$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is
$\alpha x + \left( {1 + \beta } \right)y + z = 0$
$\alpha x + \beta y + 2z = 0$
$D = \left| {\begin{array}{*{20}{c}}
{1 + \alpha }&\beta &1\\
\alpha &{1 + \beta }&1\\
\alpha &\beta &2
\end{array}} \right|$
${C_1} \to {C_1} + {C_2} + {C_3}$
$D = \left( {\alpha + \beta + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
1&{1 + \beta }&1\\
1&\beta &2
\end{array}} \right|$
${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$
$D = \left( {\alpha + \beta + 2} \right)\left| {\begin{array}{*{20}{c}}
1&\beta &1\\
0&1&0\\
0&0&1
\end{array}} \right| = \alpha + \beta + 2$
For unique solution $\alpha + \beta + 2 \ne 0 \Rightarrow \alpha + \beta \ne - 2$
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