Determinant – JEE Main Previous Year Question with Solutions

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Q. 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

[AIEEE – 2009]

Sol. (1)

Q. 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

[AIEEE – 2010]

Sol. (4)

Here $D=0$ $\& \quad D_{1} \neq 0$ so we can say

no solution

Q. 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

[AIEEE – 2011]

Sol. (4)

Q. 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}

[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 $\mathrm{R}-\{-3,2\}$

Q. 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

[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

Q. 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

[JEE(Main)-2014]

Sol. (3)

$\therefore k=1$

Q. 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

[JEE(Main)-2015]

Sol. (1)

Q. 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 $\lambda$

[JEE(Main)-2016]

Sol. (1)

$\left|\begin{array}{ccc}{1} & {\lambda} & {-1} \\ {\lambda} & {-1} & {-1} \\ {1} & {1} & {-\lambda}\end{array}\right|=0 \quad \Rightarrow \quad \lambda=0,1,-1$

Q. 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

[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

Q. 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)

[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)

Q. 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

[JEE(Main)-2018]

Sol. (1)

Q. 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) $a=-1, b \neq 9$

[JEE(Main)-2018]

Sol. (4)

Q. 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

[JEE(Main)-2018]

Sol. (2)