Question

If $p{x}^4+q{x}^3+r{x}^2+sx+t=\left|\begin{array}{ccc}{x}^2+3x& x-1& x+3\\ x+1& 2-x& x-3\\ x-3& x+4& 3x\end{array}\right|$ then $t$ is equal to

• A. 33
• B. 0
• C. 21
• D. none

Answer 1

The correct option is C 21

Given, $p{x}^4+q{x}^3+r{x}^2+sx+t=\left|\begin{array}{ccc}{x}^2+3x& x-1& x+3\\ x+1& 2-x& x-3\\ x-3& x+4& 3x\end{array}\right|$
Put $x=0$

$\Rightarrow t=\left|\begin{array}{ccc}0& -1& 3\\ 1& 2& -3\\ -3& 4& 0\end{array}\right|$

On expanding the above determinant we get,

$\therefore t=21$

Hence, option C.