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(C) For the system to have a unique solution,the determinant of the coefficient matrix must be non-zero: $\Delta = \begin{vmatrix} 3 & -4 & k \ 1 & 2

Vedclass · Accepted Answer

-$2$

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(C) For the system to have a unique solution,the determinant of the coefficient matrix must be non-zero: $\Delta = \begin{vmatrix} 3 & -4 & k \ 1 & 2 & -1 \ k & -1 & 3 \end{vmatrix} 
eq 0$. Expanding the determinant: $3(6 - 1) + 4(3 + k) + k(-1 - 2k) 
eq 0$. $15 + 12 + 4k - k - 2k^2 
eq 0 \implies -2k^2 + 3k + 27 
eq 0 \implies 2k^2 - 3k - 27 
eq 0$. Factoring the quadratic: $(2k - 9)(k + 3) 
eq 0$,so $k 
eq \frac{9}{2}$ and $k 
eq -3$. Given $2\beta - \gamma = 8$,we use the second equation $x + 2y - z = 9$. Substituting $x = \alpha, y = \beta, z = \gamma$: $\alpha + 2\beta - \gamma = 9 \implies \alpha + 8 = 9 \implies \alpha = 1$. Since $k 
eq m$,$m$ represents the values for which the system does not have a unique solution,i.e.,$m \in \{\frac{9}{2}, -3\}$. If $m = -3$,then $\alpha + m = 1 + (-3) = -2$.

If the system of simultaneous linear equations $3x - 4y + kz + 13 = 0$,$x + 2y - z - 9 = 0$,and $kx - y + 3z + 7 = 0$ has a unique solution $x = \alpha, y = \beta, z = \gamma$ for $k \neq m$ and $2\beta - \gamma = 8$,then $\alpha + m =$

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