Question

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

• A. $(-4,5)$
• B. $(4,5)$
• C. $(-4,-5)$
• D. $(-4,3)$

Answer 1

Answer: (A)

$(-4,5)$

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

Consider $D=\left|\begin{array}{ccc}x-4& 2x& 2x\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|$

Applying $R_1\to R_1+R_2+R_3$

$D=\left|\begin{array}{ccc}5x-4& 5x-4& 5x-4\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|$

$D=(5x-4)\left|\begin{array}{ccc}1& 1& 1\\ 2x& x-4& 2x\\ 2x& 2x& x-4\end{array}\right|$

Applying $R_2\to R_2-R_3$

$D=(5x-4)\left|\begin{array}{ccc}1& 1& 1\\ 0& -(x+4)& x+4\\ 2x& 2x& x-4\end{array}\right|$

$D=(5x-4)\left[1\right(-(x+4)(x-4)-2x(x+4))-1(-2x(x+4))+1(2x(x+4)\left)\right]$

$D=(5x-4)[-x^2+16-2x^2-8x+2x^2+8x+2x^2+8x]$

$D=(5x-4)[x^2+8x+16]$

$D=(5x-4)(x+4{)}^2$

$D=(-4+5x)(x-(-4){)}^2$

Hence, $A=-4,B=5$