Determinant - JEE Main Previous Year Question with Solutions

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[esquestion] Let $a, b, c$ be such that $\mathrm{b}(\mathrm{a}+\mathrm{c}) \neq 0 . \mathrm{If}$ If $\left|\begin{array}{ccc}{a} & {a+1} & {a-1} \\ {-b} & {b+1} & {b-1} \\ {c} & {c-1} & {c+1}\end{array}\right|+\left|\begin{array}{ccc}{a+1} & {b+1} & {c-1} \\ {a-1} & {b-1} & {c+1} \\ {-1} & {a} & {(-1)^{n+1} b} & {(-1)^{n} c}\end{array}\right|=0$ then the value of n is :- (1) Any odd integer (2) Any integer (3) Zero (4) Any even integer #tag# [AIEEE - 2009] #sol# (1) [/esquestion] [esquestion] Consider the system of linear equations : $\mathrm{x}_{1}+2 \mathrm{x}_{2}+\mathrm{x}_{3}=3$ $2 \mathrm{x}_{1}+3 \mathrm{x}_{2}+\mathrm{x}_{3}=3$ $3 x_{1}+5 x_{2}+2 x_{3}=1$ The system has (1) Infinite number of solutions (2) Exactly 3 solutions (3) A unique solution (4) No solution #tag# [AIEEE - 2010] #sol# (4) Here $D=0$ $\& \quad D_{1} \neq 0$ so we can say no solution [/esquestion] [esquestion] The number of values of k for which the linear equations 4x + ky + 2z = 0 kx + 4y + z = 0 2x + 2y + z = 0 possess a non-zero solution is :- (1) 1 (2) zero (3) 3 (4) 2 #tag# [AIEEE - 2011] #sol# (4) [/esquestion] [esquestion] If the trivial solution is the only solution of the system of equations x – ky + z = 0 kx + 3y – kz = 0 3x + y – z = 0 Then the set of all values of k is: (1) {2, –3}      (2) R – {2, –3}     (3) R – {2} (4) R – {–3} #tag# [AIEEE - 2011] #sol# (2) Here for trival solution $D \neq 0$ So $\mathrm{D}=\left|\begin{array}{ccc}{1} & {-\mathrm{k}} & {1} \\ {\mathrm{k}} & {3} & {-\mathrm{k}} \\ {3} & {1} & {-1}\end{array}\right|=0$ $\Rightarrow \mathrm{D}=2 \mathrm{k}^{2}-12+2 \mathrm{k}=0 \Rightarrow \mathrm{k}=-3,2$ so $\mathrm{R}-\{-3,2\}$ [/esquestion] [esquestion] The number of values of k, for which the system of equations : (k + 1)x + 8y = 4k kx + (k + 3)y = 3k – 1 has no solution, is – (1) infinite (2) 1 (3) 2 (4) 3 #tag# [JEE(Main)-2013] #sol# (2) $\frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1}$ (1) = (2) $\Rightarrow \quad k^{2}-4 k+3=0$ k = 1, 3 for k = 1 (2) = (3) for $\mathrm{k}=3 \quad(2) \neq(3)$ k = 3 [/esquestion] [esquestion] If $\alpha, \beta \neq 0,$ and $f(\mathrm{n})=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$ and $\left|\begin{array}{ccc}{3} & {1+f(1)} & {1+f(2)} \\ {1+f(1)} & {1+f(2)} & {1+f(3)} \\ {1+f(2)} & {1+f(3)} & {1+f(4)}\end{array}\right|=\mathrm{K}(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$ then K is equal to : (1) $\alpha \beta$ (2) $\frac{1}{\alpha \beta}$ (3) 1 (4) –1 #tag# [JEE(Main)-2014] #sol# (3) $\therefore k=1$ [/esquestion] [esquestion] The set of all values of $\lambda$ for which the system of linear equations : $2 \mathrm{x}_{1}-2 \mathrm{x}_{2}+\mathrm{x}_{3}=\lambda \mathrm{x}_{1}$ $2 \mathrm{x}_{1}-3 \mathrm{x}_{2}+2 \mathrm{x}_{3}=\lambda \mathrm{x}_{2}$ $-\mathrm{x}_{1}+2 \mathrm{x}_{2}=\lambda \mathrm{x}_{3}$ has a non-trivial solution (1) contains two elements (2) contains more than two elements (3) is an empty set (4) is a singleton #tag# [JEE(Main)-2015] #sol# (1) [/esquestion] [esquestion] The system of linear equations $\mathrm{x}+\lambda \mathrm{y}-\mathrm{z}=0$ $\lambda \mathrm{x}-\mathrm{y}-\mathrm{z}=0$ $\mathrm{x}+\mathrm{y}-\lambda \mathrm{z}=0$ has a non-trivial solution for : (1) exactly three values of $\lambda$ (2) infinitely many values of $\lambda$ (3) exactly one value of $\lambda$ (4) exactly two values of $\lambda$ #tag# [JEE(Main)-2016] #sol# (1) $\left|\begin{array}{ccc}{1} & {\lambda} & {-1} \\ {\lambda} & {-1} & {-1} \\ {1} & {1} & {-\lambda}\end{array}\right|=0 \quad \Rightarrow \quad \lambda=0,1,-1$ [/esquestion] [esquestion] If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is : (1) a singleton (2) an empty set (3) an infinite set (4) a finite set containing two or more elements #tag# [JEE(Main)-2017] #sol# (1) $D=\left|\begin{array}{lll}{1} & {1} & {1} \\ {1} & {a} & {1} \\ {a} & {b} & {1}\end{array}\right|=0 \Rightarrow a=1$ and at a = 1 $\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$ But at a = 1 and b = 1 $\left.\begin{array}{ll}{\text { First two equations are }} & {x+y+z=1} \\ {\text { and third equation is }} & {x+y+z=0}\end{array}\right] \Rightarrow$ There is nosolution. $\mathrm{b}=\{1\} \Rightarrow$ it is a singleton set [/esquestion] [esquestion] If $\left|\begin{array}{ccc}{x-4} & {2 x} & {2 x} \\ {2 x} & {x-4} & {2 x} \\ {2 x} & {2 x} & {x-4}\end{array}\right|=(A+B x)(x-A)^{2},$ then the ordered pair $(A, B)$ is equal to : (1) (–4, 3)            (2) (–4, 5)             (3) (4, 5)               (4) (–4, –5) #tag# [JEE(Main)-2018] #sol# (2) $\left|\begin{array}{ccc}{\mathrm{x}-4} & {2 \mathrm{x}} & {2 \mathrm{x}} \\ {2 \mathrm{x}} & {\mathrm{x}-4} & {2 \mathrm{x}} \\ {2 \mathrm{x}} & {2 \mathrm{x}} & {\mathrm{x}-4}\end{array}\right|=(\mathrm{A}+\mathrm{Bx})(\mathrm{x}-\mathrm{A})^{2}$ Put $x=0 \Rightarrow\left|\begin{array}{ccc}{-4} & {0} & {0} \\ {0} & {-4} & {0} \\ {0} & {0} & {-4}\end{array}\right|=A^{3} \Rightarrow A=-4$ $\left|\begin{array}{ccc}{\mathrm{x}-4} & {2 \mathrm{x}} & {2 \mathrm{x}} \\ {2 \mathrm{x}} & {\mathrm{x}-4} & {2 \mathrm{x}} \\ {2 \mathrm{x}} & {2 \mathrm{x}} & {\mathrm{x}-4}\end{array}\right|=(\mathrm{Bx}-4)(\mathrm{x}+4)^{2}$ $\left|\begin{array}{ccc}{1-\frac{4}{\mathrm{x}}} & {2} & {2} \\ {2} & {1-\frac{4}{\mathrm{x}}} & {2} \\ {2} & {2} & {1-\frac{4}{\mathrm{x}}}\end{array}\right|=\left(\mathrm{B}-\frac{4}{\mathrm{x}}\right)\left(1+\frac{4}{\mathrm{x}}\right)^{2}$ Put $\mathrm{x} \rightarrow \infty \quad \Rightarrow \quad\left|\begin{array}{lll}{1} & {2} & {2} \\ {2} & {1} & {2} \\ {2} & {2} & {1}\end{array}\right|=\mathrm{B} \Rightarrow \mathrm{B}=5$ ordered pair (A, B) is (–4, 5) [/esquestion] [esquestion] If the system of linear equations $x+k y+3 z=0$ $3 x+k y-2 z=0$ $2 x+4 y-3 z=0$ has a non-zero solution $(\mathrm{x}, \mathrm{y}, \mathrm{z}),$ then $\frac{\mathrm{xz}}{\mathrm{y}^{2}}$ is equal to : (1) 10 (2) – 30 (3) 30 (4) –10 #tag# [JEE(Main)-2018] #sol# (1) [/esquestion] [esquestion] If the system of linear equations : $x+a y+z=3$ $\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}=6$ $x+5 y+3 z=b$ has no solution, then :- (1) $a=-1, b=9$ (2) $a \neq-1, b=9$ (3) $a=1, b \neq 9$ (4) $a=-1, b \neq 9$ #tag# [JEE(Main)-2018] #sol# (4) [/esquestion] [esquestion] The number of values of k for which the system of linear equations, (k+2)x + 10y = k kx + (k+3) y = k – 1 has no solution is : (1) infinitely many (2) 1 (3) 2 (4) 3 #tag# [JEE(Main)-2018] #sol# (2) [/esquestion]

This is one chapter among many, so students can use the JEE Mains PYQ chapter wise index to continue their Maths PYQ practice.

Determinants is only one part of the exam — see exactly how it fits into a complete paper on our JEE Main previous year question paper page.