Solution

MATRIX & DETERMINANT

Class 12 - Mathematics
Section A
4 1 4
⎡ ⎤

1. (a) ⎢ 3 −2 5 ⎥
⎣ ⎦
1 3 −1

Explanation:
4 3 1
⎡ ⎤
Let A = ⎢ 1 −2 3⎥
⎣ ⎦
4 5 −1
′
4 3 1 4 1 4
⎡ ⎤ ⎡ ⎤
Then, A ′
= ⎢1 −2 3⎥ = ⎢3 −2 5⎥ [interchanging the elements of rows and columns]
⎣ ⎦ ⎣ ⎦
4 5 −1 1 3 −1

2.
(b) A = [aij]n where aij = { 0 if i ≠ j
k if i = j for k ∈ R.

Explanation:
A scalar matrix is a type of diagonal matrix if
A = [aij]n where aij = { k for i = j
0 for i ≠ j

3.
(c) A + B = B + A
Explanation:
∵ A + B = B + A [by properties matrix addition, is communtative]

4. (a) 16A
Explanation:
2 0 0
⎡ ⎤

A= ⎢0 2 0⎥
⎣ ⎦
0 0 2

1 0 0
⎡ ⎤

⇒ A= 2⎢0 1 0⎥
⎣ ⎦
0 0 1

We can write as
A = 2I
Hence, A5 = (2I)5
A5 = 32I
A5 = 16 × 2I = 16A
5.
(d) 9
Explanation:
9

6. (a) Both A and R are true and R is the correct explanation of A.

Explanation:
Since, A is a skew symmetric matrix, then
A = -A'
⇒ A2 = (-A')2

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 ⇒ A2 = (A2)'
⇒ A2 is symmetric matrix.
Section B
7. We know that, the notation, namely A = [a ij ]m×n indicates that A is a matrix of order m × n, also
1 ≤ i ≤ m, 1 ≤ j ≤ n; i, j ∈ N .
Here, A = [a ij ]2×2
2
(i−2j)
where, a ij =
2
, 1 ≤ i ≤ 2; 1 ≤ j ≤ 2 ...(i)
2
(1−2)
1
∴ a11 = =
2 2
2
(1−2×2)
9
a12 = =
2 2
2
(2−2×1)
a21 = = 0
2
2
(2−2×2)
a22 = = 2
2

1 9

Thus, A = [ 2 2
]
0 2
2×2

8. Given,
3 4
⎡ ⎤
−1 2 1
AT = ⎢ −1 2⎥ ,B=[ ]
1 2 3
⎣ ⎦
0 1

Now
−1 1
⎡ ⎤

BT = ⎢ 2 2⎥
⎣ ⎦
1 3

3 4 −1 1
⎡ ⎤ ⎡ ⎤

AT - BT = ⎢ −1 2⎥ − ⎢ 2 2⎥
⎣ ⎦ ⎣ ⎦
0 1 1 3

3 + 1 4 − 1
⎡ ⎤
AT - BT = ⎢ −1 − 2 2 − 2⎥
⎣ ⎦
0 − 1 1 − 3

4 3
⎡ ⎤

AT - BT = ⎢ −3 0 ⎥
⎣ ⎦
−1 −2

Section C
9. Given, A = B
2
2x + 1 2y x + 3 y + 2
[ ]= [ ]
2
0 y − 5y 0 −6

Since equal matrices has all corresponding entries equal,

So, 2x + 1 = x + 3 ...(i)
2y = y2 + 2 ...(ii)
y2 - 5y = - 6 ...(iii)
Solving equation (i) we get,
2x + 1 = x + 3
2x - x = 3 - 1
x=2
Solving equation (ii) we get,
2y = y2 + 2
y2 - 2y + 2 = 0
D = b2 - 4ac
= (-2)2 - 4(1)(2)
= 4 - 8 = -4
So, there is no real value of y from equation (ii).
Solving equation (iii),

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 y2 - 5y = - 6
y2 - 5y + 6 = 0
y2 - 3y - 2y + 6 = 0
y (y - 3) - 2 (y - 2) = 0
y = 3 or y = 2
From solution of equation (i), (ii) and (iii),
we can say that A and B can not be equal for any value of y.
10. We may write the given matrix as
A = IA
2 0 −1 1 0 0
⎡ ⎤ ⎡ ⎤
∴ ⎢5 1 0⎥ = ⎢0 1 0⎥A
⎣ ⎦ ⎣ ⎦
0 1 3 0 0 1

[Applying R2 → R2 - (5/2) R1]

