Determinants Test

Result

Determinants Test - 4

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Weekly Quiz Competition

Question 1
1 / -0.25

is equal to

A
2

B
1

C
-2

D
0

Solution

Apply , C1 →C1 - C3, C₂ →C₂ -C₃

= 10 - 12 = -2
Question 2
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If A+B+C = π, then the value of

A
2

B
1

C
0

D
none of these

Solution
Question 3
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If A is a non singular matrix of order 3 , then |adj(A³ )| =

A
|A|⁹

B
|A|⁶

C
|A|⁸

D
none of these

Solution

If A is anon singular matrix of order , then
Question 4
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If A and B are any 2 × 2 matrices , then det. (A+B) = 0 implies

A
det A = 0 and det B = 0

B
det A = 0 or det B = 0

C
detA + det B = 0

D
none of these

Solution

Det.(A+B) ≠ Det.A + Det.B.
Question 5
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A
λ= 2

B
λ≠2

C
λ≠–2

D
None of these

Solution

Correct option is D.
Question 6
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If A B be two square matrices such that AB = O, then

A
Det.B = 0

B
BA

C
Det. A = 0 or Det. B = 0

D
Det. = A = 0

Solution

If A B be two square matrices such that AB = O, then,
Question 7
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A
–2

B
0

C
x+2

D
none of these

Solution

Apply , C₁ →C₁ - C₂ , C₂ →C₂ - C₃ ,

Because here row 1 and 2 are identical
Question 8
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If , then equals

A
0

B
–abc

C
2abc

D
none of these.

Solution

Because , the determinant of a skew symmetric matrix of odd order is always zero and of even order is a non zero perfect square.
Question 9
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If I₃ is the identity matrix of order 3 , then 1₃⁻¹ is

A
0

B
3I₃

C
I₃

D
none of these

Solution

Because , the inverse of an identity matrix is an identity matrix.
Question 10
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If A and B are square matrices of same order and A ’denotes the transpose of A , then

A
(AB)’= B ’A ’

B
AB = O ⇒|A| = 0 and |B|=0

C
AB = O ⇒ A = 0 or B = 0

D
(AB)’= A ’B ’

Solution

By the property of transpose of a matrix ,(AB)’= B ’A ’.
Question 11
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A square matrix A is invertible iff det A is equal to

A
0

B
–1

C
1

D
non zero

Solution

Only non-singular matrices possess inverse.
Question 12
1 / -0.25

A
0

B
(x+3a) (x −a)³

C
a⁴

D
none of these

Solution

Apply, C₁ →C₁ + C₂ +C₃ +C₄ ,

Apply, R1 →R₁ - R₂ ,

Apply, R₁ → R₁ - R₂ , R₂ →R₂ - R₃

=(x+3a) (a -x)³ (1) = (x+3a)(a-x)³
Question 13
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If the entries in a 3 x 3 determinant are either 0 or 1 , then the greatest value of this determinant is :

A
2

B
1

C
0

D
none of these.

Solution

Greatest value = 2
Question 14
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The roots of the equation are

A
1 , –2

B
–1 , 2

C
1 , 2

D
–1, –2

Solution

Operate,

Apply R₃ →R₃ - R₁ , R₂ → R₂ -R₁ ,

⇒-6(5x² - 20) +15(2x-4) = 0

⇒(x- 2)(x+1) = 0 ⇒x=2, -1

⇒-6(5x² - 20) + 15(2x-4) = 0

⇒(x-2)(x+1) = 0 ⇒x=2, -1
Question 15
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In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into n determinants, where n has value

A
1

B
9

C
24

D
16

Solution

N = 2 ×3 × 4 = 24.