MATRICES

Exercise 3.1

Q1:
In the matrix
[ 2  5  19  –7
A =   35 –2  5/2  12
  √3  1  –5  17 ], write:

(i) The order of the matrix
(ii) The number of elements
(iii) The elements a13, a21, a33, a24, a23

Solution:

The given matrix has 3 rows and 4 columns.

(i) Order of the matrix:
Number of rows × Number of columns = 3 × 4

(ii) Number of elements:
Total elements = 3 × 4 = 12

(iii) Required elements:
a13 = element in 1st row, 3rd column = 19
a21 = element in 2nd row, 1st column = 35
a33 = element in 3rd row, 3rd column = –5
a24 = element in 2nd row, 4th column = 12
a23 = element in 2nd row, 3rd column = 5/2

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Q2. If a matrix has 24 elements, what are the possible orders it can have? What if it has 13 elements?

Solution:
An order of a matrix is (rows) × (columns). Possible orders are all factor pairs of the number of elements.

• For 24 elements, factor pairs of 24 are:
  1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1.
  (Usually we list unique row×column pairs: 1×24, 2×12, 3×8, 4×6, 6×4, 8×3, 12×2, 24×1.)

• For 13 elements (13 is prime), the only factor pairs are:
  1 × 13 and 13 × 1.

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Q3. If a matrix has 18 elements, what are the possible orders it can have? What if it has 5 elements?

Solution:

• For 18 elements, factor pairs of 18 are:
  1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1.

• For 5 elements (5 is prime), the only orders are:
  1 × 5 and 5 × 1.

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Q 4. Construct a 2 × 2 matrix A = [aij], whose elements are given by:

(i) aij = (i + j)² / 2
(ii) aij = i / j
(iii) aij = (i + 2j)² / 2

(Here i and j take values 1 and 2.)

Solution:

(i) aij = (i + j)² / 2

• a11 = (1 + 1)² / 2 = 4 / 2 = 2

• a12 = (1 + 2)² / 2 = 9 / 2

• a21 = (2 + 1)² / 2 = 9 / 2

• a22 = (2 + 2)² / 2 = 16 / 2 = 8

Matrix A = [ 2  9/2
9/2  8 ]

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(ii) aij = i / j

• a11 = 1 / 1 = 1

• a12 = 1 / 2 = 1/2

• a21 = 2 / 1 = 2

• a22 = 2 / 2 = 1

Matrix A =
[ 1  1/2
2   1 ]

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(iii) aij = (i + 2j)² / 2

• a11 = (1 + 2·1)² / 2 = 3² / 2 = 9/2

• a12 = (1 + 2·2)² / 2 = 5² / 2 = 25/2

• a21 = (2 + 2·1)² / 2 = 4² / 2 = 16/2 = 8

• a22 = (2 + 2·2)² / 2 = 6² / 2 = 36/2 = 18

Matrix A =
[ 9/2 25/2
8   18 ]

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Q5

Construct a 3 × 4 matrix whose elements are given by
(i) aᵢⱼ = (1/2) × | -3i + j |
(ii) aᵢⱼ = 2i − j
where i = 1,2,3 and j = 1,2,3,4.

Solution (i)

Formula: aᵢⱼ = (1/2)| -3i + j |

Row i = 1:
j = 1: -3 + 1 = -2 → | -2 | = 2 → 2/2 = 1
j = 2: -3 + 2 = -1 → | -1 | = 1 → 1/2
j = 3: -3 + 3 = 0 → 0/2 = 0
j = 4: -3 + 4 = 1 → 1/2

Row i = 2:
j = 1: -6 + 1 = -5 → 5/2
j = 2: -6 + 2 = -4 → 4/2 = 2
j = 3: -6 + 3 = -3 → 3/2
j = 4: -6 + 4 = -2 → 1

Row i = 3:
j = 1: -9 + 1 = -8 → 8/2 = 4
j = 2: -9 + 2 = -7 → 7/2
j = 3: -9 + 3 = -6 → 6/2 = 3
j = 4: -9 + 4 = -5 → 5/2

Matrix:
1 1/2 0 1/2
5/2 2 3/2 1
4 7/2 3 5/2

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Solution (ii)

Formula: aᵢⱼ = 2i − j

Row i = 1:
1, 0, -1, -2

Row i = 2:
3, 2, 1, 0

Row i = 3:
5, 4, 3, 2

Matrix:
1 0 -1 -2
3 2 1 0
5 4 3 2

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Q6 Find x, y, z.

