The value of a for which the system of equati | vedclass

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(A) The system of linear equations has a non-zero solution if and only if the determinant of the coefficient matrix is zero,i.e.,$\Delta = 0$. $\Delta

Vedclass · Accepted Answer

$-1$

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(A) The system of linear equations has a non-zero solution if and only if the determinant of the coefficient matrix is zero,i.e.,$\Delta = 0$. $\Delta = \begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \ a & a+1 & a+2 \ 1 & 1 & 1 \end{vmatrix} = 0$. Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$: $\Delta = \begin{vmatrix} a^3 & (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \ a & 1 & 1 \ 1 & 0 & 0 \end{vmatrix} = 0$. Expanding along the third row $(R_3)$: $1 \cdot \begin{vmatrix} (a+1)^3 - a^3 & (a+2)^3 - (a+1)^3 \ 1 & 1 \end{vmatrix} = 0$. $((a+1)^3 - a^3) - ((a+2)^3 - (a+1)^3) = 0$. $(3a^2 + 3a + 1) - (3(a+1)^2 + 3(a+1) + 1) = 0$. $(3a^2 + 3a + 1) - (3a^2 + 6a + 3 + 3a + 3 + 1) = 0$. $(3a^2 + 3a + 1) - (3a^2 + 9a + 7) = 0$. $-6a - 6 = 0$. $-6a = 6 \Rightarrow a = -1$.

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,$x + y + z = 0$ has a non-zero solution is

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