Question & Answer
QUESTION

Expand the following determinant:

   $| {\begin{array}{*{20}{c}}

  1&{ - 3}&4  

  3&5&{ - 3}  

  2&{ - 5}&0 

\end{array}} \right| $


ANSWER Verified Verified
\[
  {\text{Let }}\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|{\text{ be a general determinant }} \\
  {\text{As we know that }}\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|{\text{ is expanded as,}} \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|{\text{ }} = a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|{\text{ }} \\
  {\text{This can be reduced as,}} \\
   \Rightarrow a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|{\text{ = }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ }} \\
   \Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|{\text{ }} = {\text{ }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ (1)}} \\
  {\text{Now we know we have to expand }}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|{\text{ }} \\
  {\text{So, to expand }}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|{\text{ first we have to compare its elements with the elements of }}\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right|{\text{ }}. \\
  {\text{So on comparing we get }}a = 1,b = - 3,c = 4,d = 3,e = 5,f = - 3,g = 2,h = - 5{\text{ and }}i = 0 \\
  {\text{Now putting values of }}a,b,c,d,e,f,g,h{\text{ and }}i{\text{ in equation 3 we get,}} \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|{\text{ }} = {\text{ }}1\left( {5*0 - ( - 5)*( - 3)} \right) + 3\left( {3*0 - 2*( - 3)} \right) + 4\left( {3*( - 5) - 2*5} \right) \\
   \Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|{\text{ = }} - 15 + 18 - 100 = - 97 \\
  {\text{Hence }}\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|{\text{ }} = - 97{\text{ }} \\
  {\text{NOTE: - Whenever you came up to expand a determinant then better way is to expand using cofactors}}{\text{.}} \\
  {\text{While expanding calculations should be carefully done}}{\text{.}} \\
\]