(i)

4 = y → y = 4
3 = z → z = 3
x = 1
5 = 5 (true)

Answer: x = 1, y = 4, z = 3

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(ii)

x + y = 6
5 + z = 5 → z = 0
xy = 8

Possible Solutions: (x, y) = (2, 4) or (4, 2)

Answer: x = 2, y = 4, z = 0 (or x = 4, y = 2, z = 0)

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(iii)

x + y + z = 9
x + z = 5
y + z = 7

From x + z = 5, and y + z = 7:
Subtract from x + y + z = 9 → y = 4
Then 4 + z = 7 → z = 3
Then x + 3 = 5 → x = 2

Answer: x = 2, y = 4, z = 3

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Q7

Find a, b, c, d from:

a − b = −1
2a + c = 5
2a − b = 0
3c + d = 13

2a − b = 0 → b = 2a
Substitute in a − b = −1:
a − 2a = −1 → a = 1 → b = 2
2a + c = 5 → 2 + c = 5 → c = 3
3c + d = 13 → 9 + d = 13 → d = 4

Answer: a = 1, b = 2, c = 3, d = 4

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Q8:

A = [aij]m×n is a square matrix, if
(A) m < n
(B) m > n
(C) m = n
(D) None of these

Solution:
A square matrix has the same number of rows and columns. Therefore the condition is m = n.

Answer: (C) m = n

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Q9. Which of the given values of x and y make the following pair of matrices equal?

Left matrix:
[ 3x + 7 5
y + 1 2 − 3x ]

Right matrix:
[ 0 y − 2
8 4 ]

Options:
(A) x = −1/3, y = 7
(B) Not possible to find
(C) y = 7, x = −2/3
(D) x = −1/3, y = −2/3

Solution:

Two matrices are equal only if corresponding entries are equal. So equate entries:

1. Top-left: 3x + 7 = 0
   ⇒ 3x = −7 ⇒ x = −7/3.

2. Top-right: 5 = y − 2
   ⇒ y − 2 = 5 ⇒ y = 7.

3. Bottom-left: y + 1 = 8
   ⇒ y + 1 = 8 ⇒ y = 7 (consistent with step 2).

4. Bottom-right: 2 − 3x = 4
   ⇒ −3x = 2 ⇒ x = −2/3.

Compare results for x from steps 1 and 4: x would need to be both −7/3 and −2/3, which is impossible.

Therefore there is no single pair (x, y) that satisfies all four corresponding equalities. Hence the matrices cannot be made equal for any x, y.

Answer: (B) Not possible to find

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Q10. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512

Solution:
A 3 × 3 matrix has 9 entries. Each entry can be either 0 or 1, so there are 2 choices per entry. Total number of distinct matrices = 2^9 = 512.

Answer: (D) 512

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Exercise 3.2

Q1:

Let

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

Find each of the following:

(i) A + B  (ii) A − B  (iii) 3A − C  (iv) AB  (v) BA

Solution

(i) A + B

A + B = [ 2 4
    3 2 ] + [ 1 3
      −2 5 ]

= [ (2+1) (4+3)
 (3−2) (2+5) ]

= [ 3 7
 1 7 ]

(ii) A − B

A − B = [ 2 4
    3 2 ] − [ 1 3
       −2 5 ]

= [ (2−1) (4−3)
 (3−(−2)) (2−5) ]

= [ 1 1
 5 −3 ]

(iii) 3A − C

3A = 3 × [ 2 4
    3 2 ] = [ 6 12
     9 6 ]

Now,
3A − C = [ 6 12
     9 6 ] − [ −2 5
      3 4 ]

= [ (6−(−2)) (12−5)
  (9−3)  (6−4) ]

= [ 8 7
 6 2 ]

(iv) AB

AB = [ 2 4
  3 2 ] × [ 1 3
    −2 5 ]

= [ (2×1 + 4×(−2)) (2×3 + 4×5)
 (3×1 + 2×(−2)) (3×3 + 2×5) ]

= [ −6 26
  −1 19 ]

(v) BA

BA = [ 1 3
  −2 5 ] × [ 2 4
    3 2 ]

= [ (1×2 + 3×3) (1×4 + 3×2)
  (−2×2 + 5×3) (−2×4 + 5×2) ]

= [ 11 10
  11 2 ]

Final Answers

A + B = [ 3 7
    1 7 ]

A − B = [ 1 1
   5 −3 ]

3A − C = [ 8 7
   6 2 ]

