The constant term in the expansion of left|b | vedclass

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(B) To find the constant term in the expansion of the determinant,we set $ x = 0 $.Substituting $ x = 0 $ into the determinant,we get:$ \Delta = \left

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$ 06 $

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(B) To find the constant term in the expansion of the determinant,we set $ x = 0 $.Substituting $ x = 0 $ into the determinant,we get:$ \Delta = \left|\begin{array}{ccc} 3(0)+1 & 2(0)-1 & 0+2 \ 5(0)-1 & 3(0)+2 & 0+1 \ 7(0)-2 & 3(0)+1 & 4(0)-1 \end{array}ight| $$ \Delta = \left|\begin{array}{ccc} 1 & -1 & 2 \ -1 & 2 & 1 \ -2 & 1 & -1 \end{array}ight| $Expanding along the first row:$ \Delta = 1 \cdot \left|\begin{array}{cc} 2 & 1 \ 1 & -1 \end{array}ight| - (-1) \cdot \left|\begin{array}{cc} -1 & 1 \ -2 & -1 \end{array}ight| + 2 \cdot \left|\begin{array}{cc} -1 & 2 \ -2 & 1 \end{array}ight| $$ \Delta = 1 \cdot (-2 - 1) + 1 \cdot (1 - (-2)) + 2 \cdot (-1 - (-4)) $$ \Delta = 1 \cdot (-3) + 1 \cdot (3) + 2 \cdot (3) $$ \Delta = -3 + 3 + 6 = 6 $Thus,the constant term is $ 6 $.

The constant term in the expansion of $ \left|\begin{array}{ccc} 3x+1 & 2x-1 & x+2 \\ 5x-1 & 3x+2 & x+1 \\ 7x-2 & 3x+1 & 4x-1 \end{array}\right| $ is

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