Question

If $y=\sin px$, then $\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$ is equal to

• A. $0$
• B. $1$
• C. $y$
• D. $-y$

Answer 1

The correct option is A $0$

The given information is: $y=\sin px$

On successive differentiation we get $y_1$, $y_2$ and other terms. They are $y_1=p\cos px$

$y_2=-p^2\sin px$

$y_3=-p^3\cos px$

$y_4=p^4\sin px$

$y_5=p^5\cos px$

$y_6=-p^6\sin px$

$y_7=-p^7\cos px$

$y_8=p^8\sin px$

Substituting these values in the determinant we get,

$\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$ = $\left|\begin{array}{ccc}\sin px& p\cos px& -p^2\sin px\\ -p^3\cos px& p^4\sin px& p^5\cos px\\ -p^6\sin px& -p^7\cos px& p^8\sin px\end{array}\right|$

Now using the property of determinants we take $p^3$ common from the second row and $-p^6$ from the third row we get,

$\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$ = $-p^9\left|\begin{array}{ccc}\sin px& p\cos px& -p^2\sin px\\ -\cos px& p\sin px& p^2\cos px\\ \sin px& p\cos px& -p^2\sin px\end{array}\right|$

Using the property of determinants, we know if two rows or columns of a determinant are equal, then its value is 0. Here row no. 1 and 3 are qual and hence the value of this determinant is 0 i.e.

$\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$ = $0$