DeterminantHard
Question
If the system of equations
$${3x + y + 4z = 3 }{2x + \alpha y - z = - 3 }{X + 2y + z = 4 }$$has no solution, then the value of $\alpha$ is equal to:
Options
A.19
B.4
C.13
D.23
Solution
for no solution $\Delta = 0$
$${\left| \begin{matrix} 3 & 1 & 4 \\ 2 & \alpha & - 1 \\ 1 & 2 & 1 \end{matrix} \right| = 0 }{\Rightarrow 3(a + 2) + 1( - 1 - 2) + 4(4 - \alpha) = 0 }{\Rightarrow 19 - \alpha = 0 \Rightarrow \alpha = 19 }$$& for $\alpha = 19$
$${\Delta_{x} = \left| \begin{matrix} 3 & 1 & 4 \\ - 3 & 19 & - 1 \\ 4 & 2 & 1 \end{matrix} \right| = 3(21) + 1( - 1) + 4( - 82) }{\neq 0 }$$∴ no solution for $\alpha = 19$
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