)
We know on exchanging two rows a minus came outside and we know that,
$\Delta = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&5 \\ 3&6&{12} \end{array}} \right|$
On applying the transformation in the determinant above we get,
${R_1} \Rightarrow {R_1} + {R_3}$
We get the determinant as,
$\Delta = \left| {\begin{array}{*{20}{c}} 4&8&{15} \\ 2&3&5 \\ 3&6&{12} \end{array}} \right|$………(2)
From equation (1) and (2)

We get, $\Delta = - \Delta '$

So, the correct answer is “Option C”.

Note: Whenever we face such a type of question of determinants we have to exchange or add or subtract rows and columns to get the required answer. And also remember properties of determinants. We need to know that applying the row or column operation in any of the determinants within itself doesn’t change the value of the determinants. Taking this in consideration we have solved the above problem. Knowing this will solve your above problem and will give you the right answer.