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

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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 & 0 & 0 \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 \Rightarrow k = -2$.

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

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