JEE Advanced 2023 · Paper 1 · Mathematics Q14

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MathematicsList match+3 / -1Question 14 of 51

Let $\alpha$ , $\beta$ and $\gamma$ be real numbers. Consider the following system of linear equations

$\begin{aligned} x + 2y + z &= 7\\ x + \alpha z &= 11\\ 2x - 3y + \beta z &= \gamma \end{aligned}$

Match each entry in List-I to the correct entries in List-II.

The correct option is:

List-I

P
If $\beta = \frac12(7\alpha - 3)$ and $\gamma = 28$ , then the system has
Q
If $\beta = \frac12(7\alpha - 3)$ and $\gamma \neq 28$ , then the system has
R
If $\beta \neq \frac12(7\alpha - 3)$ where $\alpha = 1$ and $\gamma \neq 28$ , then the system has
S
If $\beta \neq \frac12(7\alpha - 3)$ where $\alpha = 1$ and $\gamma = 28$ , then the system has

List-II

1
a unique solution
2
no solution
3
infinitely many solutions
4
$x = 11$ , $y = -2$ and $z = 0$ as a solution
5
$x = -15$ , $y = 4$ and $z = 0$ as a solution

P → 3; Q → 2; R → 1; S → 4

P → 3; Q → 2; R → 5; S → 4

P → 2; Q → 1; R → 4; S → 5

P → 2; Q → 1; R → 1; S → 3

Answer and worked solution

Official answer: A

Determinant.
$\Delta=\begin{vmatrix}1&2&1\\1&0&\alpha\\2&-3&\beta\end{vmatrix}=7\alpha-2\beta-3,$
which is zero exactly when $\beta=\tfrac12(7\alpha-3)$ .

When $\Delta=0$ . The third left-hand side is a combination of the first two: $-\frac32(\text{eq}_1)+\frac72(\text{eq}_2)$ . This gives coefficients $(2,-3,\beta)$ and right-hand side $-\frac{21}2+\frac{77}2=28$ . So:

(P) $\gamma=28$ : the third equation is redundant, giving infinitely many solutions → 3.
(Q) $\gamma\ne28$ : the equations are inconsistent, giving no solution → 2.

When $\Delta\ne0$ there is a unique solution:

(R) → 1.
(S) With $\alpha=1$ , $\gamma=28$ : $(11,-2,0)$ satisfies $11-4=7$ , $11=11$ and $22+6=28$ , so it is the solution → 4.

P→3, Q→2, R→1, S→4.

Topics:Systems of linear equations Determinants and properties

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JEE Advanced 2023 · Paper 1, Mathematics question 14