Question

If $\left|\begin{array}{ccc}x+2& 2x+3& 3x+4\\ 2x+3& 3x+4& 4x+5\\ 3x+5& 5x+8& 10x+17\end{array}\right|=0$ then $x$ is equal to

• A. $-1,-2$
• B. $1,2$
• C. $1,-2$
• D. $-1,2$

Answer 1

Answer: (A)

$(-1,-2)$

$\left|\begin{array}{ccc}x+2& 2x+3& 3x+4\\ 2x+3& 3x+4& 4x+5\\ 3x+5& 5x+8& 10x+17\end{array}\right|=0$
From $R_2\to R_2-2R_1,R_3\to R_3-3R_1$
$\Rightarrow \left|\begin{array}{ccc}x+2& 2x+3& 3x+4\\ -1& -(x+2)& -2x-3\\ -1& -(x+1)& x+5\end{array}\right|=0$
From $R_1\to R_1+R_2$
$\Rightarrow \left|\begin{array}{ccc}x+1& x+1& x+1\\ -1& -(x+2)& -(2x+3)\\ -1& -(x+1)& x+5\end{array}\right|=0$
Takin $(x+1)$ as a common from first row and also taking $-1$ as a common from both second and third row's.
$\Rightarrow (x+1)\left|\begin{array}{ccc}1& 1& 1\\ 1& x+2& 2x+3\\ 1& x+1& -(x+5)\end{array}\right|=0$
$R_2\to R_2-R_1$
$\Rightarrow (x+1)\left|\begin{array}{ccc}1& 1& 1\\ 0& (x+1)& 2(x+1)\\ 1& x+1& -(x+5)\end{array}\right|=0$
Again taking $(x+1)$ as a common from second row .
$\Rightarrow (x+1{)}^2\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 1& x+1& -(x+5)\end{array}\right|=0$
$R_3\to R_3-R_1$
$\Rightarrow {(x+1)}^2\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& x& -(x+6)\end{array}\right|=0$
Expand determinant along first column, we get
$\Rightarrow (x+1{)}^2[-x-6-2x]=0$
$\Rightarrow (x+1{)}^2(-3x-6)=0$
$\Rightarrow (x+1{)}^2=0$ and $-3x-6=0$
$\Rightarrow x+1=0$ and $-3x=6$
$\Rightarrow x=-1$ and $x=\frac{6}{-3}$
$\therefore x=-1,-2$