Solution

MATRICES AND DETERMINANT TEST

Class 12 - Mathematics
Section A
1
1
1. (a) [ 2
]
2 1

Explanation:
a11 a12
In general, the matrix A of order 2 × 2 is given by A = [ ] ,
a21 a22
i
Now, aij = ; i = 1, 2 and j = 1, 2
j

∴ a11 = 1, a 12 =
1

2
, a21 = 2; a22 = 1
1
1
Thus, matrix A is [ 2
]
2 1

2.
π
(d) 3

Explanation:
cos α − sin α
Given A = [ ]
sin α cos α

cos α sin α
Therefore, A ′
= [ ]
− sin α cos α

Also given that A + A' = I ...(1)

(Putting the values in equation (1))
cos α − sin α cos α sin α 1 0
[ ]+ [ ]= [ ]
sin α cos α − sin α cos α 0 1

cos α + cos α − sin α + sin α 1 0

⇒ [ ]= [ ]
sin α − sin α cos α + cos α 0 1

2 cos α 0 1 0
⇒ [ ]= [ ]
0 2 cos α 0 1

We know the two matrices are equal only when all their corresponding elements or entries are equal i.e. if A = B, then aij and
bij for all i and j.
This implies,
2 cos α = 1
1
⇒ cos α =
2

π π 1
⇒ cos α = cos . . . (∵ cos = )
3 3 2

π
⇒ α =
3

3.
(d) A is a zero matrix
Explanation:
If a matrix A is both symmetric and skew-symmetric,
A’ = A & A’ = -A
Comparing both the equations,
A = -A
A+A=0
2A = 0
A=0
then A is a zero matrix.

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4.
(b) x = 2
Explanation:
0 1 −2
⎡ ⎤

Given, A = ⎢ −1 0 3⎥
⎣ ⎦
x −3 0

We know that, if A is a skew-symmetric matrix, then

A = -AT ...(i)
From Eq. (i) We, get
0 1 −2 0 −1 x
⎡ ⎤ ⎡ ⎤

⎢ −1 0 3⎥ = −⎢ 1 0 −3 ⎥
⎣ ⎦ ⎣ ⎦
x −3 0 −2 3 0

0 1 −2 0 1 −x
⎡ ⎤ ⎡ ⎤
⇒ ⎢ −1 0 3 ⎥ = ⎢ −1 0 3⎥
⎣ ⎦ ⎣ ⎦
x −3 0 2 −3 0

On comparing the corresponding element, we get

-2 = -x ⇒ x = 2

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

6.
(c) (-6, 11)
Explanation:
(-6, 11)

7.
(b) 26
Explanation:
|A| = d
|adj A| = |A|n - 1
Here, n = 3, |A| = 8
|adj A| = 82
|adj A| = (23)2 = 26

8.
−1
x 0 0
⎡ ⎤

(d) ⎢ 0 y
−1
0 ⎥

⎣ −1 ⎦
0 0 z

Explanation:
x 0 0
⎡ ⎤

Here, A = ⎢ 0 y 0⎥
⎣ ⎦
0 0 z

Clearly, we can see that

yz 0 0
⎡ ⎤

adjA= ⎢ 0 xz 0 ⎥ and |A| = xyz

⎣ ⎦
0 0 xy

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 yz 0 0
⎡ ⎤
adjA 1
−1
∴ A = = ⎢0 xz 0 ⎥
|A| xyz

⎣ ⎦
0 0 xy
−1
x 0 0
⎡ ⎤
−1
= ⎢0 y 0 ⎥
⎣ −1 ⎦
0 0 z

9.
(b) |A| = 22|B|
Explanation:
2 2 1 1
Let A = [ ] and B = [ ]
4 0 2 0

Now, |A| = 0 - 8 = -8
and |B| = 0 - 2 = -2
Observe that |A| = 4(-2) = 22 |B|

10. (a) Idempotent

Explanation:
clearly for given matrix A 2
= A

Therefore idempotent
4 2 −2 1
11. Given matrices are, A = [ ] and B = [ ]
1 3 3 2

