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This mock test of Competition Level Test: Matrices & Determinants for Class 12 helps you for every Class 12 entrance exam.
This contains 30 Multiple Choice Questions for Class 12 Competition Level Test: Matrices & Determinants (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The number of different orders of a matrix having 12 elements is

Solution:

(1,12),(2,6),(3,4(,(4,3),(6,2),(12,1)

6 different matrices.

QUESTION: 2

then x is equal to

Solution:

x²+x=0

x(x+1)=0

therefore x=0 or -1

3-x+1=5

x=-1.

Hence, x = -1

QUESTION: 3

If and , then

Solution:

No, we cannot multiply 3*1 by 3*3, So it doesnt exist.

QUESTION: 4

If , and , then B equal to

Solution:

QUESTION: 5

If A and B are square matrices of order 2, then (A + B)^{2} equal to

Solution:

(A+B)^{2} = (A+B) (A+B)

= A(A+B) +B(A+B)

A^{2} + AB + BA + B^{2}

QUESTION: 6

If A is a skew – symmetric matrix, then trace of A is equal to

Solution:

We know that for a skew-symmetric matrix the sum of diagonal elements is zero.

a[ij] = 0∀i = j

So,tr(A) = 0

QUESTION: 7

Solution:

QUESTION: 8

Which of the following is a vector?

Solution:

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

QUESTION: 9

If A is a square matrix such that A^{2} = then A^{–1 }equal to

Solution:

As we know that, A^{2} = I

On multiplying both sides by A^{-1}

(A^{2})A^{-1} = I A^{-1}

A = I A-^{1}

A = A^{-1}

QUESTION: 10

If A and B are square matrices of order 3 such that |A| = –1, |B| = 3, then |3AB| is equal to

Solution:

∣AB∣=∣A∣∣B∣

Also for a square matrix of order 3

∣KA∣=K^{3}∣A∣ because each element of the matrix A is multiplied by k and hence in this case we will have k^{3} common.

∴∣3AB∣ = (3)^{3}

∣A∣∣B∣ = 27(−1)(3)

= −81

QUESTION: 11

If , then value of A^{-1} is equal to

Solution:

QUESTION: 12

If the order of matrix A is m×p. And the order of B is p×n. Then the order of matrix AB is?

Solution:

If we are given 2 matrices of with order a×b and c×d, in order to be multiplicable, b must be equal to c and the resultant matrix will have order a×d. So using this formula in this question, We have p = p and so the resultant matrix will have m×n order. So, order of matrix AB is m×n.

QUESTION: 13

The system of equation –2x + y + z = 1, x – 2y + z = –2, x + y + λz = 4 will have no solution if

Solution:

D = 0

{(-2,1,1) (1,-2,1) ( 1,1, λ) = 0

= (4λ +1 +1) - (-2 - 2 + λ) = 0

4λ + 2 + 4 - λ = 0

3λ + 6 = 0

λ = -2

QUESTION: 14

The system of the linear equations x + y – z = 6, x + 2y – 3z = 14 and 2x + 5y – λz = 9 (λ ∈ R) has a unique solution if

Solution:

For a system of linear equations having unique solution, value of the determinant should not be 0

⇒ {(1,1,-1) (1,2,-3) (2,5,-λ)} not equal to o

⇒ [−2λ+15−1(−λ+6)−(5−4)] ≠ 0

⇒ −λ+8 ≠ 0

⇒ λ ≠ 8

QUESTION: 15

Solution:

QUESTION: 16

If A = diag (2, –1, 3), B = diag (–1, 3, 2), then A^{2} B equal to

Solution:

QUESTION: 17

and then B^{T}A^{T} is

Solution:

A = {(2,-1) (-7,4)} B = {(4,1) (7,2)}

A’ = {(2,-7) (-1,4)} B’ = {(4,7) (1,2)}

B’A’ = {(1,0) (0,1)}

Therefore, identity matrix.

QUESTION: 18

If the matrix AB is a zero matrix, then

Solution:

QUESTION: 19

Which relation true for

Solution:

QUESTION: 20

If AB = A and BA = B, then B^{2} is equal to

Solution:

AB = A and BA = B

B^{2} = B.B

= (BA).B

= B.(AB)

= B.A

= B

QUESTION: 21

If A and B are symmetric matrices, then ABA is

Solution:

As A and B are symmetric matrices, therefore

A’ = A & B’ = B

So we have

(ABA)’ = A’ (AB)’= A’B’A’

= ABA ie symmetric matrix

QUESTION: 22

If A is a skew – symmetric matrix and n is an even positive integer, then A^{n} is

Solution:

A → skew symmetric matrix

n → positive integer

An → is symmetric if n is an odd integer and is skew symmetric if n is an even integer.

QUESTION: 23

If A is a non–singular matrix and A^{T} denotes the transpose of A, then

Solution:

As we know that, |A| is not equal to zero

|A’| is also not equal to zero

So |A| + |A’| will not be equal to zero. Hence option d is correct.

QUESTION: 24

Which of the following is incorrect

Solution:

QUESTION: 25

If A is square matrix of order 3, then the true statement is (where l is unit matrix).

Solution:

If A is an n×n square matrix and n is odd, then

det(−A)=−det(A)

Proof:

det(−A)=(−1)n det(A) = −det(A)

QUESTION: 26

The transpose of a column matrix is a :

Solution:

QUESTION: 27

From the matrix equation AB = AC, we conclude B = C provided

Solution:

Let ∣A∣ is not equal to 0

∴A ^{−1} exists.

Given AB=AC

Now, pre multiplying by A^{−1} on both sides. we get,

A^{−1}(AB) = A^{−1}(AC)

⇒ IB = IC

⇒ B = C

So, A is non singular

QUESTION: 28

Which of the following property of matrix multiplication is correct:

Solution:

Matrix multiplication is not commutative in general i.e., AB is not equal to BA.

Matrix multiplication is distributive over matrix addition i.e.,

(i) A(B+C)=AB+AC

(ii) (A+B)C=AB+AC, whenever both sides of equality are defined.

Matrix multiplication is associative i.e., (AB)C=A(BC), whenever both sides are defined.

QUESTION: 29

If satisfies the equation x^{2} – (a + d) x + k = 0, then

Solution:

Let λI = {(a,b) (c,d)}

{(λ,0) (0,λ)} - {(a,b) (c,d)}

= {(λ-a,0) (0,λ-d)} = 0

A = {(λ-a,b) (c,λ-d)}

= λ^{2} - (a+d)λ + (ad+bc)

= A^{2} - (a+d)A + (ad+bc)

⇒ k = ad - bc

QUESTION: 30

Which of the following is a nilpotent matrix

Solution:

An n×n matrix A is called nilpotent if Ak=O, where O is the n×n zero matrix.

(a) The matrix A is nilpotent if and only if all the eigenvalues of A is zero.

(b) The matrix A is nilpotent if and only if An=O.

Therefore, option c gives zero matrix.

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