AB = [ −6 26
   −1 19 ]

BA = [ 11 10
   11 2 ]

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Q2: Compute the following:

(i)
A =
| a  b |
| –b a |

B =
| a b |
| b a |

Find: A + B

Solution:
Add corresponding elements:

| a + a  b + b |
| –b + b  a + a |

=
| 2a 2b |
| 0 2a |

Answer:
A + B = | 2a 2b |
| 0 2a |

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(ii)
Add:

| a² + b² b² + c² |
| a² + c² a² + b² |
+
| 2ab 2bc |
| –2ac –2ab |

Solution:
| (a² + b² + 2ab) (b² + c² + 2bc) |
| (a² + c² – 2ac) (a² + b² – 2ab) |

=
| (a + b)² (b + c)² |
| (a – c)² (a – b)² |

Answer: | (a + b)² (b + c)² |
| (a – c)² (a – b)² |

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(iii)
Add the two matrices:

| –1 4 –6 | + | 12 7 6 |
| 8  5 16 | + | 8  0 5 |
| 2  8  5 | + | 3  2 4 |

Solution:

| (–1+12) (4+7) (–6+6) |
| (8+8)  (5+0)  (16+5) |
| (2+3)  (8+2)  (5+4) |

=
| 11 11 0 |
| 16 5 21 |
| 5 10 9 |

Answer:
| 11 11 0 |
| 16 5 21 |
| 5  10 9 |

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(iv)
Add:

| cos²x sin²x | + | sin²x cos²x |
| sin²x cos²x | + | cos²x sin²x |

Solution:
| (cos²x + sin²x) (sin²x + cos²x) |
| (sin²x + cos²x) (cos²x + sin²x) |

Since cos²x + sin²x = 1,

=
| 1 1 |
| 1 1 |

Answer:
| 1 1 |
| 1 1 |

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Q3: Compute the indicated products.

(i) Multiply
[ a b ]
[ −b a ]
by
[ a −b ]
[ b a ]

Solution (elementwise product):
Top-left = a·a + b·b = a² + b²
Top-right = a·(−b) + b·a = −ab + ab = 0
Bottom-left = (−b)·a + a·b = −ab + ab = 0
Bottom-right = (−b)(−b) + a·a = b² + a² = a² + b²

Answer:
[ a² + b² 0 ]
[ 0 a² + b² ]

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(ii) Multiply column [ 1 ] [ 2 ] [ 3 ] by row [ 2 3 4 ]

This is an outer product. Rows are multiples of the row vector:

Answer (3 × 3):
[ 1·2 1·3 1·4 ] = [ 2 3 4 ]
[ 2·2 2·3 2·4 ] = [ 4 6 8 ]
[ 3·2 3·3 3·4 ] = [ 6 9 12 ]

So the product is:
[ 2 3 4 ]
[ 4 6 8 ]
[ 6 9 12 ]

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(iii) Multiply
[ 1 −2 ]
[ 2 3 ]
by
[ 1 2 3 ]
[ 2 3 1 ]

Compute row×column:

Row1·Col1 = 1·1 + (−2)·2 = 1 − 4 = −3
Row1·Col2 = 1·2 + (−2)·3 = 2 − 6 = −4
Row1·Col3 = 1·3 + (−2)·1 = 3 − 2 = 1

Row2·Col1 = 2·1 + 3·2 = 2 + 6 = 8
Row2·Col2 = 2·2 + 3·3 = 4 + 9 = 13
Row2·Col3 = 2·3 + 3·1 = 6 + 3 = 9

Answer: (2 × 3 matrix)
[ −3 −4 1 ]
[ 8 13 9 ]

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(iv) Multiply
[ 2 3 4 ]
[ 3 4 5 ]
[ 4 5 6 ]
by
[ 1 −3 5 ]
[ 0 2 4 ]
[ 3 0 5 ]

Compute (row by column):

Row1:
(2·1 + 3·0 + 4·3) = 2 + 0 + 12 = 14
(2·(−3) + 3·2 + 4·0) = −6 + 6 + 0 = 0
(2·5 + 3·4 + 4·5) = 10 + 12 + 20 = 42

Row2:
(3·1 + 4·0 + 5·3) = 3 + 0 + 15 = 18
(3·(−3) + 4·2 + 5·0) = −9 + 8 + 0 = −1
(3·5 + 4·4 + 5·5) = 15 + 16 + 25 = 56

