Determinants Test

Result

Determinants Test - 10

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Weekly Quiz Competition

Question 1
1 / -0.25

Inverse of a matrix A exists, if

A
| A | = 0

B
Matrix A is non-singular

C
Matrix A is singular

D
A^T = A

Solution

Only non-singular matrices have inverses.
Suppose that A is invertible. This means that we have the inverse matrix A⁻¹ of A.
Consider the equation Ax=0. We show that this equation has only zero solution.
Multiplying it by 　A⁻¹ on the left, we obtain
A⁻¹ Ax = A^-1 0
⇒x = 0
Question 2
1 / -0.25

If A = , and |3A| = k|A|, then the value of k is

A
27

B
1

C
25

D
4

Solution

|kA| = kⁿ |A|
|3A| = (3)ⁿ |A|
= 27|A| = k|A|
k = 27
Question 3
1 / -0.25

Find the value of

A
-1

B
1

C
2

D
0

Solution

Correct Answer :- d
Explanation:- 3(6-6) -2(6-9) +3(4-6)
= 3(0) + 6 - 6
= 0
Question 4
1 / -0.25

If A =, then adj A is:

A

B

C

D
Question 5
1 / -0.25

The cofactor of an element 9 of the determinant is :

A
48

B
–48

C
27

D
46

Solution

Co factor of element 9 will be {(6 * 8) -(-5 * 0)}
= 48
Question 6
1 / -0.25

The minor M_ij of an element a_ij of a determinant is defined as the value of the determinant obtained after deleting the

A
j^th row of the determinant

B
i^th column and jth row of the determinant

C
i^th row and j^th column of the determinant

D
i^th row of the determinant

Solution

A minor, Mij, of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column.
Question 7
1 / -0.25

The cofactor of 5 in

A
3

B
29

C
-3

D
5

Solution

A21 = {(0, -1) (3,4)}
⇒(0 - (-3)) = 3
A21 = (-1)⁽²⁺¹⁾ (3)
= -3
Question 8
1 / -0.25

If , then the relation between x and y is

A
y = -3x

B
x = 3y

C
x = -3y

D
y = 3x

Solution

½{(0,0,1) (1,3,1) (x,y,1)} = 0
{(0,0,1) (1,3,1) (x,y,1)} = 0/(½)
{(0,0,1) (1,3,1) (x,y,1)} = 0
0(3-y) -0(1-x) +1(y-3x) = 0
=>y - 3x = 0
=>y = 3x
Question 9
1 / -0.25

If matrix A = and A² + aA + b = O, then the values of a and b are:

A
a = –6, b = 2

B
a = 6, b = –2

C
a = 6, b = 2

D
a = –6, b = –2
Question 10
1 / -0.25

If and A_ij are cofactors of a_ij , then

A
Δ= a₁₁ A₁₁ + a₂₁ A₂₂ + a₃₁ A₃₃

B
Δ= a₁₁ A₃₁ + a₁₂ A₃₂ + a₁₃ A₃₃

C
Δ= a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃

D
Δ= a₁₁ A₁₁ + a₂₂ A₂₂ + a₃₃ A₃₃

Solution

Δ= Sum of products of element of row(or column) with their corresponding co-factors.
Δ= a11 A11 + a21 A21 + a31 A31