The determinant left| {begin{array}{*{20}{c}} | vedclass

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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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

By using properties of determinants,show that:
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

By using properties of determinants,show that:
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

$\left| {\begin{array}{ccc} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right| = $

The determinant $\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} \right|$ is a divisor of:

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