Discussion Forum

If  $\begin{bmatrix}x-4 & 2x & 2x\\2x & x-4 & 2x\\ 2x & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$  then the ordered pair ( A,B) is equal to 

Answer:

Explanation:

Given

$\begin{bmatrix}x-4 & 2x & 2x\\2x & x-4 & 2x\\ 2x & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

$\Rightarrow$ Apply  C1$\rightarrow$  C₁+ C₂+C₃

$\begin{bmatrix}5x-4 & 2x & 2x\\5x-4 & x-4 & 2x\\ 5x-4 & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Taking common (5x-4) from C_{1 ,}we get  

$\left(5x-4\right)\begin{bmatrix}1 & 2x & 2x\\1 & x-4 & 2x\\ 1 & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Apply  R₂ $\rightarrow$ R₂ - R₁ and  R₃ $\rightarrow$ R₃ - R₁

$\left(5x-4\right)\begin{bmatrix}1 & 2x & 0\\0 & -x-4 & 0\\ 0 & 0 & -x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Expanding Along C₁ , we get

   ( 5x-4) ( x+4)² = ( A+Bx)  ( x -A)²

Equaring We get,

   A= -4, B=5