Determinants Test

Result

Determinants Test - 5

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Weekly Quiz Competition

Question 1
1 / -0.25

If the value of the determinant , then the value of is

A
-24

B
32

C
24

D
-32

Solution

Determinant = -2(-12 + 12) -2(-6 - 2) +1(-12 - 4)
= - 2(0) - 2(-8) + 1(16)
= 0 + 16 + 16
= 32
Question 2
1 / -0.25

The value of x, if = 0 , is

A
1/5

B
-1/5

C
-1/3

D
1/3

Solution
Question 3
1 / -0.25

If the value of , then the value of

A
-7

B
14

C
7

D
-14

Solution

If you interchange (switch) two rows (or columns) of a matrix A to get B, then det(B) = –det(A). Hence, the determinant is -(-7)
= 7
Question 4
1 / -0.25

The operation that will simplify the Determinant
by taking (a+b+c) common from the first Row is

A
R₂ →R₂ +R₁

B
R₁ →R₁ +R₂

C
R₁ →R₁ -R₂

D
R₁ →R₂ -R₁

Solution

R₁ ------>R₁ + R₂ , taking (a+b+c) common from the first row, we get the resultant matrix.
Question 5
1 / -0.25

The value of the determinant is

A
4

B
0

C
-1

D
1

Solution

|A|=3((4)-(4))-2((-4)-(2))-1((-8)-(-4))
|A|= -2(-6)-(-6)
|A|= -2(0)
=0
Question 6
1 / -0.25

The value of the Δ

A
1

B
-1

C
2

D
none
Question 7
1 / -0.25

If the value of the determinant is equal to 14, then the value of the determinant is equal to

A
-42

B
42

C
-14

D
14

Solution

Taking 3 common from the column I in second determinant, we get the 14 * 3 = 42
Question 8
1 / -0.25

The value of is

A
(5x + 2) (2 - x)²

B
(5x + 2)² (2 - x)

C
(5x + 2)² (2 - x)²

D
(5x + 2) (2 - x)

Solution

A = {(5x+2, 2x, 2x) (5x+2, x+2, 2x) (5x+2, 2x, x+2)}
= (5x+2) {(1, 2x, 2x) (1, x+2, 2x) (1, 2x, x+2)}
= (5x+2) [(1(x+2)^2 - 4x^2) - 2x(x+2 - 2x) + 2x(2x - x - 2)]
= (5x+2) [(x+2 - 2x)(x+2+2x)-2x(2 - x) + 2x(x-2)]
= (5x+2) [(2-x)(3x+2) -2x(2-x) -2x(2-x)]
= (5x+2) (2-x)[3x+2-2x-2x]
= (5x+2) (2-x)²
Question 9
1 / -0.25

Given

A
-2

B
1

C
2

D
xyz+1=0

Solution
Question 10
1 / -0.25

+ Δ. then Δis

A

B

C

D