2 0 −1 1 0 0
⎡ ⎤ ⎡ ⎤

⇒ ⎢0 1 5/2 ⎥ = ⎢ −5/2 1 0⎥A

⎣ ⎦ ⎣ ⎦
0 1 3 0 0 1

[Applying R3 → R3 - R2]
2 0 −1 1 0 0
⎡ ⎤ ⎡ ⎤

⇒ ⎢0 1 5/2 ⎥ = ⎢ −5/2 1 0⎥A

⎣ ⎦ ⎣ ⎦
0 0 1/2 5/2 −1 1

[Applying R2 → R2 - 5R3]
2 0 −1 1 0 0
⎡ ⎤ ⎡ ⎤

⇒ ⎢0 1 0 ⎥ = ⎢ −15 6 −5 ⎥ A

⎣ ⎦ ⎣ ⎦
0 0 1/2 5/2 −1 1

[Applying R1 → R1 + 2R3]
2 0 0 6 −2 2
⎡ ⎤ ⎡ ⎤

⇒ ⎢0 1 0 ⎥ = ⎢ −15 6 −5 ⎥ A

⎣ ⎦ ⎣ ⎦
0 0 1/2 5/2 −1 1

[Applying R1 → 1

2
R1 and R3 → 2R3]
1 0 0 3 −1 1
⎡ ⎤ ⎡ ⎤
⇒ ⎢0 1 0 ⎥ = ⎢ −15 6 −5 ⎥ A
⎣ ⎦ ⎣ ⎦
0 0 1 −5 −2 2

3 −1 1
⎡ ⎤

Hence, ⎢ −15 6 −5 ⎥ is the inverse of given matrix A.

⎣ ⎦
−5 −2 2

Section D
11. i. Number of items purchased by shopkeepers A, B and C can be written in matrix form as
N otebooks pens pencils

144 60 72 A
⎡ ⎤

X= ⎢ 120 72 84 ⎥ B
⎣ ⎦
132 156 96 C

40 N ote book
⎡ ⎤

ii. Since, Y = ⎢ 12 ⎥ P en
⎣ ⎦
3 P encil

144 60 72 40
⎡ ⎤⎡ ⎤

∴ XY = ⎢ 120 72 84 ⎥ ⎢ 12 ⎥
⎣ ⎦⎣ ⎦
132 156 96 3

5760 + 720 + 216 6696

⎡ ⎤ ⎡ ⎤

=⎢ 4800 + 864 + 252 ⎥ = ⎢ 5916 ⎥

⎣ ⎦ ⎣ ⎦
5280 + 1872 + 288 7440

iii. (A + I)2 = A2 + 2A + I = 3A + I
⇒ (A + I)3 = (3A + I) (A + I)
= 3A2 + 4A + I = 7A + I

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 ∴(A + I)3 - 7A = I
OR
A2 - B2 = (A - B) (A + B) = A2 + AB - BA - B2
∴ AB = BA
Section E
–
cos α + sin α √2 sin α
12. Given A = [ – ] .
− √2 sin α cos α − sin α
–
cos nα + sin nα √2 sin nα
We need to prove that An = [ – ] .
− √2 sin nα cos nα − sin nα

We will prove this result using the principle of mathematical induction.

Step 1: When n = 1, we have An = A1
–
cos α + sin α √2 sin α
n
⇒ A = [ ]
–
− √2 sin α cos α − sin α

∴ An = A
Hence, the equation is true for n = 1.
Step 2: Let us assume the equation true for some n = k, where k is a positive integer.
–
cos kα + sin kα √2 sin kα
k
⇒ A = [ – ]
− √2 sin kα cos kα − sin kα

To prove the given equation is true using mathematical induction, we have to show that the equation is true for n = k+1.Which in
turns proves that the given equation is true for n=k, which was our assumption.
–
cos(k + 1)α + sin(k + 1)α √2 sin(k + 1)α
Ak+1 = [ – ] .
− √2 sin(k + 1)α cos(k + 1)α − sin(k + 1)α

We know Ak+1 = Ak × A.
– –
cos kα + sin kα √2 sin kα cos α + sin α √2 sin α
k+1
⇒ A = [ – ][ – ]
− √2 sin kα cos kα − sin kα − √2 sin α cos α − sin α

We evaluate each value of this matrix independently.

(a) The value at index (1, 1)
– –
A
11
= (cos kα + sin kα)(cos α + sin α) + (√2 sin kα)(−√2 sin α)
k+1

⇒A = cos kα cos α + cos kα sin α + sin kα cos α + sin kα sin α −2 sin kα sin α
k+1

11

k+1
⇒ A = cos kα cos α + cos kα sin α + sin kα cos α − sin kα sin α
11

k+1
⇒ A = cos kα cos α − sin kα sin α + sin kα cos α + cos kα sin α
11
k+1
⇒ A = cos(kα + α) + sin(kα + α)
11
k+1
∴ A = cos(k + 1)α + sin(k + 1)α
11

