Question

If a,b,c,d are complex numbers ,then the value of the determinant \Delta = \left| {\begin{array}{*{20}{c}} 2&{a + b + c + d}&{ab + cd} \\\ {a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)} \\\ {ab + cd}&{ab(c + d) + cd(a + b)}&{2abcd} \end{array}} \right|equals
A) 0
B) $abcd$
C) $a + b + c + d$
D) $2abcd$

Answer 1

Hint: We can split the determinant into two determinants where we can write the elements of the determinant as a sum of two determinants.

{{a_1} + {a_2}}&{{a_3} + {a_4}}&{{a_5} + {a_6}} \\\ {{b_1} + {b_2}}&{{b_3} + {b_4}}&{{b_5} + {b_6}} \\\ {{c_1} + {c_2}}&{{c_3} + {c_4}}&{{c_5} + {c_6}} \end{array}} \right|$$ Now we can split the above determinant as follows, $$\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}&{{a_5}} \\\ {{b_1}}&{{b_3}}&{{b_5}} \\\ {{c_1}}&{{c_3}}&{{c_5}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_4}}&{{a_6}} \\\ {{b_2}}&{{b_4}}&{{b_6}} \\\ {{c_2}}&{{c_4}}&{{c_6}} \end{array}} \right|$$ And let, $$\Delta = {\Delta _1} + {\Delta _2}$$ Where $${\Delta _1} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_3}}&{{a_5}} \\\ {{b_1}}&{{b_3}}&{{b_5}} \\\ {{c_1}}&{{c_3}}&{{c_5}} \end{array}} \right|$$and$${\Delta _2} = \left| {\begin{array}{*{20}{c}} {{a_2}}&{{a_4}}&{{a_6}} \\\ {{b_2}}&{{b_4}}&{{b_6}} \\\ {{c_2}}&{{c_4}}&{{c_6}} \end{array}} \right|$$ _Complete step by step answer:_ It is given in the question that, $$\Delta = \left| {\begin{array}{*{20}{c}} 2&{a + b + c + d}&{ab + cd} \\\ {a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)} \\\ {ab + cd}&{ab(c + d) + cd(a + b)}&{2abcd} \end{array}} \right|$$ Let us rewrite the elements of the equation as follows, $$\Delta = \left| {\begin{array}{*{20}{c}} {1 + 1}&{a + b + c + d}&{ab + cd} \\\ {a + b + c + d}&{(a + b)(c + d) + (a + b)(c + d)}&{ab(c + d) + cd(a + b)} \\\ {ab + cd}&{ab(c + d) + cd(a + b)}&{abcd + abcd} \end{array}} \right|$$ Now let us split the determinant into sum of two determinants, $$\Delta = \left| {\begin{array}{*{20}{c}} 1&{c + d}&{cd} \\\ {a + b}&{(a + b)(c + d)}&{cd(a + b)} \\\ {ab}&{ab(c + d)}&{abcd} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&{a + b}&{ab} \\\ {c + d}&{(a + b)(c + d)}&{ab(c + d)} \\\ {cd}&{cd(a + b)}&{abcd} \end{array}} \right|$$ Now let us take the common element in each of the above two determinants, then we get, $$\Delta = (c + d).cd\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ {a + b}&{(a + b)}&{(a + b)} \\\ {ab}&{ab}&{ab} \end{array}} \right| + (a + b).ab\left| {\begin{array}{*{20}{c}} 1&1&1 \\\ {c + d}&{(c + d)}&{(c + d)} \\\ {cd}&{cd}&{cd} \end{array}} \right|$$ From the first determinant we have taken (c+d) in common from the second column and cd from the third column, Similarly we have taken (a+b) in common from the second column and ab from the third column of the second determinant. Let us apply the following operations in both the determinants mentioned above, $${C_2}' = {C_2} - {C_1}$$ and $${C_3}' = {C_3} - {C_1}$$ where $${C_1}$$is the first column, $${C_2}$$is the second column, and $${C_3}$$is the third column. By applying the above process we get, $$\Delta = (c + d).cd\left| {\begin{array}{*{20}{c}} 1&0&0 \\\ {a + b}&0&0 \\\ {ab}&0&0 \end{array}} \right| + (a + b).ab\left| {\begin{array}{*{20}{c}} 1&0&0 \\\ {c + d}&0&0 \\\ {cd}&0&0 \end{array}} \right|$$ On finding the determinant value we get, $$\Delta = (c + d).cd \times 0 + (a + b).ab \times 0$$ On further simplification we arrive at the value of the determinant, $$\Delta = 0 + 0 = 0$$ Hence we have found the value of determinant as zero, Hence the determinant $$\Delta = \left| {\begin{array}{*{20}{c}} 2&{a + b + c + d}&{ab + cd} \\\ {a + b + c + d}&{2(a + b)(c + d)}&{ab(c + d) + cd(a + b)} \\\ {ab + cd}&{ab(c + d) + cd(a + b)}&{2abcd} \end{array}} \right|$$equals 0. _So, Option (A) is the correct answer._ Note: Additional information: If each row of any column of a determinant is a multiple of the same constant then we can take it outside the determinant taking common from each row.

\vartriangle = \left| {\begin{array}{{20}{c}}
{c{a_1}}&{{a_2}}&{{a_3}} \\
{c{b_1}}&{{b_2}}&{{b_3}} \\
{c{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right| \\
= c\left| {\begin{array}{{20}{c}}
{{a_1}}&{{a_2}}&{{a_3}} \\
{{b_1}}&{{b_2}}&{{b_3}} \\
{{c_1}}&{{c_2}}&{{c_3}}
\end{array}} \right| \\

$$By the property of determinant we have whenever two columns of a determinant are 0 then, the value of the determinant is 0.$$