Matrices and Determinants jee mains questions | JEE Main Maths Matrices and Determinants Previous Year Question

Q 11 :
Let $α β γ = 45; α, β, γ \in R .$ If $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$ for some $x, y, z \in R, x y z \neq 0,$ then $6 α + 4 β + γ$ is equal to _______. [2024]

0%

(55)

We have, $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$

$\Rightarrow α x + y + 2 z = 0; x + β y + 3 z = 0; 2 x + 2 y + γ z = 0$

Since, $x y z \neq 0$ so the system of equations has non-trivial solution.

Now, $| \begin{matrix} α & 1 & 2 \\ 1 & β & 3 \\ 2 & 2 & γ \end{matrix} |$ $= 0$

$\Rightarrow α (β γ - 6) - 1 (γ - 6) + 2 (2 - 2 β) = 0$

$\Rightarrow α β γ - 6 α - γ + 6 + 4 - 4 β = 0$

$\Rightarrow 45 + 10 - 6 α - γ - 4 β = 0 \Rightarrow 6 α + 4 β + γ = 55$

Q 12 :
If the system of equations, $2 x + 7 y + λ z = 3,$ $3 x + 2 y + 5 z = 4,$ $x + μ y + 32 z = - 1$ has infinitely many solutions, then $(λ - μ)$ is equal to _______. [2024]

0%

(38)

Given, $2 x + 7 y + λ z = 3$

$3 x + 2 y + 5 z = 4$

$x + μ y + 32 z = - 1$ has infinitely many solutions.

$\Rightarrow Δ = 0, Δ_{1} = 0, Δ_{2} = 0$ and $Δ_{3} = 0$

$Δ_{2} =$ $| \begin{matrix} 2 & λ & 3 \\ 3 & 5 & 4 \\ 1 & 32 & - 1 \end{matrix} |$ $= 0$ and $Δ_{3} =$ $| \begin{matrix} 2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & μ & - 1 \end{matrix} |$ $= 0$

$\Rightarrow - 266 + λ (7) + 273 = 0$ and $- 4 - 8 μ + 49 + 9 μ - 6 = 0$

$\Rightarrow 7 λ + 7 = 0$ and $μ + 39 = 0$

$\Rightarrow λ = - 1$ and $μ = - 39$

So, $λ - μ = - 1 + 39 = 38$

Q 13 :
If the system of Linear equations

$3 x + y + β z = 3$

$2 x + α y - z = - 3$

$x + 2 y + z = 4$

has infinitely many solutions, then the value of $22 β - 9 α$ is : [2025]

31
49
37
43

1

2

3

4

(1)

Since, the given system of equations has infinitely many solutions, $△ = △_{1} = △_{2} = △_{3} = 0$

$△ =$ $| \begin{matrix} 3 & 1 & β \\ 2 & α & - 1 \\ 1 & 2 & 1 \end{matrix} |$ $= 0$

$\Rightarrow 3 α + 4 β - α β + 3 = 0$ ...(i)

$△_{3} =$ $| \begin{matrix} 3 & 1 & 3 \\ 2 & α & - 3 \\ 1 & 2 & 4 \end{matrix} |$ $= 0$

$\Rightarrow 9 α + 19 = 0 \Rightarrow α = \frac{- 19}{9}$

From (i), we get

$β = \frac{6}{11}$

$∴ 22 β - 9 α = 22 \times \frac{6}{11} - 9 \times \frac{- 19}{9} = 12 + 19 = 31$ .

