JEE Main 2022 — Mathematics Algebra

If we have three equations in three variables,

$${a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1} {a}_{2}x+{b}_{2}y+{c}_{2}z={d}_{2} {a}_{3}x+{b}_{3}y+{c}_{3}z={d}_{3}$$a1​x+b1​y+c1​z=d1​a2​x+b2​y+c2​z=d2​a3​x+b3​y+c3​z=d3​

then $\Delta =|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|$Δ=∣a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣,${\Delta }_{1}=|\begin{matrix}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{matrix}|$Δ1​=∣d1​d2​d3​​b1​b2​b3​​c1​c2​c3​​∣,${\Delta }_{2}=|\begin{matrix}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{matrix}|$Δ2​=∣a1​a2​a3​​d1​d2​d3​​c1​c2​c3​​∣,${\Delta }_{3}=|\begin{matrix}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{matrix}|$Δ3​=∣a1​a2​a3​​b1​b2​b3​​d1​d2​d3​​∣.

If the set of equations do not have a solution then $\Delta =0$Δ=0 and at least one among ${\Delta }_{1},{\Delta }_{2},{\Delta }_{3}$Δ1​,Δ2​,Δ3​ is not equal to zero.

$$\Delta =|\begin{matrix}2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda |\end{matrix}|=0$$Δ=∣211​31−1​−11∣λ∣​∣=0

$$\Rightarrow |\lambda |=7\Rightarrow \lambda =\pm 7\ldots (1)$$⇒∣λ∣=7⇒λ=±7…(1)

Case 1: $\lambda =7$λ=7

The system of equations are

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

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

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

$$2a+b=1 3a+b=-1 \Rightarrow a=-2,b=5$$2a+b=13a+b=−1⇒a=−2,b=5

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

Case 2: $\lambda =-7$λ=−7

The system of equations are

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

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

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

$$2a+b=1 3a+b=-1 \Rightarrow a=-2,b=5$$2a+b=13a+b=−1⇒a=−2,b=5

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