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.
\[\Delta = \left| {\begin{array}{*{20}{c}}
  {{a_1} + {a_2}}&{{a_3} + {a_4}}&{{a_5} + {a_6}} \\
  {{b_1} + {b_2}}&{{b_3} + {b_4}}&{{b_5} + {b_6}} \\
  {{c_1} + {c_2}}&{{c_3} + {c_4}}&{{c_5} + {c_6}}
\end{array}} \right|\]
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.