Q 14 :
If the system of equations

$2 x + λ y + 3 z = 5$

$3 x + 2 y - z = 7$

$4 x + 5 y + μ z = 9$

has infinitely many solutions, then $(λ^{2} + μ^{2})$ is equal to: [2025]

22
18
26
30

1

2

3

4

(3)

For infinitely man solutions, we have $△ = △_{1} = △_{2} = △_{3} = 0$

Now, $△ =$ $| \begin{matrix} 2 & λ & 3 \\ 3 & 2 & - 1 \\ 4 & 5 & μ \end{matrix} |$ $= 0$

$\Rightarrow 2 (2 μ + 5) - λ (3 μ + 4) + 3 (15 - 8) = 0$

$\Rightarrow 4 μ + 10 - 3 λ μ - 4 λ + 21 = 0$

$\Rightarrow 4 μ - 4 λ - 3 λ μ + 31 = 0$ ... (i)

Now, $△_{2} =$ $| \begin{matrix} 2 & 5 & 3 \\ 3 & 7 & - 1 \\ 4 & 9 & μ \end{matrix} |$ $= 0$

$\Rightarrow 2 (7 μ + 9) - 5 (3 μ + 4) + 3 (27 - 28)$

$\Rightarrow 14 μ + 18 - 15 μ - 20 - 3 = 0$

$\Rightarrow - μ + 18 - 23 = 0 \Rightarrow μ = - 5$

Substituting the value of $μ$ in (i), we get

$- 20 - 4 λ + 15 λ + 31 = 0$

$\Rightarrow 11 λ + 11 = 0 \Rightarrow λ = - 1$

$∴ λ^{2} + μ^{2} = {(- 1)}^{2} + {(- 5)}^{2} = 26$ .

Q 15 :
Let the system of equations :

$2 x + 3 y + 5 z = 9$

$7 x + 3 y - 2 z = 8$

$12 x + 3 y - (4 + λ) z = 16 - μ$

have infinitely many solutions. Then the radius of the circle centred at $(λ, μ)$ and touching the line 4x = 3y is [2025]

$\frac{17}{5}$
$\frac{7}{5}$
$\frac{21}{5}$
7

1

2

3

4

(2)

Since, the given system of equation has infinitely many solutions.

$△ = △_{1} = △_{2} = △_{3} = 0$

$△ =$ $| \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & - 2 \\ 12 & 3 & - (λ + 4) \end{matrix} |$ $= 0$

$\Rightarrow 12 (- 21) - 3 (- 39) - (λ + 4) (- 15) = 0$

$\Rightarrow - 252 + 117 + 15 (λ + 4) = 0$

$\Rightarrow 15 λ - 75 = 0 \Rightarrow λ = 5$

$△_{1} =$ $| \begin{matrix} 9 & 3 & 5 \\ 8 & 3 & - 2 \\ 16 - μ & 3 & - 9 \end{matrix} |$ $= 0$

$\Rightarrow 9 (- 21) - 3 (- 40 - 2 μ) + 5 (- 24 + 3 μ) = 0$

$\Rightarrow μ = 9$

So, centre of circle (5, 9)

Also, radius = length of $⊥$ from centre (5, 9) to 4x = 3y

$∴ r =$ $| \frac{4 (5) - 3 (9)}{\sqrt{{(4)}^{2} + {(3)}^{2}}} |$ $=$ $| \frac{20 - 27}{5} |$ $= \frac{7}{5}$ .

Q 16 :
Let the system of equations

$x + 5 y - z = 1$

$4 x + 3 y - 3 z = 7$

$24 x + y + λ z = μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

3
5
6
4

1

2

3

4

(1)

For infinitely many solutions, we have D = 0

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 4 & 3 & - 3 \\ 24 & 1 & λ \end{matrix} |$ $= 0$

$\Rightarrow 1 (3 λ + 3) - 5 (4 λ + 72) - 1 (4 - 72) = 0$

$\Rightarrow 17 λ = - 289 \Rightarrow λ = - 17$ ... (i)

Now, $D_{1} = 0$

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 7 & 3 & - 3 \\ μ & 1 & - 17 \end{matrix} |$ $= 0$

$\Rightarrow 1 (- 51 + 3) - 5 (- 119 + 3 μ) - 1 (7 - 3 μ) = 0$

$\Rightarrow - 48 + 595 - 15 μ - 7 + 3 μ = 0$

$\Rightarrow 12 μ = 540$

$\Rightarrow μ = 45$ ... (ii)

