Question

The value of the determinant $\Delta =\left|\begin{array}{ccc}x+2& x+3& x+5\\ x+4& x+6& x+9\\ x+8& x+11& x+15\end{array}\right|$, is

• A. $2$
• B. $-2$
• C. $3$
• D. $x-1$

Answer 1

Answer: (B)

$-2$

$\Delta =\left|\begin{array}{ccc}x+2& x+3& x+5\\ x+4& x+6& x+9\\ x+8& x+11& x+15\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=\left|\begin{array}{ccc}-1& -2& x+5\\ -2& -3& x+9\\ -3& -4& x+15\end{array}\right|$

$Applying-C_{1}and-C_{2}weget$
$=\left|\begin{array}{ccc}1& 2& x+5\\ 2& 3& x+9\\ 3& 4& x+15\end{array}\right|$

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$=\left|\begin{array}{ccc}-1& -1& -4\\ -1& -1& -6\\ 3& 4& x+15\end{array}\right|$

$Applying-C_{1}and-C_2$
$=\left|\begin{array}{ccc}1& 1& 4\\ 1& 1& 6\\ 3& 4& x+15\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 4\\ 1& 1& 6\\ 3& 4& x+15\end{array}\right|$
$=\left|\begin{array}{ccc}0& 0& -2\\ 1& 1& 6\\ 3& 4& x+15\end{array}\right|$
On expanding the above determinanat we get,
$=-2$