We have 3A – 2B + X = 0
So X = -(3A – 2B)
Thus,
4 2 −2 1
X = −3 [ ]− 2[ ]
1 3 3 2

3 × 4 + 2 × 2 3 × 2 − 2 × 1
X = −[ ]
3 × 1 − 2 × 3 3 × 3 − 2 × 2

−16 −4
X=[ ]
3 −5

2 4 1 3
12. A + 2B = [ ] + 2 ([ ])
3 2 −2 5

2 4 2 6
= [ ]+ [ ]
3 2 −4 10

4 10
= [ ]
−1 12

13. |A⋅ adj A| = |A|n = 4n where n is order of matrix

14. We know that, for a non-singular square matrix of order n
|adj (A) | = |A| n- 1
Here, the order of A is 3 × 3, therefore n = 3
Hence, |adj (A) |= |A|2 ...(i)
But |adj (A)| = |A| [given]....(ii)
From Eqs. (i) and (ii), we get,
k=2
Section B
a11 a12 a13
⎛ ⎞

15. A= ⎜ a 21 a22 a23 ⎟ …......(1)

⎝ ⎠
a31 a32 a33

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 b11 b12
⎛ ⎞

B = ⎜b 21 b22 ⎟ …......(2)
⎝ ⎠
b31 b32

2 3 −5 2 −1
⎛ ⎞ ⎛ ⎞

Given, A = [aij] = ⎜ 1 4 9 ⎟ B = [bij] = ⎜ −3 4 ⎟

⎝ ⎠ ⎝ ⎠
0 7 −2 1 2

Now, Comparing with equation (1) and (2)

a11 = 2, a22 = 4, b11 = 2, b22 = 4
a11 b11 + a22 b22 = 2 × 2+ 4 × 4 = 4 + 16 = 20
3 1 3 1 8 5
16. A2
= [ ][ ] = [ ]
−1 2 −1 2 −5 3

15 5 7 0
5A = [ ] , 71 = [ ]
−5 10 0 7

0 0
⇒ A2 - 5A + 7I = [ ]= 0
0 0

⇒ A-1(A2 - 5A + 7) = A-1 0
⇒ A - 5t + 7A-1 = 0
⇒ 7A-1 = 5I - A
1
5 0 3 1
−1
⇒ A = ([ ]− [ ])
7
0 5 −1 2

1
2 −1
−1
⇒ A = [ ]
7
1 3

0 0
17. Given: A = [ ]
4 0

We will find A2,

0 0 0 0
A2 = A × A = [ ][ ]
4 0 4 0

0 + 0 0 + 0
2
⇒ A = [ ]
0 + 0 0 + 0

⇒ A2 = 0
Hence, A16 = (A2)8 = (0)8 = 0
Hence A16 is a null matrix.
18. Since, A and B are square matrices of same order and B is a skew-symmetric matrix
∴ B' = -B .....(1)

Now, we have to prove that A’BA is a skew-symmetric matrix.

(A'BA)' = [A'(BA)]'=(BA)'(A')' ...[∴(AB)'=B'A']
= (A'B')A. [∴ (A')'=A]
= A'(-B)A [using (1)]
= -(A'BA)
Hence, A'BA is a skew-symmetric matrix.
19. Given,
1 2 2
⎛ ⎞
A= ⎜ 2 1 x ⎟
⎝ ⎠
−2 2 −1

1 2 −2
⎛ ⎞
′
A = ⎜2 1 2 ⎟

⎝ ⎠
2 x −1

Since, A A' = 9I
1 2 2 1 2 −2 1 0 0
⎛ ⎞⎛ ⎞ ⎛ ⎞

⎜ 2 1 x ⎟⎜2 1 2 ⎟ = 9⎜0 1 0⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
−2 2 −1 2 x −1 0 0 1

