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]

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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]

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Sol. (4)Here $D=0$ $\& \quad D_{1} \neq 0$ so we can sayno solution

Q. The number of values of k for which the linear equations4x + ky + 2z = 0kx + 4y + z = 02x + 2y + z = 0possess a non-zero solution is :- (1) 1 (2) zero (3) 3 (4) 2 [AIEEE – 2011]

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Sol. (4)

Q. If the trivial solution is the only solution of the system of equationsx – ky + z = 0kx + 3y – kz = 03x + y – z = 0Then the set of all values of k is: (1) {2, –3}      (2) R – {2, –3}     (3) R – {2} (4) R – {–3} [AIEEE – 2011]

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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$$\Rightarrow \mathrm{D}=2 \mathrm{k}^{2}-12+2 \mathrm{k}=0 \Rightarrow \mathrm{k}=-3,2 so \mathrm{R}-\{-3,2\} Q. The number of values of k, for which the system of equations :(k + 1)x + 8y = 4kkx + (k + 3)y = 3k – 1has no solution, is –(1) infinite (2) 1 (3) 2 (4) 3 [JEE(Main)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)\frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1}(1) = (2)\Rightarrow \quad k^{2}-4 k+3=0k = 1, 3for k = 1 (2) = (3)for \mathrm{k}=3 \quad(2) \neq(3)k = 3 Q. If \alpha, \beta \neq 0, and f(\mathrm{n})=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}} and \left|\begin{array}{ccc}{3} & {1+f(1)} & {1+f(2)} \\ {1+f(1)} & {1+f(2)} & {1+f(3)} \\ {1+f(2)} & {1+f(3)} & {1+f(4)}\end{array}\right|=\mathrm{K}(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2} then K is equal to :(1) \alpha \beta(2) \frac{1}{\alpha \beta}(3) 1(4) –1 [JEE(Main)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\therefore k=1 Q. The set of all values of \lambda for which the system of linear equations :2 \mathrm{x}_{1}-2 \mathrm{x}_{2}+\mathrm{x}_{3}=\lambda \mathrm{x}_{1}$$2 \mathrm{x}_{1}-3 \mathrm{x}_{2}+2 \mathrm{x}_{3}=\lambda \mathrm{x}_{2}$$-\mathrm{x}_{1}+2 \mathrm{x}_{2}=\lambda \mathrm{x}_{3}has a non-trivial solution(1) contains two elements(2) contains more than two elements(3) is an empty set(4) is a singleton [JEE(Main)-2015] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. The system of linear equations\mathrm{x}+\lambda \mathrm{y}-\mathrm{z}=0$$\lambda \mathrm{x}-\mathrm{y}-\mathrm{z}=0$$\mathrm{x}+\mathrm{y}-\lambda \mathrm{z}=0has 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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 equationsx + y + z = 1x + ay + z = 1ax + by + z = 0has 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)D=\left|\begin{array}{lll}{1} & {1} & {1} \\ {1} & {a} & {1} \\ {a} & {b} & {1}\end{array}\right|=0 \Rightarrow a=1and at a = 1\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0But 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] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... 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}$$\left|\begin{array}{ccc}{1-\frac{4}{\mathrm{x}}} & {2} & {2} \\ {2} & {1-\frac{4}{\mathrm{x}}} & {2} \\ {2} & {2} & {1-\frac{4}{\mathrm{x}}}\end{array}\right|=\left(\mathrm{B}-\frac{4}{\mathrm{x}}\right)\left(1+\frac{4}{\mathrm{x}}\right)^{2}Put \mathrm{x} \rightarrow \infty \quad \Rightarrow \quad\left|\begin{array}{lll}{1} & {2} & {2} \\ {2} & {1} & {2} \\ {2} & {2} & {1}\end{array}\right|=\mathrm{B} \Rightarrow \mathrm{B}=5ordered pair (A, B) is (–4, 5) Q. If the system of linear equations x+k y+3 z=0$$3 x+k y-2 z=0$$2 x+4 y-3 z=0has a non-zero solution (\mathrm{x}, \mathrm{y}, \mathrm{z}), then \frac{\mathrm{xz}}{\mathrm{y}^{2}} is equal to :(1) 10 (2) – 30 (3) 30 (4) –10 [JEE(Main)-2018] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1) Q. If the system of linear equations :x+a y+z=3$$\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]

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Sol. (4)

Q. The number of values of k for which the system of linear equations,(k+2)x + 10y = kkx + (k+3) y = k – 1has no solution is :(1) infinitely many (2) 1 (3) 2 (4) 3 [JEE(Main)-2018]

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Sol. (2)

• May 20, 2021 at 8:32 am

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• February 25, 2021 at 10:07 am

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• February 18, 2021 at 9:13 pm