Using (i) and (ii), we get

x + 5y – z = 1, 4x + 3y – 3z =7, 24x + y –17z = 45

$\Rightarrow z = x + 5 y - 1$

$∴ 4 x + 3 y - 3 x - 15 y + 3 = 7$

$\Rightarrow x - 12 y = 4$

$\Rightarrow x = 4 + 12 y and z = 4 + 12 y + 5 y - 1 = 3 + 17 y$

$∴ (x, y, z) = (4 + 12 k, k, 3 + 17 k)$ ( $∵$ Assume y = k)

Also, $7 \leq 7 + 30 k \leq 77$

$\Rightarrow 0 \leq 30 k < 70$

$\Rightarrow 0 \leq k \leq 2.3$

$\Rightarrow$ k = 0, 1, 2

Thus, there are three possible solutions.

Q 17 :
If the system of linear equations:

x + y + 2z = 6

2x + 3y + az = a + 1

– x – 3y + bz = 2b

where $a, b \in R$ , has infinitely many solutions, then $7 a + 3 b$ is equal to : [2025]

16
22
9
12

1

2

3

4

(1)

For infinitely many solutions, we have

$△ =$ $| \begin{matrix} 1 & 1 & 2 \\ 2 & 3 & a \\ - 1 & - 3 & b \end{matrix} |$ $= 0 \Rightarrow 2 a + b - 6 = 0$ ...(i)

Also, $△_{3} =$ $| \begin{matrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ - 1 & - 3 & 2 b \end{matrix} |$ $= 0 \Rightarrow a + b - 8 = 0$ ... (ii)

Solving (i) and (ii), we get a = –2, b = 10

$∴$ 7a + 3b = 7(–2) + 3(10) =16.

Q 18 :
If the system of equations

$(λ - 1) x + (λ - 4) y + λ z = 5$

$λ x + (λ - 1) y + (λ - 4) z = 7$

$(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$

has infinitely many solutions, then $λ^{2} + λ$ is equal to [2025]

20
12
6
10

1

2

3

4

(2)

For infinitely many solutions,

$D =$ $| \begin{matrix} (λ - 1) & (λ - 4) & λ \\ λ & (λ - 1) & (λ - 4) \\ (λ + 1) & (λ + 2) & - (λ + 2) \end{matrix} |$ $= 0$ ... (i)

On solving (i) along column (1), we get

$2 λ^{2} - 5 λ - 3 = 0$

$\Rightarrow (2 λ + 1) (λ - 3) = 0$

$\Rightarrow λ = 3 or - \frac{1}{2}$ ... (ii)

$D_{2} = 0 \Rightarrow$ $| \begin{matrix} (λ - 1) & 5 & λ \\ λ & 7 & (λ - 4) \\ (λ + 1) & 9 & - (λ + 2) \end{matrix} |$ $= 0$

$\Rightarrow 2 λ^{2} - 13 λ + 21 = 0$

$\Rightarrow (2 λ - 7) (λ - 3) = 0$

$\Rightarrow λ = 7 / 2 or λ = 3$ ... (iii)

Thus, $λ = 3$ (From (ii) and (iii))

$∴ λ^{2} + λ = {(3)}^{2} + 3 = 9 + 3 = 12$ .

Q 19 :
The system of equations

x + y + z = 6

x + 2y + 5x = 9

$x + 5 y + λ z = μ$

has no solution if [2025]

$λ = 15, μ \neq 17$
$λ \neq 17, μ \neq 18$
$λ = 17, μ = 18$
$λ = 17, μ \neq 18$

1

2

3

4

(4)

Let $A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & λ \end{matrix}] ∴$ $| A |$ $=$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & λ \end{matrix} |$ $= 0$

$\Rightarrow 1 (2 λ - 25) - (λ - 5) + 1 (5 - 2) = 0 \Rightarrow λ = 17$

$D_{z} =$ $| \begin{matrix} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & μ \end{matrix} |$ $\neq 0$ $[\begin{matrix} ∵ If | A | = 0 then atleast one D_{x}, D_{y}, D_{z} \\ is non zero for inconsistent system \end{matrix}]$

$\Rightarrow μ \neq 18$ .

