JEE Main 2022 — Determinants Question with Solution

If we have three equations in three variables, 

$$\begin{gathered} a_{1}x+b_{1}y+c_{1}z=d_1 \\ {a}_{2}x+b_{2}y+c_{2}z=d_2 \\ {a}_{3}x+b_{3}y+c_{3}z=d_3 \end{gathered}$$

then $∆=\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|$,${∆}_1=\left|\begin{array}{ccc}{d}_1& b_1& c_1\\ d_2& b_2& c_2\\ d_3& b_3& c_3\end{array}\right|$,${∆}_2=\left|\begin{array}{ccc}{a}_1& d_1& c_1\\ a_2& d_2& c_2\\ a_3& d_3& c_3\end{array}\right|$,${∆}_3=\left|\begin{array}{ccc}{a}_1& b_1& d_1\\ a_2& b_2& d_2\\ a_3& b_3& d_3\end{array}\right|$.

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.

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

$\Rightarrow \left|\lambda \right|=7\Rightarrow \lambda =\pm 7 \dots \left(1\right)$

Case 1: $\lambda =7$

The system of equations are 

$$\begin{gathered} 2x+3y-z=-2 \\ x+y+z=4 \\ x-y+7z=24 \end{gathered}$$

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

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

$$\begin{gathered} 2a+b=1 \\ 3a+b=-1 \\ \Rightarrow a=-2, b=5 \end{gathered}$$

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 

$$\begin{gathered} 2x+3y-z=-2 \\ x+y+z=4 \\ x-y-7z=-32 \end{gathered}$$

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

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

$$\begin{gathered} 2a+b=1 \\ 3a+b=-1 \\ \Rightarrow a=-2, b=5 \end{gathered}$$

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.