Discussion Forum

If   $\alpha,\beta\neq0$   and f(n)=  $\alpha^{n}+\beta^{n}$   and $\begin{bmatrix}3 & 1+f(1)&1+f(2) \\1+f(1) & 1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{bmatrix}$  = $K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$  , then K is equal to

Answer:

Explanation:

 Use the property that, two determinants can be multipied row-to row or row -to-column to 

write the given determinant as the product of two determinants and then expand.

Given,  f(n)=  $\alpha^{n}+\beta^{n}$,  f(1)=  $\alpha^{1}+\beta^{1}$, f(2)=  $\alpha^{2}+\beta^{2}$, 

 f(3)=  $\alpha^{3}+\beta^{3}$, f(4)=  $\alpha^{4}+\beta^{4}$

Let     $\triangle=\begin{bmatrix}3 & 1+f(1)&1+f(2) \\1+f(1) & 1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{bmatrix}$

       $\Rightarrow\triangle=\begin{bmatrix}3 & 1+\alpha+\beta&1+\alpha^{2}+\beta^{2} \\1+\alpha+\beta & 1+\alpha^{2}+\beta^{2}&1+\alpha^{3}+\beta^{3}\\ 1+\alpha^{2}+\beta^{2}&1+\alpha^{3}+\beta^{3}&1+\alpha^{4}+\beta^{4} \end{bmatrix}$

$\Rightarrow\triangle=\begin{bmatrix}1.1+1.1+1.1 & 1.1+1.\alpha+1.\beta&1.1+1.\alpha^{2}+1.\beta^{2} \\1.1+1.\alpha+1.\beta & 1.1+\alpha.\alpha^{}+\beta.\beta^{}&1.1+\alpha.\alpha^{2}+\beta.\beta^{2}\\ 1.1+1.\alpha^{2}+1.\beta^{2}&1+\alpha^{2}.\alpha+\beta^{2}.\beta&1.1+\alpha^{2}.\alpha^{2}+\beta^{2}.\beta^{2} \end{bmatrix}$

=$\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}=\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}^{2}$

On expanding , we get 

$\triangle=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

 But given     $\triangle=K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

 hence,   $K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

                          K=1