#### Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 16 maths

Correct option (b)

Hint:

Using properties of determinant

Given:

$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$

Solution:

$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$

Taking 52 and 53 common from R1 and R2 respectively

$\Rightarrow 5^{2}\times 5^{3}\begin{vmatrix} 1 &5 &5^{2} \\ 1 &5 &5^{2} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}$

Here R1 and R2 are identical.

Hence from the properties of determinant, if any two row or column of the matrix is identical ,then the value of determinant is equal to zero.

Hence,

$\begin{vmatrix} 5^{2} &5^{3} &5^{4} \\ 5^{3} &5^{4} &5^{5} \\ 5^{4} &5^{5} &5^{6} \end{vmatrix}=0$