Let α, β and γ be real numbers. Consider the following system of linear equations x +2 y + z =7 x+α z=11 2 x-3 y+β z=γ Match each entry in List-I to the correct entries in List-II. LIST-I LIST-II --- --- P) If β=1/2(7 α-3) and γ=28 , then the system has 1)A unique solution Q) If β=1/2(7 α-3) and γ neq 28 , then the system has 2) No solution R) If β neq 1/2(7 α-3) where α=1 and γ neq 28 , then the system has 3) Infinitely many soliutions S) If β neq 1/2(7 α-3) where α=1 and γ=28 , then the system has 4) x =11, y =-2 and z =0 as a solution 5) x =-15, y =4 and z =0 as a solution

( P ) rightarrow(3) & ( Q ) rightarrow(2) & ( R ) rightarrow(1) & ( S ) rightarrow(4)

( P ) rightarrow(3) & ( Q ) rightarrow(2) & ( R ) rightarrow(5) & ( S ) rightarrow(4)

( P ) rightarrow(2) & ( Q ) rightarrow(1) & ( R ) rightarrow(4) & ( S ) rightarrow(5)

( P ) rightarrow(2) & ( Q ) rightarrow(1) & ( R ) rightarrow(1) & ( S ) rightarrow(3)

Mathematics · Determinants

JEE Advanced 2023 — Paper 1 — Question 13

Let $\alpha, \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations $\mathrm{x}+2\mathrm{y}+\mathrm{z}=7$ $x+\alpha z=11$ $2 x-3 y+\beta z=\gamma$ Match each entry in List-I to the correct entries in List-II.

LIST-I	LIST-II
P) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$ , then the system has	1)A unique solution
Q) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$ , then the system has	2) No solution
R) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$ , then the system has	3) Infinitely many soliutions
S) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$ , then the system has	4) $\mathrm{x}=11, \mathrm{y}=-2$ and $\mathrm{z}=0$ as a solution
	5) $\mathrm{x}=-15, \mathrm{y}=4$ and $\mathrm{z}=0$ as a solution

Option A:
$(\mathrm{P}) \rightarrow(3)$ & $(\mathrm{Q}) \rightarrow(2)$ & $(\mathrm{R}) \rightarrow(1)$ & $(\mathrm{S}) \rightarrow(4)$
Correct
Option B:
$(\mathrm{P}) \rightarrow(3)$ & $(\mathrm{Q}) \rightarrow(2)$ & $(\mathrm{R}) \rightarrow(5)$ & $(\mathrm{S}) \rightarrow(4)$
Option C:
$(\mathrm{P}) \rightarrow(2)$ & $(\mathrm{Q}) \rightarrow(1)$ & $(\mathrm{R}) \rightarrow(4)$ & $(\mathrm{S}) \rightarrow(5)$
Option D:
$(\mathrm{P}) \rightarrow(2)$ & $(\mathrm{Q}) \rightarrow(1)$ & $(\mathrm{R}) \rightarrow(1)$ & $(\mathrm{S}) \rightarrow(3)$

Answer: A

Step-by-step solution

\Delta=\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{array}\right|=(7 \alpha-3)-2 \beta

\Delta_{\mathrm{x}}=\left|\begin{array}{ccc} 7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta \end{array}\right|=21 \alpha-22 \beta+2 \alpha \gamma-33

\Delta_{y}=\left|\begin{array}{ccc} 1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta \end{array}\right|=14 \alpha+4 \beta+\gamma-\alpha \gamma-22

$\Delta_{z}=\left|\begin{array}{ccc}1 & 2 & 7 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma\end{array}\right|=-2 \gamma+56$

(P) If $\beta=\frac{1}{2}(7 \alpha-3) \& \gamma=28$ , then $\Delta=\Delta_{x}=\Delta_{y}=\Delta_{z}=0$ So infinitely many solution

(Q) If $\beta=\frac{1}{2}(7 \alpha-3) \& \gamma \neq 28$ , then $\Delta=0$ but $\Delta_{\mathrm{z}} \neq 0$ so no solution.

(R) If $\beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1 \& \gamma \neq 28$ , then $\Delta \neq 0$ so unique solution.

(S) If $\beta \neq \frac{1}{2}(7 \alpha-3), \alpha=1, \gamma=28$ , then $\Delta \neq 0$ $\Delta=(7 \alpha-3)-2 \beta=4-2 \beta$

$\Delta_{\mathrm{x}}=44-22 \beta$

$\Delta_{\mathrm{y}}=4 \beta-8$ $\Delta_{\mathrm{z}}=0$

$\mathrm{x}=11, \mathrm{y}=-2, \mathrm{z}=0$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Determinants
Topic: Consistency using matrix Inversion, rank method