1 + 4 + 4 2 + 2 + 2x −2 + 4 − 2
⎛ ⎞
2
⎜ 2 + 2 + 2x 4 + 1 + x −4 + 2 − x ⎟
⎝ ⎠
−2 + 4 − 2 −4 + 2 − x 4 + 4 + 1

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 9 0 0
⎛ ⎞

= ⎜0 9 0⎟
⎝ ⎠
0 0 9

9 2x + 4 0 9 0 0
⎛ ⎞ ⎛ ⎞
2
⎜ 2x + 4 x + 5 −x − 2 ⎟ = ⎜ 0 9 0⎟
⎝ ⎠ ⎝ ⎠
0 −x − 2 9 0 0 9

Now, x2 + 5 = 9
x2 = 9 - 5
x2 = 4
–
x = √4
x = ±2
Also, 2x + 4 = 0
2x = -4
4
x=− 2

So, x = -2
20. By using the definition, on expanding the given determinant with respect to C​​​1
∣ 2 3 −2 ∣

D=∣ 1 2 3
∣
∣ ∣
∣ −2 1 −3 ∣

∣2 3∣ ∣3 −2 ∣ ∣3 −2 ∣
⇒ D = (- 1)1+1 (2) ∣ ∣ + (-1)2+1 (1) ∣ ∣ + (-1)3+1 (- 2) ∣ ∣
∣1 −3 ∣ ∣1 −3 ∣ ∣2 3∣

∣2 3∣ ∣3 −2 ∣ ∣3 −2 ∣
⇒ D= 2∣ ∣ − ∣ ∣ − 2∣ ∣
∣1 −3 ∣ ∣1 −3 ∣ ∣2 3∣

⇒ D = 2 (- 6 - 3) - (- 9 + 2) - 2 (9 + 4) = -18 + 7 - 26 = - 37
21. Matrix form of given equations is AX = B
1 −1 1 x 4
⎡ ⎤⎡ ⎤ ⎡ ⎤

⇒ ⎢2 1 −3 ⎥ ⎢ y ⎥ = ⎢ 0 ⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦
1 1 1 z 2

1 −1 1 x 4
⎡ ⎤ ⎡ ⎤ ⎡ ⎤

Here A = ⎢ 2 1 −3 ⎥ , X = ⎢ y ⎥ and B = ⎢ 0 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
1 1 1 z 2

∣1 −1 1 ∣
∣ ∣
∴ |A| = 2 1 −3
∣ ∣
∣1 1 1 ∣

= 1(1 + 3) - ( - 1)(2 + 3) + 1(2 - 1)

= 4 + 5 + 1 = 10 ≠ 0

Therefore, solution is unique and X = A −1

B=
1
(adj. A) B
|A|

x 4 2 2 4
⎡ ⎤ ⎡ ⎤⎡ ⎤
1
⇒ ⎢ y ⎥ = ⎢ −5 0 5⎥⎢0⎥
10
⎣ ⎦ ⎣ ⎦⎣ ⎦
z 1 2 3 2

16 + 0 + 4
⎡ ⎤
1
= ⎢ −20 + 0 + 10 ⎥
10
⎣ ⎦
4 + 0 + 6

20 2
⎡ ⎤ ⎡ ⎤
1
= ⎢ −10 ⎥ = ⎢ −1 ⎥
10
⎣ ⎦ ⎣ ⎦
10 1

Therefore, x = 2, y = - 1 and z = 1
22. Given system of equations:
ax + y + z = 0
x + by + z = 0
x + y + cz = 0
and, It is given that x, y, z are not all zero. This means that there are non-trivial solutions of the given system of equations.

