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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{b

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$2 \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\{{a_2}}&{{b_2}}&{{c_2}}\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} ight|$

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(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.Applying the operation $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\Delta = \left| {\begin{array}{*{20}{c}}{2(a_1+b_1+c_1)}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{2(a_2+b_2+c_2)}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{2(a_3+b_3+c_3)}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.Taking $2$ common from $C_1$:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{a_1} + {b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{a_2} + {b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{a_3} + {b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.Applying $C_1 ightarrow C_1 - C_2$:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\{{b_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\{{b_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} ight|$.Applying $C_2 ightarrow C_2 - C_3$:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1}}&{{a_1} + {b_1}}\{{b_2}}&{{c_2}}&{{a_2} + {b_2}}\{{b_3}}&{{c_3}}&{{a_3} + {b_3}}\end{array}} ight|$.Applying $C_3 ightarrow C_3 - C_1$:$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{b_1}}&{{c_1}}&{{a_1}}\{{b_2}}&{{c_2}}&{{a_2}}\{{b_3}}&{{c_3}}&{{a_3}}\end{array}} ight|$.By swapping $C_1$ and $C_2$,then $C_2$ and $C_3$ (two swaps,so sign remains positive):$\Delta = 2 \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\{{a_2}}&{{b_2}}&{{c_2}}\{{a_3}}&{{b_3}}&{{c_3}}\end{array}} ight|$.

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1} + {c_1}}&{{c_1} + {a_1}}&{{a_1} + {b_1}}\\{{b_2} + {c_2}}&{{c_2} + {a_2}}&{{a_2} + {b_2}}\\{{b_3} + {c_3}}&{{c_3} + {a_3}}&{{a_3} + {b_3}}\end{array}} \right|$ is equal to:

Similar Questions

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|=a^{3}$

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

The value of the determinant $\left| {\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$ is independent of:

If ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$,then the value of $\sum\limits_{r = 1}^n {D_r} = $

If $\left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}} \right| = k\,abc{(a + b + c)^3}$,then the value of $k$ is

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