datavorous/entrance-exam-dataset huggingface dataset:
If the system of equations

kx+y+2z=1

3x-y-2z=2

-2x-2y-4z=3

has infinitely many solutions, then k is equal to ______ .

Options:

The correct answer is:
21

Solution:
If the system of equations

P1: kx+y+2z=1

P2: 3x-y-2z=2

P3: -2x-2y-4z=3

Which are equivalent to equations of planes.

We observe 5P2-P1=3P3

So, 15-K=-6

⇒ K=21

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If the system of equations
$\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

Options:
A $\frac{11 \pi}{24}$
B $\frac{5 \pi}{24}$
C $\frac{7 \pi}{24}$
D $\frac{3 \pi}{4}$

The correct answer is:
$\frac{5 \pi}{24}$

Solution:
$\begin{aligned} & \left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \sin \alpha & -\cos \alpha \\ 1 & \cos \alpha & \sin \alpha\end{array}\right|=0 \\ & \Rightarrow 1-\sqrt{2} \sin \alpha(\sin \alpha+\cos \alpha)+\sqrt{2} \cos \alpha(\cos \alpha-\sin \alpha)=0 \\ & \Rightarrow 1+\sqrt{2} \cos 2 \alpha-\sqrt{2} \sin 2 \alpha=0 \\ & \cos 2 \alpha-\sin 2 \alpha=-\frac{1}{\sqrt{2}} \\ & \cos \left(2 \alpha+\frac{\pi}{4}\right)=-\frac{1}{2} \\ & 2 \alpha+\frac{\pi}{4}=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} \\ & \alpha+\frac{\pi}{8}=\mathrm{n} \pi \pm \frac{\pi}{3} \\ & \mathrm{n}=0 \\ & \mathrm{x}=\frac{\pi}{3}-\frac{\pi}{8}=\frac{5 \pi}{24}\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (04 Apr Shift 1)
If the system of equations x+2y+3z=3, 4x+3y-4z=4 and 8x+4y-λz=9+μ has infinitely many solutions, then the ordered pair λ,μ is equal to

Options:
A 725,215
B -725,-215
C 725,-215
D -725,215

The correct answer is:
725,-215

Solution:
Given,

The system of equations

x+2y+3z=3 ......1

4x+3y-4z=4 .........2

8x+4y-λz=9+μ .....3

Now using the operation, equation 1×4-2 we get,

5y+16z=8 ........4

Now using the operation, equation 2×2-3 we get,

2y+λ-8z=-1-μ ........5

Now using the operation, equation 4×2-3×5 we get,

72-5λz=21+5μ

Now given system of equation has infinitely many solutions, so making coefficient of z to be zero and equating 21+5μ with zero we get,

72-5λ=0⇒λ=725 and 21+5μ=0⇒μ=-215

Hence, the ordered pair λ,μ is equal to 725,-215.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If the system of equations

x-2 y+3z=9
2x+y+z=b
x-7y+az=24, has infinitely many solutions, then a-b is equal to ______

Options:

The correct answer is:
5

Solution:
For infinitely many solutions D=D1=D2=D3=0

D=1     −2      32        1         11        −7     a=0

⇒    1−2−3+71−6+a1+4=0

⇒    −5−35+5a=0

⇒  a=8

D1=9     −2       3b        1           124     −7      a=0

⇒      9     −2       3b        1           124     −7       8=0

⇒     −b−16+21+172−72−1−63+48=0

⇒     −5b+15=0

⇒     b=3

similarly D2=D3=0 for a=8,  b=3

∴ a−b=8−3=5

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (04 Sep Shift 1)
If the system of equations
x-2y+5z=3,
2x-y+z=1,
and 11x-7y+pz=q
has infinitely many solutions, then