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 ∣a 1 1∣
∣ ∣
∴
∣
1 b 1
∣
= 0 ⇒ abc - a - c - b + 2 = 0 ⇒ abc = a + b + c - 2 ...(i)
∣1 1 c ∣
(1−b)(1−c)+(1−c)(1−a)+(1−a)(1−b)
Now, 1

1−a
+
1−b
1
+
1−c
1
=
(1−a)(1−b)(1−c)

3−2(a+b+c)+(ab+bc+ca)
=
1−(a+b+c)+(ab+bc+ca)−abc

3−2(a+b+c)+(ab+bc+ca)
= [Using (i)]
1−(a+b+c)+(ab+bc+ca)−(a+b+c)+2

3−2(a+b+c)+(ab+bc+ca)
= =1
3−2(a+b+c)+(ab+bc+ca)

2 1
23. Given: A = [ ]
5 3

⇒ |A| = 1
3 −1
Adj A = [ ]
−5 2

3 −1
A–1=
adj A 1
= [ ]
|A| 1
−5 2

4 5
and, B = [ ]
3 4

|B| = – 1
4 −5
B–1=
adj A 1
= [ ]
|A| −1
−3 4

2 1 4 5 11 14
Also, AB = [ ][ ]= [ ]
5 3 3 4 29 37

|AB| = 407 – 406 = 1

37 −14
And, adj(AB) = [ ]
−29 11

adj AB 37 −14
(AB) – 1 = |AB|
=[ ]
−29 11

4 −5 3 −1
Now, B – 1A – 1 = [ ][ ]
−3 4 −5 2

37 −14
=[ ]
−29 11

Hence, (AB) – 1 = B – 1A – 1
24. The given matrix is
6 − x 4
[ ]
3 − x 1

We need to find the value of x such that the matrix is singular.

∣6 − x 4∣
i.e, ∣ ∣ = 0
∣3 − x 1∣

⇒ (6 - x) × 1 - 4 × (3 - x) = 0
⇒ (6 - x) - (12 - 4x) = 0
⇒ 6 - x - 12 + 4x = 0

⇒ 4x - x - 12 + 6 = 0

⇒ 3x - 6 = 0

⇒ x =
6

⇒ x=2
Thus, the value of x = 2 for which the matrix is singular. x = 2
Section C
−1 2 3 −4 1 −5
⎡ ⎤ ⎡ ⎤

25. A = ⎢ 5 7 9⎥ and B = ⎢ 1 2 0 ⎥
⎣ ⎦ ⎣ ⎦
−2 1 1 1 3 1

(A+B)’ = A’+B’
Explanation: We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are
equal.

