The determinant left| begin{array}{ccc} 1 & 1 | vedclass

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(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 6 \end{array} ight|$.Applying $C_1 	o C_1 + C_2$,we get $D = \left| \begin

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$\left| \begin{array}{ccc} 2 & 1 & 1 \ 2 & 2 & 3 \ 2 & 3 & 6 \end{array} ight|$

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(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 6 \end{array} ight|$.Applying $C_1 	o C_1 + C_2$,we get $D = \left| \begin{array}{ccc} 2 & 1 & 1 \ 3 & 2 & 3 \ 4 & 3 & 6 \end{array} ight|$,which is option $B$.Applying $C_2 	o C_2 + C_3$ to the original determinant,we get $D = \left| \begin{array}{ccc} 1 & 2 & 1 \ 1 & 5 & 3 \ 1 & 9 & 6 \end{array} ight|$,which is option $C$.Applying $C_1 	o C_1 + C_2 + C_3$ to the original determinant,we get $D = \left| \begin{array}{ccc} 3 & 1 & 1 \ 6 & 2 & 3 \ 10 & 3 & 6 \end{array} ight|$,which is option $D$.Option $A$ is $\left| \begin{array}{ccc} 2 & 1 & 1 \ 2 & 2 & 3 \ 2 & 3 & 6 \end{array} ight| = 2 \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & 6 \end{array} ight| = 2D 
eq D$.Thus,the determinant is not equal to option $A$.

The determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{array} \right|$ is not equal to

Similar Questions

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then

The value of the determinant $\left|\begin{array}{ccc}a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b\end{array}\right|$ is

If $a_{n} (>0)$ is the $n^{\text{th}}$ term of a $G$.$P$.,then the value of the determinant $\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|$ is equal to:

Without expanding,prove that $\Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix} = 0$.

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

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