Options:
A p+q=2
B p+q=10
C p-q=2
D p-q=5

The correct answer is:
p-q=2

Solution:
⇒ for infinite solution
D=0,D1=0,D2=0,D3=0
D=1-252-1111-7p=0⇒-p+7+22p-11+5-14+11=0
⇒-p+7+4p-22-15=0
⇒3p-30=0
⇒p=10
D1=3-251-11q-710=0⇒3-10+7+210-q+5-7+q=0
⇒-9+20-2q-35+5q=0
⇒-24+3q=0
q=8
D2=13521111q10=0 ⇒10-q-39+52q-11
⇒10+9q-27-55=0
⇒q=8
D3=1-232-1111-7q=0 ⇒-q+7+22q-11+3-3=0
⇒q=8

Topic: Mathematics, Determinants, JEE Main
If the system of equations $x+4 y-z=\lambda, 7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu+3 \lambda)$ is equal to :

Options:
A 3
B -3
C -2
D 2

The correct answer is:
-3

Solution:
$\begin{aligned} & \Delta=\left|\begin{array}{ccc}1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2\end{array}\right|=0 \\ & \Rightarrow(18-\mu)-4(14-5 \mu)-(7-45)=0 \Rightarrow \mu=0 \\ & \Delta=\Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \text { (For infinite solution) } \\ & \Delta_{\mathrm{x}}=\left|\begin{array}{ccc}\lambda & 4 & -1 \\ -3 & 9 & \mu \\ -1 & 1 & 2\end{array}\right|=0 \\ & \lambda(18-\mu)-4(-6+\mu)-1(-3+9)=0 \\ & 18 \lambda+24-6=0 \Rightarrow \lambda=-1\end{aligned}$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the system of equations x-ky+3z=0, 2x+ky-2z=0 and 3x-4y+2z=0 has non-trivial solutions, then the value of 10yx is equal to

Options:
A 3
B -152
C 57
D -57

The correct answer is:
-152

Solution:
For non-trivial solutions,
1-k 32 k-23-4 2=0⇒2k-8+k10+3-8-3k=0
⇒3k-32=0⇒k=323
Putting z=λ, we get,
3x-32y=-9λ3x-4y=-2λy=λ4x=-λ3

Topic: Mathematics, Determinants, JEE Main
If the system of equations

x+y+az=b

2x+5y+2z=6

x+2y+3z=3

has infinitely many solutions, then 2a+3b is equal to

Options:
A 25
B 20
C 23
D 28

The correct answer is:
23

Solution:
Given system of equations are

x+y+az=b.....i

2x+5y+2z=6.....ii

x+2y+3z=3.....iii

It is given that the system of equations have infinite solutions.

⇒∆=0

⇒11a252123=0

⇒115-4-16-2+a4-5=0

⇒a=7

Also, we have ⇒∆1=0

⇒b1a652323=0

⇒b15-4-118-6+712-15=0

⇒b=3

∴a=7, b=3

∴2a+3b=14+9=23

Hence, this is the correct option.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If the system of equations

x+y+z=2
2 x+4 y-z=6
3x+2y+λz=μ

has infinitely many solutions, then :

Options:
A λ+2μ=14
B 2λ-μ=5
C λ-2μ=-5
D 2λ+μ=14

The correct answer is:
2λ+μ=14

Solution:
For infinitely many solutions, all determinants should be zero.

D=11124-132λ=0⇒λ=92

D3=11224632μ=0⇒μ=5

Now check each option

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If the system of equations αx+y+z=5,x+2y+3z=4,x+3y+5z=β. Has infinitely many solutions, then the ordered pair α,β is equal to

Options:
A 1,-3
B -1,3
C 1,3
D -1,-3

The correct answer is:
1,3

Solution:
We know that, for infinitely many solutions,

Δ=0=Δx=Δy=Δz

So, Δ=α11123135=0

⇒α10-9-15-3+13-2=0

⇒α-2+1=0

⇒α=1

Now checking for Δx=511423β35=0

⇒510-9-120-3β+112-2β

⇒5-20+3β+12-2β

⇒-3+β=0

⇒β=3

Hence α,β=1,3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If the system of equations x+y+z=5, x+2y+3z=9, x+3y+αz=β has inifinitely many solutions, then β-α equals

Options:
A 8
B 21
C 5
D 18

The correct answer is:
8

Solution:
The system of linear equations a1x+b1y+c1z+d1=0, a2x+b2y+c2z+d2=0 and a3x+b3y+c3z+d3=0 have infinitely many solutions, if

D=a1b1c1a2b2c2a3b3c3=0, D1=d1b1c1d2b2c2d3b3c3=0, D2=a1d1c1a2d2c2a3d3c3=0

and D3=a1b1d1a2b2d2a3b3d3=0.

