Question
Question: Expand the following determinant: \[ \left| {\begin{array}{*{20}{c}} 1&{ - 3}&4 \\\ 3&5&{ - 3} ...
Expand the following determinant:
\left| {\begin{array}{*{20}{c}} 1&{ - 3}&4 \\\ 3&5&{ - 3} \\\ 2&{ - 5}&0 \end{array}} \right| \\\Explanation
Solution
{\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{.}} \\\