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

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

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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

If $a, b, c > 0$ and $x, y, z \in R$,then the value of the determinant $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^y + b^{-y})^2 & (b^y - b^{-y})^2 & 1 \\ (c^z + c^{-z})^2 & (c^z - c^{-z})^2 & 1 \end{array} \right|$ is equal to:

$\left|\begin{array}{ccc}1 & bc+ad & b^2c^2+a^2d^2 \\ 1 & ca+bd & c^2a^2+b^2d^2 \\ 1 & ab+cd & a^2b^2+c^2d^2\end{array}\right|=$

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

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

$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right| = $

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