Hence, for the system of equations x+y+z=5, x+2y+3z=9 and x+3y+αz=β have infinitely many solution, if

D=11112313α=0

⇒2α-9-α-3+3-2

⇒2α-9-α+3+1=0

⇒α=5.

And, D1=511923β3α=0

⇒511923β35=0

⇒510-9-45-3β+27-2β=0

⇒5-45+3β+27-2β=0

⇒β=13.

Hence, β-α=13-5=8.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If the system of equations

x+y+z=6

2x+5y+αz=β

x+2y+3z=14

has infinitely many solutions, then α+β is equal to

Options:
A 8
B 36
C 44
D 48

The correct answer is:
44

Solution:
x+y+z=6 ....1

2x+5y+αz=β ....2

x+2y+3z=14 ...3

For infinite solution

11125α123=0

⇒15-2α-6-α+4-5=0

⇒8-α=0⇒α=8

Also 11625β1214=0

⇒70-2β-28-β+64-5=0

⇒36-β=0

i.e. α=8, β=36

⇒α+β=44

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (29 Jul Shift 2)
If the system of equations
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
has a unique solution, then $\lambda$ is not equal to

Options:
A
1

B
0

C
2

D
3

The correct answer is:
3

Solution:
Given system of equations is
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
It has unique solution.
$$
\begin{aligned}
& \therefore\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 2 & \lambda
\end{array}\right| \neq 0 \\
& \Rightarrow \quad 1(2 \lambda-6)-1(\lambda-3)+1(2-2) \neq 0 \\
& \Rightarrow \quad 2 \lambda-6-\lambda+3 \neq 0 \Rightarrow \lambda-3 \neq 0 \Rightarrow \lambda \neq 3
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2012 (07 May Online)
If the system of equations x+y+z=6, x+2y+λz=10 and x+2y+3z=μ has infinite solutions, then the value of λ+2μ is equal to

Options:
A 20
B 22
C 23
D 25

The correct answer is:
23

Solution:
For infinite solutions,
Δ=Δ1=Δ2=Δ3=0
Δ=11112λ123=16-2λ-13-λ+0=0⇒λ=3
Δ1=611102λμ23=0⇒66-2λ-130-4λ+120-2μ=0
∵λ=3⇒μ=10
Δ2=16111031103=0,Δ3=11612101210=0
λ+2μ=23

Topic: Mathematics, Determinants, JEE Main
If the system of homogeneous linear equations \( x+3 b y+b z=0, x+2 a y+a z=0 \), and \( x+4 c y+c z=0 \) has a non-zero solution, then \( a, b, c \)

Options:
A Are in A.P.
B Are in G.P.
C Are in H.P.
D Satisfy \( a+2 b+3 c=0 \)

The correct answer is:
Are in H.P.

Solution:
The system of linear equations has a non-zero solution, then 12aa13bb14cc=0

Applying R2→R2-R1,R3→R3-R1,

12aa03b-2ab-a04c-2ac-a=0

⇒(3b-2a)(c-a)-(4c-2a)(b-a)=0

⇒3bc-3ba-2ac+2a2=4bc-2ab-4ac+2a2

⇒4ac-2ac=4bc-2ab-3bc+3ab

⇒2ac=bc+ab

On dividing by abc, we get

2b=1a+1c

Hence, a, b, c are in H.P.

