Question

If $y=rx$ and $z=tx$ all variable being functions of $x$. If suffixes denote the order differentiation with respect to $x$ and $\left|\begin{array}{ccc}x& y& z\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|=k\left(x\right)\left|\begin{array}{cc}{r}_1& t_1\\ r_2& t_2\end{array}\right|$, then $k\left(8\right)$

Answer 1

The given determinant is:

$\left|\begin{array}{ccc}x& y& z\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$

Substituting all the values and after differentiating we get,

$=\left|\begin{array}{ccc}x& rx& tx\\ 1& r+r_{1}x& t+t_{1}x\\ 0& r_1+r_{2}x+r_1& t_1+t_{2}x+t_1\end{array}\right|$

$=\left|\begin{array}{ccc}x& rx& tx\\ 1& r+r_{1}x& t+t_{1}x\\ 0& 2r_1+r_{2}x& 2t_1+t_{2}x\end{array}\right|$

Using the property of determinant to take x common from the 1st row we get,

$=x\left|\begin{array}{ccc}1& r& t\\ 1& r+r_{1}x& t+t_{1}x\\ 0& 2r_1+r_{2}x& 2t_1+t_{2}x\end{array}\right|$

Subtracting the 2nd row with 1st row we get,

$=x\left|\begin{array}{ccc}1& r& t\\ 0& r_{1}x& t_{1}x\\ 0& 2r_1+r_{2}x& 2t_1+t_{2}x\end{array}\right|$

Using the property of determinant to take x common from the 2nd row we get,

$=x^2\left|\begin{array}{ccc}1& r& t\\ 0& r_1& t_1\\ 0& 2r_1+r_{2}x& 2t_1+t_{2}x\end{array}\right|$

Subtracting the 3rd row with twice the second row we get,

$=x^2\left|\begin{array}{ccc}1& r& t\\ 0& r_1& t_1\\ 0& r_{2}x& t_{2}x\end{array}\right|$

Using the property of determinant to take x common from the 3rd row we get,

$=x^3\left|\begin{array}{ccc}1& r& t\\ 0& r_1& t_1\\ 0& r_2& t_2\end{array}\right|$

$=x^3\left|\begin{array}{cc}{r}_1& t_1\\ r_2& t_2\end{array}\right|$

$k\left(x\right)=x^3$

$k\left(8\right)=8^3=512$ ....Answer