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Expand the following determinant
$\left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|$

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Last updated date: 21st Feb 2024
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Answer
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Hint: To solve this problem we will use the properties of determinant to expand the determinant given in the question.

Complete step-by-step answer:
Now, given determinant $\vartriangle = \left| {\begin{array}{*{20}{c}}
  1&{ - 3}&4 \\
  3&5&{ - 3} \\
  2&{ - 5}&0
\end{array}} \right|$
Now to solve the determinant we can expand the determinant through any row or through any column. We will get the same result in both cases. We will expand the given determinant through row ${R_1}$. Expanding the determinant, we get
$ \Rightarrow $ $\vartriangle = 1\left| {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 5}&0
\end{array}} \right| - ( - 3)\left| {\begin{array}{*{20}{c}}
  3&{ - 3} \\
  2&0
\end{array}} \right| + 4\left| {\begin{array}{*{20}{c}}
  3&5 \\
  2&{ - 5}
\end{array}} \right|$
$ \Rightarrow $ $\vartriangle = 1(0 - 15) + 3(0 + 6) + 4( - 15 - 10)$
$ \Rightarrow $ $\vartriangle = - 15 + 18 - 100$
$ \Rightarrow $ $\vartriangle = - 97$
So, the value of determinant after expanding = -97.

Note: While solving the problems related to determinant, make sure that you use proper sign convention while expanding the determinant. Signs should be proper whether you expand through row or by column, it is mandatory. Also, do calculations properly to avoid any mistake.