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

Last updated date: 25th Jul 2024
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Hint: To solve this problem we will use the properties of determinant to expand the determinant given in the question.

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
$\Rightarrow$ $\vartriangle = 1\left| {\begin{array}{*{20}{c}} 5&{ - 3} \\ { - 5}&0 \end{array}} \right| - ( - 3)\left| {\begin{array}{*{20}{c}} 3&{ - 3} \\ 2&0 \end{array}} \right| + 4\left| {\begin{array}{*{20}{c}} 3&5 \\ 2&{ - 5} \end{array}} \right|$
$\Rightarrow$ $\vartriangle = 1(0 - 15) + 3(0 + 6) + 4( - 15 - 10)$
$\Rightarrow$ $\vartriangle = - 15 + 18 - 100$
$\Rightarrow$ $\vartriangle = - 97$