JEE Mathematics 2023 Question with Solution

Given: The first system is

$$\begin{aligned} x + y + kz &= 2 \\ 2x + 3y - z &= 1 \\ 3x + 4y + 2z &= k \end{aligned}$$

and it has infinitely many solutions.

The second system is

$$\begin{aligned} (k+1)x + (2k-1)y &= 7 \\ (2k+1)x + (k+5)y &= 10 \end{aligned}$$

Find: Which option is correct for the second system.

For the first system to have infinitely many solutions, the determinant of the coefficient matrix must be zero.

$$\begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix} = 0$$

Expanding along the first row,

$$\text{Det}(A)=1\begin{vmatrix}3 & -1\\4 & 2\end{vmatrix}-1\begin{vmatrix}2 & -1\\3 & 2\end{vmatrix}+k\begin{vmatrix}2 & 3\\3 & 4\end{vmatrix}$$

Now,

$$\begin{vmatrix}3 & -1\\4 & 2\end{vmatrix}=6+4=10, \qquad \begin{vmatrix}2 & -1\\3 & 2\end{vmatrix}=4+3=7, \qquad \begin{vmatrix}2 & 3\\3 & 4\end{vmatrix}=8-9=-1$$

Therefore,

$$\text{Det}(A)=10-7-k=3-k$$

So,

$$3-k=0 \implies k=3$$

Substituting $k=3$ in the second system,

$$4x+5y=7$$

and

$$7x+8y=10$$

Subtracting the first equation from the second,

$$(7x+8y)-(4x+5y)=10-7$$

which gives

$$3x+3y=3 \implies x+y=1$$

Also, the coefficient determinant of the second system is

$$\begin{vmatrix}4 & 5\\7 & 8\end{vmatrix}=32-35=-3 \ne 0$$

so the system has a unique solution.

Therefore, the second system has a unique solution satisfying $x+y=1$. The correct option is D.

Given: The first system has infinitely many solutions.

Find: The nature of the second system.

From the first system,

$$\begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix}=0$$

This gives

$$3-k=0 \implies k=3$$

Now the second system becomes

$$4x+5y=7$$ $$7x+8y=10$$

Subtracting the first equation from the second immediately gives

$$3x+3y=3$$

Hence,

$$x+y=1$$

Since the two equations are not proportional, the solution is unique.

Therefore, the correct option is D.