(b) The value at index (1, 2)

k+1 – –
A = (cosk α + sin kα)(√2 sin α) + (√2 sin kα)(cos α − sin α)
12
k+1 – – – –
⇒ A = √2 cos kα sin α + √2 sin kα sin α + √2 sin kα cos α − √2 sin kα sin α
12
k+1 – –
⇒ A = √2 cos kα sin α + √2 sin kα cos α
12
k+1 –
⇒ A = √2(cos kα sin α + sin kα cos α)
12

k+1 –
⇒ A = √2(sin kα cos α + cos kα sin α)
12

k+1 –
⇒ A = √2 sin(kα + α)
12
k+1 –
∴ A = √2 sin(k + 1)α
12

(c) The value at index (2, 1)

k+1 – –
A = (− √2 sin kα)(cos α + sin α) + (cos kα − sin kα)(− √2 sin α)
21

k+1 – – – –
⇒ A = − √2 sin kα cos α − √2 sin kα sin α − √2 cos kα sin α + √2 sin kα sin α
21

k+1 – –
⇒ A = − √2 sin kα cos α − √2 cos kα sin α
21
k+1 –
⇒ A = − √2(sin kα cos α + cos kα sin α)
21
k+1 –
⇒ A = − √2 sin(kα + α)
21

k+1 –
∴ A = − √2 sin(k + 1)α
21

(d) The value at index (2, 2)

k+1
⇒ A = −2 sin kα sin α + cos kα cos α − cos kα sin α − sin kα cos α + sin kα sin α
22
k+1
⇒ A = − sin kα sin α + cos kα cos α − cos kα sin α − sin kα cos α
22
k+1
⇒ A = cos kα cos α − sin kα sin α − sin kα cos α − cos kα sin α
22

k+1
⇒ A = cos kα cos α − sin kα sin α − (sin kα cos α + cos kα sin α)
22

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 k+1
⇒ A = cos(kα + α) − sin(kα + α)
22

k+1
∴ A = cos(k + 1)α − sin(k + 1)α
22

So, the matrix Ak+1 is

–
cos(k + 1)α + sin(k + 1)α √2 sin(k + 1)α
Ak+1 = [ – ]
− √2 sin(k + 1)α cos(k + 1)α − sin(k + 1)α

Hence, the equation is true for n = k + 1 under the assumption that it is true for n = k.
Therefore, by the principle of mathematical induction, the equation is true for all positive integer values of n.
–
cos nα + sin nα √2 sin nα
Thus, An = [ – ] for all n ∈ N.
− √2 sin nα cos nα − sin nα

13. L.H.S. = A3 - 6A2 + 7A + 2I

1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 2 1 0 0
⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢0 2 1⎥⎢0 2 1⎥⎢0 2 1 ⎥ −6 ⎢ 0 2 1⎥⎢0 2 1⎥ + 7⎢0 2 1 ⎥ +2 ⎢ 0 1 0⎥

⎣ ⎦⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
2 0 3 2 0 3 2 0 3 2 0 3 2 0 3 2 0 3 0 0 1

1 + 0 + 4 0 + 0 + 0 2 + 0 + 6 1 0 2 1 + 0 + 4 0 + 0 + 0 2 + 0 + 6
⎡ ⎤⎡ ⎤ ⎡ ⎤

= ⎢0 + 0 + 2 0 + 4 + 0 0 + 2 + 3 ⎥⎢ 0 2 1 ⎥−6 ⎢ 0 + 0 + 2 0 + 4 + 0 0 + 2 + 3⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦
2 + 0 + 6 0 + 0 + 0 4 + 0 + 9 2 0 3 2 + 0 + 6 0 + 0 + 0 4 + 0 + 9

7 0 14 2 0 0
⎡ ⎤ ⎡ ⎤

+⎢ 0 14 7 ⎥ + ⎢0 2 0⎥
⎣ ⎦ ⎣ ⎦
14 0 21 0 0 2

5 + 0 + 16 0 + 0 + 0 10 + 0 + 24 30 0 48 7 + 2 0 + 0 14 + 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= ⎢ 2 + 0 + 10 0 + 8 + 0 4 + 4 + 15 ⎥ − ⎢ 12 24 30 ⎥ + ⎢ 0 + 0 14 + 2 7 + 0 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
8 + 0 + 26 0 + 0 + 0 16 + 0 + 39 48 0 78 14 + 0 0 + 0 21 + 2

21 0 34 30 0 48 9 0 14 21 − 30 0 − 0 34 − 48 9 0 14
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ 12 8 23 ⎥ − ⎢ 12 24 30 ⎥ + ⎢ 0 16 7 ⎥ = ⎢ 12 − 12 8 − 24 23 − 30 ⎥ + ⎢ 0 16 7 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
34 0 55 48 0 78 14 0 23 34 − 48 0 − 0 55 − 78 14 0 23

−9 0 −14 9 0 14 −9 + 9 0 + 0 −14 + 14
⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= ⎢ 0 −16 −7 ⎥ + ⎢ 0 16 7 ⎥= ⎢ 0 + 0 −16 + 16 −7 + 7 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−14 0 −23 14 0 23 −14 + 14 0 + 0 −23 + 23

0 0 0
⎡ ⎤
= ⎢0 0 0⎥ = 0 (Zero matrix)
⎣ ⎦
0 0 0

= R.H.S. Proved.

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