If the system of linear equations 2 x + 3 y - z = - 2 x + y + z = 4 x - y + λ z = 4 λ - 4 where λ ∈ ℝ , has no solution, then

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If we have three equations in three variables,

$a_{1} x + b_{1} y + c_{1} z = d_{1} a_{2} x + b_{2} y + c_{2} z = d_{2} a_{3} x + b_{3} y + c_{3} z = d_{3}$

then $∆ = |\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}|$ , $∆_{1} = |\begin{matrix} d_{1} & b_{1} & c_{1} \\ d_{2} & b_{2} & c_{2} \\ d_{3} & b_{3} & c_{3} \end{matrix}|$ , $∆_{2} = |\begin{matrix} a_{1} & d_{1} & c_{1} \\ a_{2} & d_{2} & c_{2} \\ a_{3} & d_{3} & c_{3} \end{matrix}|$ , $∆_{3} = |\begin{matrix} a_{1} & b_{1} & d_{1} \\ a_{2} & b_{2} & d_{2} \\ a_{3} & b_{3} & d_{3} \end{matrix}|$ .

If the set of equations do not have a solution then $∆ = 0$ and at least one among $∆_{1}, ∆_{2}, ∆_{3}$ is not equal to zero.

$∆ = |\begin{matrix} 2 & 3 & - 1 \\ 1 & 1 & 1 \\ 1 & - 1 & |λ| \end{matrix}| = 0$

$\Rightarrow |λ| = 7 \Rightarrow λ = \pm 7 \dots (1)$

Case 1: $λ = 7$

The system of equations are

$2 x + 3 y - z = - 2 x + y + z = 4 x - y + 7 z = 24$

$a (2 x + 3 y - z + 2) + b (x + y + z - 4) = (x - y + 7 z - 24)$

Comparing the coefficients of $x$ and $y$ we get,

$2 a + b = 1 3 a + b = - 1 \Rightarrow a = - 2, b = 5$

For these values of $a$ and $b$ , coefficients of $z$ and the constant values are matching. So the set of equations will have infinitely many solutions.

Case 2: $λ = - 7$

The system of equations are

$2 x + 3 y - z = - 2 x + y + z = 4 x - y - 7 z = - 32$

$a (2 x + 3 y - z + 2) + b (x + y + z - 4) = (x - y - 7 z + 32)$

Comparing the coefficients of $x$ and $y$ we get,

$2 a + b = 1 3 a + b = - 1 \Rightarrow a = - 2, b = 5$

For these values of $a$ and $b$ , coefficients of $z$ and the constant values are not matching. So the set of equations will have no solutions.

Mathematics - Determinant Question with Solution | TestHub

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