If Delta=left|begin{array}{lll}1 & 5 & 6 0 & | vedclass

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(B) Given, $\Delta = \left|\begin{array}{lll}1 & 5 & 6 \ 0 & 1 & 7 \ 0 & 0 & 1\end{array}ight|$.Since this is an upper triangular matrix, the dete

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$(\Delta+\Delta^{\prime})^2-3(\Delta+\Delta^{\prime})+2=0$

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(B) Given, $\Delta = \left|\begin{array}{lll}1 & 5 & 6 \ 0 & 1 & 7 \ 0 & 0 & 1\end{array}ight|$.Since this is an upper triangular matrix, the determinant is the product of the diagonal elements:$\Delta = 1 	imes 1 	imes 1 = 1$.Now, $\Delta^{\prime} = \left|\begin{array}{ccc}1 & 0 & 1 \ 3 & 0 & 3 \ 4 & 6 & 100\end{array}ight|$.Expanding along the second column:$\Delta^{\prime} = -0(300-12) + 0(100-4) - 6(3-3) = 0$.Now, substitute $\Delta = 1$ and $\Delta^{\prime} = 0$ into the expression $(\Delta+\Delta^{\prime})^2-3(\Delta+\Delta^{\prime})+2$:$= (1+0)^2 - 3(1+0) + 2$$= 1^2 - 3 + 2$$= 1 - 3 + 2 = 0$.Thus, the correct option is $B$.

If $\Delta=\left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$ and $\Delta^{\prime}=\left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$, then

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