If the system of linear equations 2x + 3y - z = - 2 x + y + z = 4 x...

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 ∆ = \left|\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right|,{∆}_{1} = \left|\begin{cases}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{cases}\right|,{∆}_{2} = \left|\begin{cases}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{cases}\right|,{∆}_{3} = \left|\begin{cases}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{cases}\right|.

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.

∆ = \left|\begin{cases}2 & 3 & - 1 \\ 1 & 1 & 1 \\ 1 & - 1 & \left|\lambda\right|\end{cases}\right| = 0

$⇒ \left|\lambda\right| = 7 ⇒ \lambda = \pm 7 \ldots \left(1\right)$

Case 1: $\lambda = 7$

The system of equations are

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

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

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

$2a + b = 13a + b = - 1 ⇒ a = - 2, b = 5$

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: $\lambda = - 7$

The system of equations are

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

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

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

$2a + b = 13a + b = - 1 ⇒ a = - 2, b = 5$

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.