Question

The integral value of $x,$ for $\left|\begin{array}{ccc}{x}^2& x& 1\\ 0& 2& 1\\ 3& 1& 4\end{array}\right|=28$ is

• A. $2$
• B. $4$
• C. $8$
• D. $12$

Answer 1

The correct option is A $2$

$|A|=\left|\begin{array}{ccc}{x}^2& x& 1\\ 0& 2& 1\\ 3& 1& 4\end{array}\right|$
Expanding along the first row
$|A|=x^2\left|\begin{array}{cc}2& 1\\ 1& 4\end{array}\right|-x\left|\begin{array}{cc}0& 1\\ 3& 4\end{array}\right|+1\left|\begin{array}{cc}0& 2\\ 3& 1\end{array}\right|$
$=x^2(2\times 4-1\times 1)-x(0\times 4-1\times 3)+(0\times 1-2\times 3)$
$=x^2(8-1)-x(0-3)+1(0-6)$
$=7x^2+3x-6$
Also $|A|=28$
$\Rightarrow 7x^2+3x-6=28$
$\Rightarrow 7x^2+3x-34=0$
$\Rightarrow 7x^2+17x-14x-34=0$
$\Rightarrow x(7x+17)-2(7x+17)=0$
$\Rightarrow (x-2)(7x+17)=0$
$x=2,-\frac{17}{7}$
So integral value of $x$ is $2$