Question

If we are given a determinant $f'\left( x \right)=\left| \begin{matrix} mx & mx-p & mx+p \\\ n & n+p & n-p \\\ mx+2n & mx+2n+p & mx+2n-p \\\ \end{matrix} \right|$ then $y=f\left( x \right)$ represent
A. A straight line parallel to x-axis
B. A straight line parallel to y-axis
C. parabola
D. A straight line with negative slope.

Answer 1

In this problem, we are given a determinant $f'\left( x \right)$ and we have to represent $y=f\left( x \right)$. We can first use the elementary transformation method for the given determinant $f'\left( x \right)$, by multiplying the number 2 to the second row and we can subtract the third row, first row and the results second row and write it in the third row, we will get the third row equals to 0. Which gives $f'\left( x \right)$ equals to zero. We can then find the $y=f\left( x \right)$, as we integrate $f'\left( x \right)$, we will get $f\left( x \right)$.

Complete step by step solution:
We know that the given determinant is,

mx & mx-p & mx+p \\\ n & n+p & n-p \\\ mx+2n & mx+2n+p & mx+2n-p \\\ \end{matrix} \right|$$ We can now apply the elementary transformation method for the determinant. We should know that the elementary transformation method is, 1.Interchanging any two rows (or columns). 2.Multiplying each element in rows or columns by a non-zero scalar. 3.Addition to the elements of any row the same scalar multiples of corresponding elements of any other rows. We can multiply the number 2 to the second row and we can subtract the third row, first row and the results second row and write it in the third row. We can now write the representation of the elementary transformation for the given determinant, $$\Rightarrow {{R}_{3}}\to {{R}_{3}}-{{R}_{1}}-2{{R}_{2}}$$ . We can now apply $${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}-2{{R}_{2}}$$, we get $$\Rightarrow f'\left( x \right)=\left| \begin{matrix} mx & mx-p & mx+p \\\ n & n+p & n-p \\\ 0 & 0 & 0 \\\ \end{matrix} \right|=0$$ Since, one of the rows is zero, the determinant is also zero. $$\Rightarrow f'\left( x \right)=0$$ We know that differentiating a constant number gives zero. Therefore, $$\begin{aligned} & \Rightarrow y=f\left( x \right)=c \\\ & \Rightarrow y=c \\\ \end{aligned}$$ Now, we can say that the line passes through any of the constants in the y-axis, which will be parallel to x. **Therefore, the answer is A. a straight line parallel to x-axis.** **Note:** Students make mistakes, while solving the elementary transformation method, where we have to know which number can be multiplied to which row and which rows should be added or subtracted within each other to get a third row to be zero, if not we can further proceed the elementary transformation.