Browse Questions

# There are two values of $a$ which makes the determinant $\Delta=\begin{vmatrix}1 & -2 & 5\\2 & a & 1\\0 & 4 & 2a\end{vmatrix}= 86,$ then sum of these number is

$\begin{array}{1 1} 4 \\ 5 \\ -4 \\ 9 \end{array}$

Toolbox:
• The value of the determinant for order $3\times 3$ is given by $\Delta=\begin{vmatrix}a_{11} &a_{12} & a_{13}\\a_{21} &a_{22} &a_{23}\\a_{31} &a_{32} &a_{33}\end{vmatrix}$
• Expanding along $R_1$ we get
• $|\Delta|=a_{11}(a_{22}\times a_{33}-a_{32}\times a_{23})-a_{12}(a_{21}\times a_{33}-a_{23}\times a_{31})+a_{13}(a_{21}\times a_{32}-a_{22}\times a_{31})$
on expanding along $R_1$ we get
$|\Delta|=1(2a^2-4)-(-2)(4a-0)+5(8=0)$
$\quad=2a^2-4+8a+40$
But |$\Delta|=86$ is given
Hence $2a^2+8a+36=86\Rightarrow 2a^2+8a+36-86=0$
$\Rightarrow 2a^2+8a-50=0$
Dividing through out by 2 we get,
$a^2+4a-25=0$
This is a quadratic equation with respect to a.Where the coefficient of a represents the sum of the roots.
Coefficient of a is 4.
Therefore the sum of the roots is 4
Hence A is the correct answer.