If the system of equations (k+1)^3 x + (k+2) | vedclass

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(B) The given system of equations is: $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $ $ (k+1) x + (k+2) y = k+3 $ $ x + y = 1 $ For the system to be consistent, t

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$-2$

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(B) The given system of equations is: $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $ $ (k+1) x + (k+2) y = k+3 $ $ x + y = 1 $ For the system to be consistent, the determinant of the augmented matrix must be zero. Let $ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 & (k+3)^3 \ k+1 & k+2 & k+3 \ 1 & 1 & 1 \end{vmatrix} = 0 $. Applying column operations $ C_2 ightarrow C_2 - C_1 $ and $ C_3 ightarrow C_3 - C_1 $: $ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 - (k+1)^3 & (k+3)^3 - (k+1)^3 \ k+1 & (k+2) - (k+1) & (k+3) - (k+1) \ 1 & 1 - 1 & 1 - 1 \end{vmatrix} = 0 $ $ D = \begin{vmatrix} (k+1)^3 & 3k^2 + 9k + 7 & 6k^2 + 24k + 26 \ k+1 & 1 & 2 \ 1 & 0 & 0 \end{vmatrix} = 0 $ Expanding along the third row: $ 1 \cdot [2(3k^2 + 9k + 7) - 1(6k^2 + 24k + 26)] = 0 $ $ 6k^2 + 18k + 14 - 6k^2 - 24k - 26 = 0 $ $ -6k - 12 = 0 $ $ -6k = 12 $ $ k = -2 $.

If the system of equations $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $, $ (k+1) x + (k+2) y = k+3 $, and $ x + y = 1 $ is consistent, then the value of $ k $ is:

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