Using properties of determinants prove that: $\left|\begin{array}{lll}\mathrm{x}-3 & \mathrm{x}-4 & \mathrm{x}-\alpha \\ \mathrm{x}-2 & \mathrm{x}-3 & \mathrm{x}-\beta \\ \mathrm{x}-1 & \mathrm{x}-2 & \mathrm{x}-\gamma\end{array}\right|=0$, where $\alpha, \beta, \mathrm{y}$ are in AP.

Home » Using properties of determinants prove that: , where are in AP.

Using properties of determinants prove that: $\left|\begin{array}{lll}\mathrm{x}-3 & \mathrm{x}-4 & \mathrm{x}-\alpha \\ \mathrm{x}-2 & \mathrm{x}-3 & \mathrm{x}-\beta \\ \mathrm{x}-1 & \mathrm{x}-2 & \mathrm{x}-\gamma\end{array}\right|=0$ , where $\alpha, \beta, \mathrm{y}$ are in AP.

Using properties of determinants prove that: $\left|\begin{array}{lll}\mathrm{x}-3 & \mathrm{x}-4 & \mathrm{x}-\alpha \\ \mathrm{x}-2 & \mathrm{x}-3 & \mathrm{x}-\beta \\ \mathrm{x}-1 & \mathrm{x}-2 & \mathrm{x}-\gamma\end{array}\right|=0$ , where $\alpha, \beta, \mathrm{y}$ are in AP.

Solution:

Given that $\alpha, \beta, \gamma$ are in an $A P$ , which means $2 \beta=\alpha+\gamma$
Operating $R_{3} \rightarrow R_{3}-2 R_{2}+R_{1}$
$\begin{array}{l} =\left|\begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1-2 x+4+x-3 & x-2-2 x+6+x-4 & x-y-2 x+2 \beta+x-\alpha \end{array}\right| \\ =\left|\begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ 0 & 0 & -\gamma+2 \beta-\alpha \end{array}\right| \text { [we know that } 2 \beta=\alpha+y \end{array}$
Operating $R_{1} \rightarrow R_{1}-R_{3}, R_{2} \rightarrow R \rightarrow 2-R_{3}$
$\begin{array}{l} =\left|\begin{array}{ccc} \mathrm{x}-3 & \mathrm{x}-4 & \mathrm{x}-\alpha \\ \mathrm{x}-2 & \mathrm{x}-3 & \mathrm{x}-\beta \\ 0 & 0 & -\gamma+\alpha+\gamma-\alpha \end{array}\right| \\ =\left|\begin{array}{ccc} \mathrm{x}-3 & \mathrm{x}-4 & \mathrm{x}-\alpha \\ \mathrm{x}-2 & \mathrm{x}-3 & \mathrm{x}-\beta \\ 0 & 0 & 0 \end{array}\right| \end{array}$
[By the properties of determinants, we know that if all the elements of a row or column is 0 , then the value of the determinant is also 0]
$=0$
Hence proved