Question

If $\left[\begin{array}{ccc}1& x& 1\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 0& 5& 1\\ 0& 3& 2\end{array}\right]\left[\begin{array}{c}1\\ 1\\ x\end{array}\right]=0,$ then $x=$

• A. $\frac{-9\pm \sqrt{35}}{2}$
• B. $\frac{-7\pm \sqrt{53}}{2}$
• C. $\frac{-9\pm \sqrt{53}}{2}$
• D. $\frac{-7\pm \sqrt{35}}{2}$

Answer 1

The correct option is C $\frac{-9\pm \sqrt{53}}{2}$

Given,

$\left(\begin{array}{ccc}1& x& 1\end{array}\right)\left(\begin{array}{ccc}1& 3& 2\\ 0& 5& 1\\ 0& 3& 2\end{array}\right)\left(\begin{array}{c}1\\ 1\\ x\end{array}\right)=0$

Upon multiplying the first 2 matrices, we get,

$\left(\begin{array}{ccc}1\cdot 1+x\cdot 0+1\cdot 0& 1\cdot 3+x\cdot 5+1\cdot 3& 1\cdot 2+x\cdot 1+1\cdot 2\end{array}\right)\left(\begin{array}{c}1\\ 1\\ x\end{array}\right)=0$

Upon further simplification, we get,

$\left(\begin{array}{ccc}1& 6+5x& 4+x\end{array}\right)\left(\begin{array}{c}1\\ 1\\ x\end{array}\right)=0$

Again multiplying the above matrices, we get,

$\left(\begin{array}{c}1\cdot 1+(6+5x)\cdot 1+(4+x)x\end{array}\right)=0$

Upon further simplification, we get,

$\left(\begin{array}{c}{x}^2+9x+7\end{array}\right)=0$

Solving the above quadratic equation, we get,

$x=\frac{-9\pm \sqrt{53}}{2}$