Topic: Mathematics, Determinants, JEE Main
If the system of linear equation $x+2 a y+a z=0, \quad x+3 b y+b z=0, \quad x+4 c y+c z=0$ has a non zero solution, then $a, b, c$

Options:
A Are in A.P
B Are in G.P
C Are in H.P
D Satisfy $a+2 b+3 c=0$

The correct answer is:
Are in H.P

Solution:
$\left|\begin{array}{lll}1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c\end{array}\right|=0$ $\left[C_2 \rightarrow C_2-2 C_3\right]$
$=\left|\begin{array}{ccc}1 & 0 & a \\ 1 & b & b \\ 1 & 2 c & c\end{array}\right|=0$ $\left[R_3 \rightarrow R_3-R_2, R_2 \rightarrow R_2-R_1\right]$
$\Rightarrow\left|\begin{array}{ccc}1 & 0 & a \\ 0 & b & b-a \\ 0 & 2 c-b & c-b\end{array}\right|=0$ $b(c-b)-(b-a)(2 c-b)=0$
On simplification, $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\therefore a, b, c$ are in Harmonic progression.

Topic: Mathematics, Determinants, JEE Main
If the system of linear equations
2x+2ay+az=0
2x+3by+bz=0
2x+4cy+cz=0,
where a,b,c∈R are non-zero and distinct; has a non-zero solution, then

Options:
A 1a,1b,1c are in A.P.
B a,b,c are in G.P.
C a+b+c=0
D a,b,c are in A.P.

The correct answer is:
1a,1b,1c are in A.P.

Solution:
22aa23bb24cc=0
12aa13bb14cc=0
3bc-4bc-2ac-4ac+2ab-3ab=0
-bc+2ac-ab=0
ab+bc=2ac
a,b,c in H.P.
⇒1a,1b,1c in A.P.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If the system of linear equations $2 x+2 y+3 z=a$

$3 x-y+5 z=b$

$x-3 y+2 z=c$

where, $a, b,$ care non-zero real numbers, has more than onc solution, then

Options:
A $b-c+a=0$
B $b-c-a=0$
C $a+b+c=0$
D $b+c-a=0$

The correct answer is:
$b-c-a=0$

Solution:
$\because$ System of equations has more than one solution

$\therefore \Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$ for infinite solution

$\begin{aligned}

\Delta_{1} &=\left|\begin{array}{ccc}

a & 2 & 3 \\

b & -1 & 5 \\

c & -3 & 2

\end{array}\right|=a(13)+2(5 c-2 b)+3(-3 b+c) \\

&=13 a-13 b+13 c=0

\end{aligned}$

i.c, $a-b+c=0$

or $\quad b-c-a=0$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If the system of linear equations
2x-3y=γ+5
αx+5y=β+1, where α,β,γ∈R has infinitely many solutions, then the value of 9α+3β+5γ is equal to

Options:

The correct answer is:
58

Solution:
Given,

2x-3y=γ+5

αx+5y=β+1

Has infinite many solutions

So, α2=5-3=β+1γ+5

⇒α=-103, 5γ+25=-3β-3

⇒9α=-30, 3β+5γ=-28

Now, 9α+3β+5γ=-58

9α+3β+5γ=58

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If the system of linear equations

2x+3y-z=-2

x+y+z=4

x-y+λz=4λ-4 where λ∈ℝ,

has no solution, then

Options:
A λ=7
B λ=-7
C λ=8
D λ2=1

The correct answer is:
λ=-7

Solution:
If we have three equations in three variables,

a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3

then ∆=a1b1c1a2b2c2a3b3c3,∆1=d1b1c1d2b2c2d3b3c3,∆2=a1d1c1a2d2c2a3d3c3,∆3=a1b1d1a2b2d2a3b3d3.

If the set of equations do not have a solution then ∆=0 and at least one among ∆1,∆2,∆3 is not equal to zero.

∆=23-11111-1λ=0

⇒λ=7⇒λ=±7 …1

Case 1: λ=7

The system of equations are

2x+3y-z=-2x+y+z=4x-y+7z=24

a2x+3y-z+2+bx+y+z-4=x-y+7z-24

Comparing the coefficients of x and y we get,

2a+b=13a+b=-1⇒a=-2, b=5

For these values of a and b, coefficients of z and the constant values are matching. So the set of equations will have infinitely many solutions.

