JEE Main 2021 — Mathematics Algebra

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

We have $D=|\begin{matrix}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{matrix}|,{D}_{x}=|\begin{matrix}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{matrix}|,{D}_{y}=|\begin{matrix}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{matrix}|$D=∣a1​a2​a3​​b1​b2​b3​​c1​c2​c3​​∣,Dx​=∣d1​d2​d3​​b1​b2​b3​​c1​c2​c3​​∣,Dy​=∣a1​a2​a3​​d1​d2​d3​​c1​c2​c3​​∣ and ${D}_{z}=|\begin{matrix}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{matrix}|.$Dz​=∣a1​a2​a3​​b1​b2​b3​​d1​d2​d3​​∣.

And, the system has no solution, if $D=0$D=0 and atleast one of {D}_{x},{D}_{y}&{D}_{z} is non-zero.

Thus, for the given system of equations,

$x+y+z=6,$x+y+z=6, $3x+5y+5z=26$3x+5y+5z=26 and $x+2y+\lambda z=\mu$x+2y+λz=μ

We have $D=|\begin{matrix}1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{matrix}|=0$D=∣131​152​15λ​∣=0

$$\Rightarrow 1(5\lambda -10)-1(3\lambda -5)+1(6-5)=0$$⇒1(5λ−10)−1(3λ−5)+1(6−5)=0

$$\Rightarrow 5\lambda -10-3\lambda +5+1=0$$⇒5λ−10−3λ+5+1=0

$$\Rightarrow 2\lambda =4$$⇒2λ=4

$$\Rightarrow \lambda =2.$$⇒λ=2.

And, ${D}_{z}=|\begin{matrix}1 & 1 & 6 \\ 3 & 5 & 26 \\ 1 & 2 & \mu \end{matrix}|\neq 0$Dz​=∣131​152​626μ​∣=0

$$\Rightarrow 1(5\mu -52)-1(3\mu -26)+6(6-5)\neq 0$$⇒1(5μ−52)−1(3μ−26)+6(6−5)=0

$$\Rightarrow 5\mu -52-3\mu +26+6\neq 0$$⇒5μ−52−3μ+26+6=0

$$\Rightarrow 2\mu \neq 20$$⇒2μ=20

$$\Rightarrow \mu \neq 10.$$⇒μ=10.

$\therefore$∴ For no solution $\lambda =2$λ=2 and $\mu \neq 10.$μ=10.