Question

Solve the determinant:

$\dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ 0&7&1 \\ 5& - 5&1\end{array}\right| + \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ - 4& - 2&1 \\ 5& - 5&1\end{array}\right|$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given determinant is

$\rm \: \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ 0&7&1 \\ 5& - 5&1\end{array}\right| + \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ - 4& - 2&1 \\ 5& - 5&1\end{array}\right|$

$\boxed{\tt{ \: OP \: R_2 \: \to \: R_2 \: - \: R_1 \: \: \: \: [in \: both \: determinants]}}$

$\rm \: = \: \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ 4&2&0 \\ 5& - 5&1\end{array}\right| + \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\0& - 7&0 \\ 5& - 5&1\end{array}\right|$

$\boxed{\tt{ \: OP \: R_3 \: \to \: R_3 \: - \: R_1 \: \: \: \: [in \: both \: determinants]}}$

$\rm \: = \: \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ 4&2&0 \\ 9& - 10&0\end{array}\right| + \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\0& - 7&0 \\ 9& - 10&0\end{array}\right|$

Now, Expanding along 3rd column, we get

$\rm \: = \: \dfrac{1}{2} ( - 40 - 18)+ \dfrac{1}{2} (0 + 63)$

$\rm \: = \: \dfrac{1}{2} ( - 58)+ \dfrac{1}{2} (63)$

$\rm \: = \: \dfrac{ - 58}{2} + \dfrac{63}{2}$

$\rm \: = \: \dfrac{ - 58 + 63}{2}$

$\rm \: = \: \dfrac{5}{2}$

Hence,

$\boxed{\tt{ \rm \:\dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ 0&7&1 \\ 5& - 5&1\end{array}\right| + \dfrac{1}{2} \left | \begin{array}{ccc} - 4 &5&1 \\ - 4& - 2&1 \\ 5& - 5&1\end{array}\right| = \: \dfrac{5}{2}}}$

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ADDITIONAL INFORMATION

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.