The value of the determinant left| begin{arra | vedclass

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(D) Let $\Delta = \left| \begin{array}{ccc} 4 & -6 & 1 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight|$.Applying the row operation $R_1 	o R_1 + R

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$-25$

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(D) Let $\Delta = \left| \begin{array}{ccc} 4 & -6 & 1 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight|$.Applying the row operation $R_1 	o R_1 + R_2$:$\Delta = \left| \begin{array}{ccc} 4-1 & -6-1 & 1+1 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight| = \left| \begin{array}{ccc} 3 & -7 & 2 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight|$.Alternatively,applying $R_1 	o R_1 + R_3$:$\Delta = \left| \begin{array}{ccc} 4-4 & -6+11 & 1-1 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight| = \left| \begin{array}{ccc} 0 & 5 & 0 \ -1 & -1 & 1 \ -4 & 11 & -1 \end{array} ight|$.Expanding along the first row $(R_1)$:$\Delta = 0 \cdot ((-1)(-1) - (1)(11)) - 5 \cdot ((-1)(-1) - (1)(-4)) + 0 \cdot ((-1)(11) - (-1)(-4))$$\Delta = -5 \cdot (1 + 4) = -5 \cdot 5 = -25$.

The value of the determinant $\left| \begin{array}{ccc} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{array} \right|$ is

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