Solution

SCREENING TEST (DUPLICATE)

Class 12 - Mathematics

1.
1 0
(b) [ ]
0 1

0 i
Explanation: A = [ ]
i 0

0 i 0 i 0 1 0 i
A4n = A4 = [ ]× [ ]×[ ]× [ ]
i 0 i 0 i 0 i 0

{n = 1, so the exponent comes out to be 4 and if n = 2, which will turn the exponent to 8, and the same cycle will repeat.}
4
i 0
=[ 4
]
0 i
2
(−1) 0
=[ 2
]
0 (−1)

1 0
=[ ]
0 1

2. (a) aij = aji

Explanation: According to definition symmetric matrix, aij = aji

3.
(c) π

cosa sin a cos a − sin a

Explanation: L.H.S.: A + A ′
= ( ) + ( )
− sin a cos a sin a cos a

cos a + cos a sin a − sin a

= ( )
− sin a + sin a cos a + cos a

2 cos a 0
= ( )
0 2 cos a

1 0
This will be equal to ( )
0 1

When 2 cos a = 1
1
cos a =
2
π
a =
3

4.
4 −6
(b) [ ]
6 4

2 −1
Explanation: Given, A = [ ]
1 2

2 −1 2 −1 4 − 1 −2 − 2
Now, A 2
= [ ][ ]= [ ]
1 2 1 2 2 + 2 −1 + 4

3 −4
2
⇒ A = [ ]
4 3

Now, A2 + 2A - 3I
3 −4 2 −1 1 0
= [ ]+ 2[ ]− 3[ ]
4 3 1 2 0 1

3 + 4 − 3 −4 − 2 − 0 4 −6
= [ ]= [ ]
4 + 2 − 0 3 + 4 − 3 6 4

5.
(b) A2 = A
∣1 0∣ ∣1 0∣∣1 0∣ ∣1 0∣
Explanation: A = ∣ ∣ , then A2 = ∣ ∣∣ ∣ = ∣ ∣ =A
∣0 0∣ ∣0 0∣∣0 0∣ ∣0 0∣

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*Screening Test*
 6. (a) 0
Explanation: A is skew-symmetric means At = −A. Taking determinant both sides
t
Δ (A ) = Δ(−A)

3
⇒ Δ(A) = (−1) Δ(A)

⇒ Δ(A) = −Δ(A)

Which is only possible when Δ(A) = 0

7.
(b) A2 = I
∣0 −1 ∣ ∣ 0 −1 ∣ ∣1 0∣
Explanation: A2 = ∣ ∣∣ ∣ = ∣ ∣ =I
∣ −1 0 ∣ ∣ −1 0 ∣ ∣0 1∣

8.
(b) -6
Explanation: -6
9.
(c) 3 × n
Explanation: A3 × m and B3 × n are two matrices. If m = n then A and B same orders as 3 × n each so the order of (5A - 2B)
should be same as 3 × n.
10.
(d) skew-symmetric matrix
Explanation: We have matrices A and B of same order.
Let P = (AB' − BA')
Then, P′ = (AB′ − BA′)'
= (AB′)′ − (BA′)′
= (B′)′ (A)′ − (A′)′B′ = BA′ − AB′ = −(AB′ − BA′) = −P
Therefore, the given matrix (AB - BA') is a skew-symmetric matrix.
11.
(b) Det(A) ∈ [2, 4]
1 sin θ 1
⎡ ⎤
Explanation: A = ⎢ − sin θ 1 sin θ ⎥
⎣ ⎦
−1 − sin θ 1

|A| = 1 (1 × 1 – sin θ × (-sin θ)) – sin θ (-sin θ+sinθ) + 1 [(- sin θ) × (−sinθ) − (−1) × 1]

|A| = 1 + sin θ + sin θ + 1

2 2

|A| = 2 + 2 sin θ 2

|A| = 2(1 + sin2θ)

Now, 0 ≤ θ ≤ 2π
⇒ sin 0 ≤ sin θ ≤ sin 2π

⇒ 0 ≤ sin2θ ≤ 1
⇒ 1 + 0 ≤ 1 + sin2θ ≤ 1 + 1
⇒ 2 ≤ 2(1 + sin2θ) ≤ 4
∴ Det (A) ∈ [2, 4]

12.
(b) 10
Explanation: We know that
A × adjA = |A| Inxn , where I is the unit matrix of order nxn.--------------------[1]
10 0
A(adj A) = [ ] Using the above property of matrices (1), we get
0 10

1 0
A(adj A) = 10 [ ]
0 1

A(adj A) = (10) I2x2

|A| I2x2 = 10 I2x2
|A| = 10

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*Screening Test*
13.
(c) 6
Explanation: For a non-zero solution
3 5
[ ]= 0
k 10

⇒ 30 - 5k = 0 ⇒ k = 6
14.
(c)
|A|I

Explanation: Since, we know that

adjA
−1
A =
|A|

pre multiply by A,
AadjA
−1
AA =
|A|

(since AA-1=I)
AadjA
I = ⇒ AadjA = |A| I
|A|

15.
(d) 64
−2 0 0
⎡ ⎤

Explanation: A = ⎢ 0 −2 0 ⎥
⎣ ⎦
0 0 −2

|A| = -2[-4 - 0] - 0 + 0
= -8
Now, |adj A| = |A|n-1 ...(where n is the order of matrix n)
= (-8)3-1
= (-8)2
= 64
16.
(c) 144
1 2 −1
⎡ ⎤

Explanation: A = ⎢ −1 1 2⎥
⎣ ⎦
2 −1 1

|A| = 14 det( adjA)= det(A)3-1 = det(A)2.Here the operation is done two times.so,
2
(n−1)
det (adj(adj A)) = |A|
2

det (adj(adj A)) = 14 (3−1)

= 14
4

17.
(b) A
Explanation: A2 = I so (A - I)3 + (A + I)3 - 7A
(A − I ) + (A + I ) - 7A = A − 3A I + 3AI - 7A
3 3 3 2 2 3 3 2 2 3
⋅ I + A + 3A I + 3AI + I

= A A − 3A + 3A − I + A A + 3A + 3A + I − 7A
2 2 2 2

= I A − 3I + 3A − I + I A + 3I + 3A + I − 7A
= A - 3I + 3A - I + A + 3I + 3A + I - 7A = A
18.
(d) 36
Explanation: 36
|adj A|= |A|n-1
=(6)3-1
= 36
19. (a) 1

det(A)

Explanation: Solution.

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*Screening Test*
 Since we know that ∣∣A−1
∣
∣ =
1

|A|

20.
(b) 2
Explanation: |adj A| = |A|k (Given)
But |adj A| = |A|2
⇒ k = 2

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*Screening Test*