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 −1 2 3 −4 1 −5
⎡ ⎤ ⎡ ⎤

So, A + B = ⎢ 5 7 9⎥ + ⎢ 1 2 0 ⎥
⎣ ⎦ ⎣ ⎦
−2 1 1 1 3 1

−1 + (−4) 2 + 1 3 + (−5)
⎡ ⎤

⇒ A+ B= ⎢ 5 + 1 7 + 2 9 + 0 ⎥

⎣ ⎦
−2 + 1 1 + 3 1 + 1

−5 3 −2
⎡ ⎤

⇒ A+ B= ⎢ 6 9 9 ⎥
⎣ ⎦
−1 4 2

−5 6 −1
⎡ ⎤
Therefore, (A + B) ′
= ⎢ 3 9 4 ⎥ ...(1)
⎣ ⎦
−2 9 2

−1 5 −2 −4 1 1
⎡ ⎤ ⎡ ⎤

Noe, A ′
= ⎢ 2 7 1 ⎥ and B ′
= ⎢ 1 2 3⎥

⎣ ⎦ ⎣ ⎦
3 9 1 −5 0 1

−1 5 −2 −4 1 1
⎡ ⎤ ⎡ ⎤

So, A' + B' = ⎢ 2 7 1 ⎥ + ⎢ 1 2 3⎥

⎣ ⎦ ⎣ ⎦
3 9 1 −5 0 1

−1 + (−4) 5 + 1 −2 + 1
⎡ ⎤
′ ′
⇒ A + B = ⎢ 2 + 1 7 + 2 1 + 3 ⎥

⎣ ⎦
3 + (−5) 9 + 0 1 + 1

−5 6 −1
⎡ ⎤
′
⇒ A + B = ⎢
′
3 9 4 ⎥ ...(2)
⎣ ⎦
−2 9 2

From equation (1) &( 2), we see that

(A+B)’ = A’+B’. Hence verified.
2 3 1
⎡ ⎤

26. The given matrix is ⎢ 3 4 1⎥

⎣ ⎦
3 7 2

∣4 1∣ ∣3 1∣ ∣3 4∣
|A| = 2 ∣ ∣ − 3∣ ∣ + 1∣ ∣
∣7 2∣ ∣3 2∣ ∣3 7∣

= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)

=2–9+9
=2
Hence, A – 1 exists
Cofactors of A are:
C11 = 1 C21 = 1 C31 = – 1
C12 = – 3 C22 = 1 C32 = 1
C13 = 9 C23 = – 5 C33 = – 1
T
C11 C12 C13
⎡ ⎤

adj A = ⎢ C 21 C22 C23 ⎥

⎣ ⎦
C31 C32 C33
T
1 −3 9
⎡ ⎤

=⎢ 1 1 −5 ⎥
⎣ ⎦
−1 1 −1

1 1 −1
⎡ ⎤
So, adj A = ⎢ −3 1 1 ⎥
⎣ ⎦
9 −5 −1

1 1 −1
⎡ ⎤

Now, A – 1 =
1
⎢ −3 1 1 ⎥
2
⎣ ⎦
9 −5 −1

1 1 −1 2 3 1
⎡ ⎤⎡ ⎤

Also, A – 1.A = 1

2
⎢ −3 1 1 ⎥⎢3 4 1⎥
⎣ ⎦⎣ ⎦
9 −5 −1 3 7 2

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 2 + 3 − 3 3 + 4 − 7 1 + 1 − 2
⎡ ⎤