Q 20 :
If the system of equations

2x – y + z = 4

$5 x + λ y + 3 z = 12$

$100 x - 47 y + μ z = 212$

has infinitely many solutions, then $μ - 2 λ$ is equal to [2025]

55
59
56
57

1

2

3

4

(4)

As the given system of equations has infinitely many solutions, then

$△ = △_{1} = △_{2} = △_{3} = 0$

$△_{2} =$ $| \begin{matrix} 2 & 4 & 1 \\ 5 & 12 & 3 \\ 100 & 212 & μ \end{matrix} |$ $= 0$

$\Rightarrow 2 (12 μ - 636) - 4 (5 μ - 300) + 1 (1060 - 1200) = 0$

$\Rightarrow 4 μ - 212 = 0$

$\Rightarrow μ = 53$

$△_{3} =$ $| \begin{matrix} 2 & - 1 & 4 \\ 5 & λ & 12 \\ 100 & - 47 & 212 \end{matrix} |$ $= 0$

$\Rightarrow 2 (212 λ + 564) + 1 (1060 - 1200) + 4 (- 235 - 100 λ) = 0$

$\Rightarrow 24 λ + 48 = 0 \Rightarrow λ = - 2$

$∴ μ - 2 λ = 53 - 2 (- 2) = 57$ .

Topic Question Set

Q 11 :
Let $α β γ = 45; α, β, γ \in R .$ If $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$ for some $x, y, z \in R, x y z \neq 0,$ then $6 α + 4 β + γ$ is equal to _______. [2024]

0%

Q 12 :
If the system of equations, $2 x + 7 y + λ z = 3,$ $3 x + 2 y + 5 z = 4,$ $x + μ y + 32 z = - 1$ has infinitely many solutions, then $(λ - μ)$ is equal to _______. [2024]

0%

Q 13 :
If the system of Linear equations

$3 x + y + β z = 3$

$2 x + α y - z = - 3$

$x + 2 y + z = 4$

has infinitely many solutions, then the value of $22 β - 9 α$ is : [2025]

Q 14 :
If the system of equations

$2 x + λ y + 3 z = 5$

$3 x + 2 y - z = 7$

$4 x + 5 y + μ z = 9$

has infinitely many solutions, then $(λ^{2} + μ^{2})$ is equal to: [2025]

Q 15 :
Let the system of equations :

$2 x + 3 y + 5 z = 9$

$7 x + 3 y - 2 z = 8$

$12 x + 3 y - (4 + λ) z = 16 - μ$

have infinitely many solutions. Then the radius of the circle centred at $(λ, μ)$ and touching the line 4x = 3y is [2025]

Q 16 :
Let the system of equations

$x + 5 y - z = 1$

$4 x + 3 y - 3 z = 7$

$24 x + y + λ z = μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

Q 17 :
If the system of linear equations:

x + y + 2z = 6

2x + 3y + az = a + 1

– x – 3y + bz = 2b

where $a, b \in R$ , has infinitely many solutions, then $7 a + 3 b$ is equal to : [2025]

Q 18 :
If the system of equations

$(λ - 1) x + (λ - 4) y + λ z = 5$

$λ x + (λ - 1) y + (λ - 4) z = 7$

$(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$

has infinitely many solutions, then $λ^{2} + λ$ is equal to [2025]

Q 19 :
The system of equations

x + y + z = 6

x + 2y + 5x = 9

$x + 5 y + λ z = μ$

has no solution if [2025]

Q 20 :
If the system of equations

2x – y + z = 4

$5 x + λ y + 3 z = 12$

$100 x - 47 y + μ z = 212$

has infinitely many solutions, then $μ - 2 λ$ is equal to [2025]

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Topic Question Set

Q 11 : Let αβγ=45; α,β,γ∈R. If x(α,1,2)+y(1,β,2)+z(2,3,γ)=(0,0,0) for some x,y,z∈R, xyz≠0, then 6α+4β+γ is equal to _______. [2024]

0%

Q 12 : If the system of equations, 2x+7y+λz=3,3x+2y+5z=4,x+μy+32z=-1 has infinitely many solutions, then (λ-μ) is equal to _______. [2024]