Row3:
(4·1 + 5·0 + 6·3) = 4 + 0 + 18 = 22
(4·(−3) + 5·2 + 6·0) = −12 + 10 + 0 = −2
(4·5 + 5·4 + 6·5) = 20 + 20 + 30 = 70

Answer (3 × 3):
[ 14 0 42 ]
[ 18 −1 56 ]
[ 22 −2 70 ]

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(v) Multiply
[ 2 1 ]
[ 3 2 ]
[ −1 1 ]
by
[ 1 0 1 ]
[ −1 2 1 ]

Compute:

Row1:
2·1 + 1·(−1) = 2 − 1 = 1
2·0 + 1·2 = 0 + 2 = 2
2·1 + 1·1 = 2 + 1 = 3

Row2:
3·1 + 2·(−1) = 3 − 2 = 1
3·0 + 2·2 = 0 + 4 = 4
3·1 + 2·1 = 3 + 2 = 5

Row3:
(−1)·1 + 1·(−1) = −1 − 1 = −2
(−1)·0 + 1·2 = 0 + 2 = 2
(−1)·1 + 1·1 = −1 + 1 = 0

Answer (3 × 3):
[ 1 2 3 ]
[ 1 4 5 ]
[ −2 2 0 ]

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(vi) Multiply
[ 3 −1 3 ]
[ −1 0 2 ]
by
[ 2 −3 ]
[ 1 0 ]
[ 3 1 ]

Compute:

Row1·Col1 = 3·2 + (−1)·1 + 3·3 = 6 − 1 + 9 = 14
Row1·Col2 = 3·(−3) + (−1)·0 + 3·1 = −9 + 0 + 3 = −6

Row2·Col1 = (−1)·2 + 0·1 + 2·3 = −2 + 0 + 6 = 4
Row2·Col2 = (−1)·(−3) + 0·0 + 2·1 = 3 + 0 + 2 = 5

Answer (2 × 2):
[ 14 −6 ]
[ 4 5 ]

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Q4: 4. If

A = [ 1  2  –3
   5  0  2
   1 –1 1 ], 
B = [ 3 –1 2
    4 2  5
    2 0  3 ], 
C = [ 4 1 2
    0 3 2
   1 –2 3 ],

then compute (A + B) and (B – C). Also, verify that
A + (B – C) = (A + B) – C.

Solution:

(i) A + B

A + B = [ (1+3) (2+–1) (–3+2)
    (5+4) (0+2) (2+5)
    (1+2) (–1+0) (1+3) ]

= [ 4 1 –1
  9 2 7
  3 –1 4 ]

(ii) B – C

B – C = [ (3–4) (–1–1) (2–2)
    (4–0) (2–3)  (5–2)
    (2–1) (0–(–2)) (3–3) ]

= [ –1 –2 0
  4 –1 3
  1 2  0 ]

(iii) A + (B – C)

A + (B – C) = [ (1+–1) (2+–2) (–3+0)
        (5+4) (0+–1)  (2+3)
        (1+1) (–1+2)  (1+0) ]

= [ 0 0 –3
  9 –1 5
  2 1 1 ]

(iv) (A + B) – C

(A + B) – C = [ (4–4) (1–1)  (–1–2)
        (9–0) (2–3)  (7–2)
        (3–1) (–1–(–2)) (4–3) ]

= [ 0 0 –3
  9 –1 5
  2 1 1 ]

Hence verified that
A + (B – C) = (A + B) – C

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Q5: If
A = [ 2/3 1  5/3
   1/3 2/3 4/3
   7/3 2  2/3 ]
and
B = [ 2/5 3/5 1
   1/5 2/5 4/5
   7/5 6/5 2/5 ],

compute 3A − 5B.

Work (multiply each matrix by scalar):

3A = [ 2 3 5
   1 2 4
   7 6 2 ]

5B = [ 2 3 5
   1 2 4
   7 6 2 ]

Therefore

3A − 5B = [ 2−2 3−3 5−5
      1−1 2−2 4−4
      7−7 6−6 2−2 ]
= [ 0 0 0
  0 0 0
  0 0 0 ].

Answer: 3A − 5B is the 3×3 zero matrix.