1 ⎢ −6 + 3 + 3 −9 + 4 + 7 −3 + 1 + 2 ⎥
⎣ ⎦
18 − 15 − 3 27 − 20 − 7 9 − 5 − 2

2 0 0 1 0 0
⎡ ⎤ ⎡ ⎤
1
⎢0 2 0⎥ = ⎢0 1 0⎥
2
⎣ ⎦ ⎣ ⎦
0 0 2 0 0 1

Hence, A – 1.A = I
Section D
27. We have,
2 3
2X + 3Y = [ ] ....(i)
4 0

−2 2
And 3X + 2Y = [ ] ...(ii)
1 −5

On subtracting Eq. (i) from Eq. (ii), we get

−2 − 2 2 − 3
∴ (3X + 2Y ) − (2X + 3Y ) = [ ]
1 − 4 −5 − 0

−4 −1
(X − Y ) = [ ] ....(iii)
−3 −5

On adding Eqs. (i) and (ii), we get

0 5
(5X + 5Y ) = [ ]
5 −5

0 5 0 1
⇒ (X + Y ) =
1

5
[ ]= [ ] ....(iv)
5 −5 1 −1

On adding Eqs. (iii) and (iv), we get

−4 0
⇒ (X − Y )(X + Y ) = [ ]
−2 −6

−2 0
⇒ 2X = 2 [ ]
−1 3

−2 0
∴ X = [ ]
−1 3

From Eq. (iv),

−2 0 0 1
[ ]+ Y = [ ]
−1 −3 1 −1

0 1 −2 0 0 + 2 1 − 0 2 1
∴ Y = [ ]− [ ] = [ ]= [ ]
1 −1 −1 −3 1 + 1 −1 + 3 2 2

28. Here, we have:

1 0 −2
⎡ ⎤

A = ⎢ −2 −1 2 ⎥
⎣ ⎦
3 4 1

A3 = A2.A
1 0 −2 1 0 −2
⎡ ⎤⎡ ⎤
2
A = ⎢ −2 −1 2 ⎥ ⎢ −2 −1 2 ⎥
⎣ ⎦⎣ ⎦
3 4 1 3 4 1

1 + 0 − 6 0 + 0 − 8 −2 + 0 − 2 −5 −8 −4
⎡ ⎤ ⎡ ⎤
= ⎢ −2 + 2 + 6 0 + 1 + 8 4 − 2 + 2 ⎥ = ⎢ 6 9 4 ⎥
⎣ ⎦ ⎣ ⎦
3 − 8 + 3 0 − 4 + 4 −6 + 8 + 1 −2 0 3

−5 −8 −4 1 0 −2
⎡ ⎤⎡ ⎤

A2.A =⎢ 6 9 4 ⎥ ⎢ −2 −1 2 ⎥
⎣ ⎦⎣ ⎦
−2 0 3 3 4 1
2
−5 + 16 − 12 0 − 8 + 16 10 − 16 − 4
⎡ ⎤

=⎢ 6 − 18 + 12 0 − 9 + 16 −12 + 18 + 4 ⎥
⎣ ⎦
−2 − 0 + 9 0 − 0 − 12 4 + 0 + 3

−1 −8 −10
⎡ ⎤
=⎢ 0 7 10 ⎥
⎣ ⎦
7 12 7

Now, A3 – A2 – 3A – I

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 −1 −8 −10 −5 −8 −4 1 0 −2 1 0 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

=⎢ 0 7 10 ⎥ − ⎢ 6 9 4 ⎥ − 3 ⎢ −2 −1 2 ⎥ − ⎢0 1 0⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
7 12 7 −2 0 3 3 4 1 0 0 1

−1 + 5 −8 + 8 −10 + 4 −3 − 1 −0 − 0 6 − 0
⎡ ⎤ ⎡ ⎤

=⎢ 0 − 6 7 − 9 10 − 4 ⎥ + ⎢ 6 − 0 +3 − 1 −6 − 0 ⎥
⎣ ⎦ ⎣ ⎦
7 + 2 12 − 0 7 − 3 −9 − 0 −12 + 0 −3 − 1

4 0 −6 −4 0 6
⎡ ⎤ ⎡ ⎤
= ⎢ −6 −2 6 ⎥ + ⎢ 6 2 −6 ⎥
⎣ ⎦ ⎣ ⎦
9 12 4 −9 −12 −4

0 0 0
⎡ ⎤

= ⎢0 0 0⎥

⎣ ⎦
0 0 0

Thus, A3 – A2 – 3A – I = 0
Multiply both sides by A −1
, we get
−1 3 −1 2 −1 −1
A A –A A – 3A A– I A = 0

A2 – A – 3I = A −1
...(since A −1
A= I )
⇒ A–1= (A2 - A - 3I)
−5 −8 −4 1 0 −2 1 0 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤

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

−5 −8 −4 1 0 −2 3 0 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
= ⎢ 6 9 4 ⎥ − ⎢ −2 −1 2 ⎥ − ⎢0 3 0⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−2 0 3 3 4 1 0 0 3

−5 − 1 − 3 −8 − 0 − 0 −4 + 2 − 0
⎡ ⎤

=⎢ 6 + 2 − 0 7 + 1 − 3 4 − 2 − 0 ⎥

⎣ ⎦
−2 − 3 − 0 0 − 4 − 0 3 − 1 − 3

−9 −8 −2
⎡ ⎤

=⎢ 8 7 2 ⎥
⎣ ⎦
−5 −4 −1

−9 −8 −2
⎡ ⎤
Hence, A – 1 = ⎢ 8 7 2 ⎥
⎣ ⎦
−5 −4 −1

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