0%

Q 13 : If the system of Linear equations 3x+y+βz=3 2x+αy–z=–3 x+2y+z=4 has infinitely many solutions, then the value of 22β-9α is : [2025]

Q 14 : If the system of equations 2x+λy+3z=5 3x+2y–z=7 4x+5y+μz=9 has infinitely many solutions, then (λ2+μ2) is equal to: [2025]

Q 15 : Let the system of equations : 2x+3y+5z=9 7x+3y–2z=8 12x+3y–(4+λ)z=16–μ have infinitely many solutions. Then the radius of the circle centred at (λ, μ) and touching the line 4x = 3y is [2025]

Q 16 : Let the system of equations x+5y–z=1 4x+3y–3z=7 24x+y+λz=μ λ,μ∈R, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7≤x+y+z≤77, is : [2025]

Q 17 : If the system of linear equations: x + y + 2z = 6 2x + 3y + az = a + 1 – x – 3y + bz = 2b where a,b∈R, has infinitely many solutions, then 7a+3b is equal to : [2025]

Q 18 : If the system of equations (λ–1)x+(λ–4)y+λz=5 λx+(λ–1)y+(λ–4)z=7 (λ+1)x+(λ+2)y–(λ+2)z=9 has infinitely many solutions, then λ2+λ is equal to [2025]

Q 19 : The system of equations x + y + z = 6 x + 2y + 5x = 9 x+5y+λz=μ has no solution if [2025]

Q 20 : If the system of equations 2x – y + z = 4 5x+λy+3z=12 100x–47y+μz=212 has infinitely many solutions, then μ–2λ is equal to [2025]

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Q 11 :
Let $α β γ = 45; α, β, γ \in R .$ If $x (α, 1, 2) + y (1, β, 2) + z (2, 3, γ) = (0, 0, 0)$ for some $x, y, z \in R, x y z \neq 0,$ then $6 α + 4 β + γ$ is equal to _______. [2024]

Q 12 :
If the system of equations, $2 x + 7 y + λ z = 3,$ $3 x + 2 y + 5 z = 4,$ $x + μ y + 32 z = - 1$ has infinitely many solutions, then $(λ - μ)$ is equal to _______. [2024]

Q 13 :
If the system of Linear equations

$3 x + y + β z = 3$

$2 x + α y - z = - 3$

$x + 2 y + z = 4$

has infinitely many solutions, then the value of $22 β - 9 α$ is : [2025]

Q 14 :
If the system of equations

$2 x + λ y + 3 z = 5$

$3 x + 2 y - z = 7$

$4 x + 5 y + μ z = 9$

has infinitely many solutions, then $(λ^{2} + μ^{2})$ is equal to: [2025]

Q 15 :
Let the system of equations :

$2 x + 3 y + 5 z = 9$

$7 x + 3 y - 2 z = 8$

$12 x + 3 y - (4 + λ) z = 16 - μ$

have infinitely many solutions. Then the radius of the circle centred at $(λ, μ)$ and touching the line 4x = 3y is [2025]

Q 16 :
Let the system of equations

$x + 5 y - z = 1$

$4 x + 3 y - 3 z = 7$

$24 x + y + λ z = μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

Q 17 :
If the system of linear equations:

x + y + 2z = 6

2x + 3y + az = a + 1

– x – 3y + bz = 2b

where $a, b \in R$ , has infinitely many solutions, then $7 a + 3 b$ is equal to : [2025]

Q 18 :
If the system of equations

$(λ - 1) x + (λ - 4) y + λ z = 5$

$λ x + (λ - 1) y + (λ - 4) z = 7$

$(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$

has infinitely many solutions, then $λ^{2} + λ$ is equal to [2025]

Q 19 :
The system of equations

x + y + z = 6

x + 2y + 5x = 9

$x + 5 y + λ z = μ$

has no solution if [2025]

Q 20 :
If the system of equations

2x – y + z = 4

$5 x + λ y + 3 z = 12$

$100 x - 47 y + μ z = 212$

has infinitely many solutions, then $μ - 2 λ$ is equal to [2025]