Case 2: λ=-7

The system of equations are

2x+3y-z=-2x+y+z=4x-y-7z=-32

a2x+3y-z+2+bx+y+z-4=x-y-7z+32

Comparing the coefficients of x and y we get,

2a+b=13a+b=-1⇒a=-2, b=5

For these values of a and b, coefficients of z and the constant values are not matching. So the set of equations will have no solutions.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (28 Jun Shift 1)
If the system of linear equations

2x+y-z=3

x-y-z=α

3x+3y+βz=3

has infinitely many solutions, then α+β-αβ is equal to __________.

Options:

The correct answer is:
7

Solution:
Given,

2x+y-z=3

x-y-z=α

3x+3y+βz=3

Let A= 21-11-1-133β & B=3α3

For Equation to have infinitely many solutions A=0 & Adj. A.B=0

A=21-11-1-133β=0

⇒-3β-3=0

⇒β=-1

Now,

Minor of A=2β+3β+36β+32β+33-2-1-3

Cofactor of A=-2β+3-β+3-6-β+3-2β+3-3213

Adjoint of A=-2β+3-β+32-β+3-2β+31-6-33

Now, adj. A.B=0

⇒-2β+3-β+32-β+3-2β+31-6-333α3=0

-6β-9-αβ-3α+6-3β-9-2αβ-3α+3-18-3α+9=0

⇒-18-3α+9=0

⇒α=-3

Therefore,α+β-αβ=-3-1-3=-7=7

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If the system of linear equations
2x+y-z=7
x-3y+2z=1
x+4y+δz=k, where δ,k∈R
has infinitely many solutions, then δ+k is equal to

Options:
A -3
B 3
C 6
D 9

The correct answer is:
3

Solution:
For infinitely many solutions, ∆=21-11-3214δ=0

⇒δ=-3

and ∆1=71-11-32K4-3=0⇒k=6

⇒δ+K=3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (29 Jun Shift 1)
If the system of linear equations
7x+11y+αz=13
5x+4y+7z=β
175x+194y+57z=361
has infinitely many solutions, then α+β+2 is equal to

Options:
A 4
B 3
C 5
D 6

The correct answer is:
4

Solution:
Given,

System of linear equations,

7x+11y+αz=13
5x+4y+7z=β
175x+194y+57z=361

Now we know that, for infinite solution D=D1=D2=D3=0

Now solving D=0 we get,

711α54717519457=0

Now applying operation R3→R3-25R1 we get,

711α5470-8157-25α=0

⇒81(49-5α)+(57-25α)(-27)=0

⇒270α=-2430⇒α=-9

And now solving D1=0 we get,

1311-9β4736119457=0

⇒134×57-7×194-β11×57+9×194+36111×7+9×4=0

⇒13-1130-β2373+361113=0

⇒13-10-β21+361=0

⇒21β=231

⇒β=11

∴α+β+2=4

Topic: Mathematics, Determinants, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If the system of linear equations.

8x+y+4z=-2

x+y+z=0

λx-3y=μ

has infinitely many solutions, then the distance of the point λ,μ,-12 from the plane 8x+y+4z+2=0 is:

Options:
A 35
B 4
C 269
D 103

The correct answer is:
103

Solution:
Given the system of linear equations.

