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This mock test of Test: Matrices And Determinants (Competition Level) for Class 12 helps you for every Class 12 entrance exam.
This contains 20 Multiple Choice Questions for Class 12 Test: Matrices And Determinants (Competition Level) (mcq) to study with solutions a complete question bank.
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QUESTION: 1

If B is a non–singular matrix and A is a square matrix, then det (B^{–1} AB) is equal to

Solution:

A is a square matrix and B is non-singular matrix, so B is invertible and a square matrix.

det(B^{−1} AB)

=det(B^{−1} BA)

=det(I_{n}A) (where I_{n} is an identity matrix of order n )

=det(A)

QUESTION: 2

If the system of equations x + 2y + 3z = 4, x + λy + 2z = 3, x + 4y + μz = 3 has an infinite anumber of solutions then

Solution:

**Correct Answer :- d**

**Explanation :** x+2y+3z=4

x+py+2z=3

x+4y+z=3

x = Δx/Δ, y = Δy/Δ, z = Δz/Δ

Δ = 0, Δx = Δy = Δz = 0

As we know that Δy = 0

A = {(1,4,3) (1,3,2) (1,3,μ)} = 0

3μ + 9 + 8 - 9 - 6 - 4μ = 0

=> μ = 2

As Δz = 0

= {(1,2,4) (1,p,3) (1,4,3)} = 0

=> 3p + 16 - 6 - 4p - 12 -6 = 0

=> p = 4

QUESTION: 3

If a, b, c are non zeros, then the system of equations (α + a) x + αy + αz = 0

αx + (α + b)y + αz = 0

αx + αy + (α + c) z = 0

has a non–trivial solution if

Solution:

For non trival solution, D = 0

{(α + a, α, α)(α, α + b, α)(α, α, α + c)} = 0

R1→ R1 - R2, R2 → R2 - R3

{(a, -b ,0)(0, b ,-c)(α, α, α + c)}

abα + abc + acα + bcα = 0

⇒ abc = −α(ab + bc + ca)

⇒ 1/α = −(1/c+ 1 / a + 1 / b)

α^{−1} = −(a^{−1}+b^{−1}+c^{−1})

QUESTION: 4

Let and . If B is the inverse of matrix A, then α is

Solution:

Since, B is the inverse of matrix A.

So, 10A^{−1} = {(4,2,2) (-5,0,α) (1,-2,3)}

⇒ 10A^{-1} . A = {(4,2,2) (-5,0,α) (1,-2,3)} . {(1,-1,1) (2,1,-3) (1,1,1)}

⇒ 10I = {(10,0,0) (-5+α, 5+α, -5+α) (0,0,10)}

⇒ {(10,0,0) (0,10,0) (0,0,10)} = {(10,0,0) (-5+α, 5+α, -5+α) (0,0,10)}

−5+α=0

⇒ α=5

QUESTION: 5

The value of a for which system of equations,

a^{3}x + (a + 1)^{3} y + (a + 2)^{3}z = 0,

ax + (a + 1) y + (a + 2) z = 0,

x + y + z = 0, has a non–zero solution is

Solution:

QUESTION: 6

Two matrices A and B are multiplied to get AB, if

Solution:

Two matrices A and B are multiplied to get AB if the number of columns of matrix A is equals to the number of rows of matrix B.

The product of two matrices A and B is defined if the number of columns of matrix A is equal to number of rows of matrix B. Let A = [a_{ij}] be an mxn matrix & B = [b_{jk}] be an nxp matrix. Then the product of A and B is order of mxp.

Hence, the answer is no of columns of A is equal to rows of B.

QUESTION: 7

Let , then A^{-1 }exists if

Solution:

QUESTION: 8

Let where 0 ≤ θ < 2π, then

Solution:

QUESTION: 9

Solution:

QUESTION: 10

then which one of the following holds for all n ≥ 1, by the principle of mathematical induction ?

Solution:

QUESTION: 11

is the unit matrix of order 2 and a, b are arbitrary constants, then is equal to

Solution:

QUESTION: 12

Identity the correct statement(s)

Solution:

QUESTION: 13

If a, b, c > 0 & x, y, z ∈ R then the determinant

Solution:

QUESTION: 14

IF x, y , z ∈ R Δ = = -16 then value of x is

Solution:

QUESTION: 15

The determinant is

Solution:

QUESTION: 16

If A, B, C are angles of a triangle ABC, then

is less than or equal to

Solution:

**ANSWER :- b**

**Solution :- sin(A+B+C) = cos(A+B+C)/2 = 0**

**Hence, sinA/2 sinB/2 sinC/2 ≤ 1/8**

QUESTION: 17

Let f(x) = then the maximum value of f(x) is

Solution:

QUESTION: 18

Value of the D = is

Solution:

QUESTION: 19

If f(x) = , then

Solution:

f(x) ={(a^{-x}, e^{x/na}, x^2) (a^{-3x}, e^{3x}/na, x^{4}) (a^{-5x}, e^{5x/na}, 1)}

f(-x) = - {(a^{-x}, e^{x/na}, x^{2}) (a^{-3x}, e^{3x/na}, x^{4}) (a^{-5x}, e^5x/na, 1)}

Adding (1) and (2) we get

f(x) + f(-x) = 0

QUESTION: 20

D = is (where a, b, c are the sides opposite to angles A, B, C respectively in a triangle)

Solution:

a/sinA = b/sinB = c/sinC = 2R

A = 2RsinA, b = 2RsinB, c = 2RsinC

{(1, 4sinB/b, cosA) (2a, 8sinA, 1) (3a, 12sinA, cosB)}

= {(1, 4sinB/(2RsinB), cosA) (4RsinA, BsinA, 1) (6RsinA, 12sinA, cosB)}

= 1(8sinAcosB- 12sinA) - 2/R(4RsinAcosB - 6RsinA) + cosA(48Rsin^2A - 48Rsin^2A)

= 8sinAcosB - 12sinA -8sinAcosB + 12sinA

= 0

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