Question

Expand the following determinant:

$| {\begin{array}{*{20}{c}}

1&{ - 3}&4

3&5&{ - 3}

2&{ - 5}&0

\end{array}} \right| $

Answer 1

{\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{.}} \\\