8x+y+4z=-2

x+y+z=0

λx-3y=μ

has infinitely many solutions, so

D=814111λ-30=0⇒λ=4

Also D1=D2=D3=0

D1=-214011μ-30=0⇒-6-3μ=0

So μ=-2

Hence the point will be 4,-2,-12

Now distance of point 4,-2,-12 from plane 8x+y+4z+2=0

=32-2-2+282+12+42=309=103

Topic: Mathematics, Determinants, JEE Main, JEE Main 2022 (26 Jul Shift 1)
If the system of linear equations :
$$
\begin{aligned}
& x_1+2 x_2+3 x_3=6 \\
& x_1+3 x_2+5 x_3=9 \\
& 2 x_1+5 x_2+a x_3=b
\end{aligned}
$$
is consistent and has infinite number of solutions, then :

Options:
A
$a=8, b$ can be any real number

B
$b=15, a$ can be any real number

C
$a \in R-\{8\}$ and $b \in R-\{15\}$

D
$a=8, b=15$

The correct answer is:
$a=8, b=15$

Solution:
Given system of equations can be written in matrix form as $\mathrm{AX}=\mathrm{B}$ where $\mathrm{A}=\left(\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{l}6 \\ 9 \\ b\end{array}\right)$
Since, system is consistent and has infinitely many solutions
$\therefore(\operatorname{adj} . \mathrm{A}) \mathrm{B}=0$
$$
\begin{aligned}
& \Rightarrow\left(\begin{array}{ccc}
3 a-25 & 15-2 a & 1 \\
10-a & a-6 & -2 \\
-1 & -1 & 1
\end{array}\right)\left(\begin{array}{l}
6 \\
9 \\
b
\end{array}\right)=\left(\begin{array}{l}
0 \\
0 \\
0
\end{array}\right) \\
& \Rightarrow-6-9+b=0 \Rightarrow b=15 \\
& \text { and } 6(10-a)+9(a-6)-2(b)=0 \\
& \Rightarrow 60-6 a+9 a-54-30=0 \\
& \Rightarrow 3 a=24 \Rightarrow a=8
\end{aligned}
$$
Hence, $a=8, b=15$.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2013 (22 Apr Online)
If the system of linear equations
$x+2 a y+a z=0$ $x+3 b y+b z=0$
$x+4 c y+c z=0$
has a non-zero solution, then $\mathrm{a}, \mathrm{b}, \mathrm{c}$

Options:
A
satisfy $a+2 b+3 c=0$

B
are in A.P.

C
are in G.P.

D
are in H.P.

The correct answer is:
are in H.P.

Solution:
$\left|\begin{array}{lll}1 & 2 \mathrm{a} & \mathrm{a} \\ 1 & 3 \mathrm{~b} & \mathrm{~b} \\ 1 & 4 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{C}_2 \rightarrow \mathrm{C}_2-2 \mathrm{C}_3$
$\left|\begin{array}{ccc}1 & 0 & \mathrm{a} \\ 1 & \mathrm{~b} & \mathrm{~b} \\ 1 & 2 \mathrm{c} & \mathrm{c}\end{array}\right|=0 \quad \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$
$\left|\begin{array}{ccc}1 & 0 & a \\ 0 & b & b-a \\ 0 & 2 c-b & c-b\end{array}\right|=0$
$b(c-b)-(b-a)(2 c-b)=0$
On simplification,
$$
\frac{2}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}}
$$
$\therefore \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in Harmonic Progression.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2003
If the system of linear equations
x-2y+kz=1
2x+y+z=2
3x-y-kz=3
has a solution x,y,z,z≠0, then x,y lies on the straight line whose equation is:

Options:
A 4x-3y-4=0
B 3x-4y-4=0
C 3x-4y-1=0
D 4x-3y-1=0

The correct answer is:
4x-3y-4=0

Solution:
x-2y+kz=1 .........(1)
2x+y+z=2 .........(2)
3x-y-kz=3 ........(3)

Hence,

1-2k2113-1-kxyz=123

For infinite solution,

1-2k2113-1-k=0

⇒-k+1+2-2k-3+k-2-3=0

⇒k=-12

Put this value in the equation 1, we get

2x-4y-z=2....4

Adding the equation 4 & 2, we get

4x-3y=4

Hence, the equation of straight line is 4x-3y-4=0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If the system of linear equations

x-2y+z=-4

2x+αy+3z=5

3x-y+βz=3

has infinitely many solutions, then 12α+13β is equal to

Options:
A 60
B 64
C 54
D 58

The correct answer is:
58

Solution:
Given:

