Question

The value of the determinant of $\left| \begin{matrix} {{a}_{1}} & m{{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & m{{a}_{2}} & {{b}_{2}} \\\ {{a}_{3}} & m{{a}_{3}} & {{b}_{3}} \\\ \end{matrix} \right|$ :
(a)0
(b)$m{{a}_{1}}{{a}_{2}}{{a}_{3}}$
(c)$m{{a}_{1}}{{b}_{2}}{{a}_{2}}$
(d)$m{{b}_{1}}{{b}_{2}}{{b}_{3}}$

Answer 1

Hint: There are some rules in determinant that we can take a common number from any row or column and another rule is that if any two rows of column are same then the value of determinant is 0. We will try to simplify use these two rules to find the value of determinant.

Complete step-by-step answer:
First we are going to use the rule that we can take common number from any row or column, in the 2nd column and take the number m as common.
Then the determinant becomes,
m$\left| \begin{matrix} {{a}_{1}} & {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{a}_{2}} & {{b}_{2}} \\\ {{a}_{3}} & {{a}_{3}} & {{b}_{3}} \\\ \end{matrix} \right|$
Now we will use the second rule of determinant that if any two rows of column are same then the value of determinant is 0.
We can see that the 1st column and the 2nd column are the same and hence the value of determinant will become 0.
Hence, the correct answer to this question is option (a).

Note: We have solved this question by using the two rules of determinant, but there is another method to solve this question, one can open the determinant and expand it then try to rearrange it and one can see that answer from both the methods will same, but the second method will take a lot of time and also will require some calculations, so it’s recommended to go with the method that is given in the solution.