JEE Mathematics 2023 Question with Solution

Given:

$$7x + 11y + \alpha z = 13$$ $$5x + 4y + 7z = \beta$$ $$175x + 194y + 57z = 361$$

Find: $\alpha + \beta + 2$, given that the system has infinitely many solutions.

For infinitely many solutions, one equation must be a linear combination of the other two.

Eliminate $x$ and $y$ by taking $10$ times the first equation and $21$ times the second equation, then comparing with the third equation:

$$10(7x + 11y + \alpha z) + 21(5x + 4y + 7z) - (175x + 194y + 57z) = 0$$

Expanding,

$$70x + 110y + 10\alpha z + 105x + 84y + 147z - 175x - 194y - 57z = 130 + 21\beta - 361$$

So,

$$(10\alpha + 90)z = 21\beta - 231$$

For infinitely many solutions, this relation must hold identically. Hence,

$$10\alpha + 90 = 0$$

which gives

$$\alpha = -9$$

Also,

$$21\beta - 231 = 0$$

so

$$\beta = 11$$

Therefore,

$$\alpha + \beta + 2 = -9 + 11 + 2 = 4$$

So the computed value is $4$. This matches option D.

However, the provided the solution labels the correct option as B, which disagrees with the working and with the listed options. Based on the extracted working, the correct option is D.

Given: the third equation must be dependent on the first two for the system to have infinitely many solutions.

Observe that

$$175 = 10 \cdot 7 + 21 \cdot 5$$

and

$$194 = 10 \cdot 11 + 21 \cdot 4$$

So the third equation is obtained by

$$10 \times \text{(equation 1)} + 21 \times \text{(equation 2)}$$

for the coefficients of $x$ and $y$.

Therefore the same must hold for the coefficient of $z$ and the constant term:

$$10\alpha + 21 \cdot 7 = 57$$ $$10\alpha + 147 = 57$$ $$10\alpha = -90$$ $$\alpha = -9$$

Also,

$$10 \cdot 13 + 21\beta = 361$$ $$130 + 21\beta = 361$$ $$21\beta = 231$$ $$\beta = 11$$

Hence,

$$\alpha + \beta + 2 = -9 + 11 + 2 = 4$$

Therefore, the correct option is D.