The system of linear equations

x-2y+z=-4,

2x+αy+3z=5 and

3x-y+βz=3 has infinity many solutions,

So, △=△1=△2=△3

Now, finding,

△2=1-4125333β=0

⇒5β-9+8β-36-9=0

⇒13β=54

And, △3=1-2-42α53-13=0

⇒3α+5+12-30+8+12α=0

⇒15α=5

⇒12α=4

Hence, the value of 12α+13β=4+54=58

Topic: Mathematics, Determinants, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If the system of linear equations x-4y+7z=g; 3y-5z=h; -2x+5y-9z=k is consistent, then:

Options:
A g+h+2k=0
B g+2h+k=0
C 2g+h+k=0
D g+h+k=0

The correct answer is:
2g+h+k=0

Solution:
x-4y+7z=g ...(i)
3y-5z=h ...(ii)
-2x+5y-9z=k ...(iii)
(i) + (ii), we get
x-y+2z=g+h ...(iv)
(ii) - (iii), we get
2x-2y+4z=h-k
⇒ x-y+2z=h-k2 ...(v)
Since, coefficient of x,y,z in (iv) and (v) are same. Hence, for system to be consistent
g+h=h-k2
⇒ 2g+2h=h-k
⇒ 2g+h+k=0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If the system of linear equations
$$
\begin{aligned}
&x+a y+z=3 \\
&x+2 y+2 z=6 \\
&x+5 y+3 z=b
\end{aligned}
$$
has no solution, then

Options:
A
$a=1, b \neq 9$

B
$a \neq-1, b=9$

C
$a=-1, b=9$

D
$a=-1, b \neq 9$

The correct answer is:
$a=-1, b \neq 9$

Solution:
As the system of equations has no solution then $\Delta$ should be zero and at least one of $\Delta_1$, $\Delta_2$ and $\Delta_3$ should not be zero.
$$
\begin{aligned}
&\therefore \Delta=\left|\begin{array}{lll}
1 & a & 1 \\
1 & 2 & 2 \\
1 & 5 & 3
\end{array}\right|=0 \\
&\Rightarrow-a-1=0 \Rightarrow a=-1 \\
&\Delta_2=\left|\begin{array}{lll}
1 & 3 & 1 \\
1 & 6 & 2 \\
1 & b & 3
\end{array}\right| \neq 0 \\
&\Rightarrow b \neq 9
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If the system of linear equations

x+ky+3z=0
3x+ky-2z=0
2x+4y-3z=0

has a non-zero solution x, y, z, then xzy2 is equal to:

Options:
A 30
B -10
C 10
D -30

The correct answer is:
10

Solution:
1k33k-224-3=0
⇒-3k+8-3-3k-12+2-5k=0
⇒-4k+44=0 ⇒ k=11
⇒x+11y+3z=0 ...1
⇒3x+11y-2z=0 ...2
⇒2x+4y-3z=0 ...3
On solving equations1 & 2, we get
2x-5z=0
⇒x=5z2
Put it in equation 3, we get
2z+4y=0
⇒z=-2y
∴xzy2=-2y×-5yy2=10

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (08 Apr)
If the system of linear equations
x+y+3z=0
x+3y+k2z=0
3x+y+3z=0
has a non-zero solution (x, y, z) for some k∈R, then x+yz is equal to :

Options:
A -3
B 9
C 3
D -9

The correct answer is:
-3

Solution:
So D=0→11313k2313=0 ⇒ k2=9

x+y+3 z=0

x+3 y+9 z=0

3 x+y+3 z=0

(1)-(3)
x=0⇒ y+3z=0

⇒yz=-3
so, x+yz=-3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the system of linear equations x+y+z=5, x+2y+2z=6, x+3y+λz=µ, (λ, µ ∈R) , has infinitely many solutions, then the value of λ+µ is:

Options:
A 7
B 10
C 12
D 9

The correct answer is:
10

Solution:
∆=11112213λ=λ-3
∆1=511622μ3λ=4λ-3
∆3=1511621μλ=λ-μ+4
∆4=11512613μ=μ-7
For infinite many solution ∆=∆1=∆2=∆3=0
⇒λ=3, & μ=7 from ∆=0 & ∆4=0
& λ=3, μ=7 Satisfies ∆3=0=∆1
∴ λ+μ=10

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If the system of linear equations,
x+y+z=6
x+2y+3z=10
3x+2y+λz=μ
has more than two solutions, then μ-λ2, is equal to.

Options:

The correct answer is:
13

Solution:
x+y+z=6 ...1
x+2y+3z=10 ...2
3x+2y+λz=μ …3

From 1 & 2,
If z=0⇒x+y=6 and x+2y=10, ⇒y=4, x=2
If y=0⇒x+z=6 and x+3z=10, ⇒z=2, x=4

So, 3x+2y+λz=μ, must pass through (2,4,0) and (4,0,2).

6+8=μ⇒μ=14
And 12+2λ=μ ⇒λ=1

So, μ-λ2=14-1=13

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If the tangent at 1,7 to the curve x2=y-6 touch the circle x2+y2+16x+12y+c=0 then the value of c is:

Options:
A 95
B 195
C 185
D 85

The correct answer is:
95

Solution:
Differentiating x2=y-6 with respect to x,
2x=dydx
dydx1,7=2×1=2
Hence, equation of tangent is y-7=2(x-1)
⇒y=2x+5
Distance from centre =
-6-2-8-512+22
⇒55=5
Radius of circle = 82+62-c
⇒100-c= 5
⇒c=95

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (08 Apr)
If the tangent at a point on the ellipse x227+y23=1 meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is

Options:
A 33
B 92
C 9
D 93

The correct answer is:
9

Solution:
x227+y23=1

Equation of tangent to ellipse

x27cosθ+y3sinθ=1

Area bounded by the line and coordinate axis

Δ=12·27cosθ·3sinθ=9sin2θ

Δ will be minimum when sin2θ=1

Δmin =9 sq. units

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2016 (09 Apr Online)
If the tangent at a point P on the parabola y2=3x is parallel to the line x+2y=1 and the tangents at the points Q and R on the ellipse x24+y21=1 are perpendicular to the line x-y=2, then the area of the triangle PQR is:

Options:
A 95
B 53
C 325
D 35

The correct answer is:
35

Solution:
Given equation of parabola is

y2=3x

⇒2ydydx=3

⇒dydxslope of tangent at P=32y

Tangent at Px1,y1 is parallel to the line x+2y=1, so slope of tangent is

m=-12

⇒32y1=-12

⇒y1=-3

So, coordinates of P≡3,-3.

Given equation of ellipse is

x24+y21=1

⇒x2+4y2=4

⇒2x+8ydydx=0

⇒dydxslope of tangent=-x4y

Also, tangent to ellipse re perpendicular to line x-y=2, so slope of tangent

m=-1

⇒-x4y=-1

⇒x=4y

Putting in equation of ellipse, we get

16y24+y21=1

⇒y=±15

So,

Q43,15,R-45,-15

Area of ∆PQR is

=123-3145151-45-151

=12325+385+0

=3025=35 sq. units

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2023 (29 Jan Shift 2)
If the tangent at a point P, with parameter t, on the curve x=4t2+3, y=8t3-1, t ∈R, meets the curve again at a point Q, then the coordinates of Q are :

Options:
A 16t2+3,Â -64t3-1
B 4t2+3,Â -8t3-1
C t2+3,Â t3-1
D t2+3, -t3-1

The correct answer is:
t2+3, -t3-1

Solution:
Given P4t2+3, 8t3-1
dydtdxdt=dydx=3t (slope of tangent at P)
Let Q=4λ2+3, 8λ3-1
Slope of PQ=3t
8t3-8λ34t2-4λ2=3t
⇒ t2+tλ-2λ2=0
t-λ t+2λ=0
t=λ or λ=-t2
∴ Qt2+3, -t3-1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (09 Apr Online)