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Q6: 6. Simplify

cos θ [ cos θ sin θ
–sin θ cos θ ] + sin θ [ sin θ –cos θ
cos θ sin θ ]

Solution:

Multiply each matrix by its scalar:

= [ cos²θ  cos θ sin θ
 –sin θ cos θ  cos²θ ] + [ sin²θ  –sin θ cos θ
 sin θ cos θ  sin²θ ]

Now add corresponding elements:

= [ cos²θ + sin²θ  cos θ sin θ – sin θ cos θ
 –sin θ cos θ + sin θ cos θ  cos²θ + sin²θ ]

Simplify each term:

cos²θ + sin²θ = 1
cos θ sin θ – sin θ cos θ = 0
–sin θ cos θ + sin θ cos θ = 0

Hence the result is:

= [ 1 0
 0 1 ]

Final Answer:
The simplified matrix is the Identity Matrix,
[ 1 0; 0 1 ].

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Q7:
Find X and Y if

(i) X + Y = [ 7 0
      2 5 ] and X − Y = [ 3 0
          0 3 ]

(ii) 2X + 3Y = [ 2 3
       4 0 ] and 3X + 2Y = [ 2 –2
        –1 5 ]

Solution:

(i)

We have
X + Y = [ 7 0
   2 5 ]
X − Y = [ 3 0
    0 3 ]

Add both equations:
(X + Y) + (X − Y) = [ 7+3 0+0
         2+0 5+3 ]

⇒ 2X = [ 10 0
    2 8 ]

So,
X = ½ [ 10 0
    2 8 ]
= [ 5 0
  1 4 ]

Now subtract (X − Y) from (X + Y):
(X + Y) − (X − Y) = [ 7−3 0−0
         2−0 5−3 ]

⇒ 2Y = [ 4 0
    2 2 ]

So,
Y = ½ [ 4 0
   2 2 ]
= [ 2 0
 1 1 ]

Hence:
X = [ 5 0
  1 4 ], Y = [ 2 0
    1 1 ]

(ii)

2X + 3Y = [ 2 3
     4 0 ]
3X + 2Y = [ 2 –2
     –1 5 ]

To eliminate Y, multiply the first equation by 3 and the second by 2:

3(2X + 3Y) = [ 6 9
      12 0 ]
2(3X + 2Y) = [ 4 –4
      –2 10 ]

Subtract:
(6X + 9Y) – (6X + 4Y) = [ 6–4 9–(–4)
       12–(–2) 0–10 ]

⇒ 5Y = [ 2  13
    14 –10 ]

Y = 1/5 [ 2 13
    14 –10 ]

= [ 2/5 13/5
  4/5  –2 ]

Now substitute Y in 2X + 3Y = [ 2 3
        4 0 ]

2X = [ 2 3
  4 0 ] – 3Y

= [ 2 3
 4 0 ] – 3 [ 2/5 13/5
      14/5 –2 ]

= [ 2–6/5  3–39/5
  4–42/5  0–(–6) ]

= [ (10–6)/5  (15–39)/5
 (20–42)/5  6 ]

= [ 4/5  –24/5
 –22/5  6 ]

Therefore,
X = ½ [ 4/5  –24/5
   –22/5   6 ]

= [ 2/5  –12/5
 –11/5  3 ]

Final Answers:

(i) X = [ 5 0
     1 4 ],

Y = [ 2 0
   1 1 ]

(ii) X = [ 2/5 –12/5
    –11/5 3 ]

 Y = [ 2/5 13/5
   14/5 –2 ]

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Q8: Find X, if

Y = [ 3 2
  1 4 ] and 2X + Y = [ 1 0
        –3 2 ]

Solution:

We have
2X + Y = [ 1 0
    –3 2 ]

⇒ 2X = [ 1 0
   –3 2 ] – Y

= [ 1–3  0–2
  3–1 2–4 ]

= [ –2 –2
 –4 –2 ]

Now divide by 2:

X = ½ [ –2 –2
    –4 –2 ]

= [ –1 –1
  –2 –1 ]

∴ X = [ –1 –1
   –2 –1 ]

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9. Find x and y, if

2 [ 1 3
  0 x ] + [ y 0
    1 2 ] = [ 5 6
  1 8 ]

Solution:

First multiply the first matrix by 2:

[ 2  6
0 2x ] + [ y 0
  1 2 ] = [ 5 6
 1 8 ]

Add the two matrices on the left:

[ 2 + y 6 + 0
0 + 1 2x + 2 ] = [ 5 6
1 8 ]

Now equate corresponding elements:

From (1,1): 2 + y = 5 ⇒ y = 3
From (2,2): 2x + 2 = 8 ⇒ 2x = 6 ⇒ x = 3

∴ x = 3, y = 3

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Q10: Solve the equation for x, y, z, and t if

2 ⎡ x z ⎤
⎣ y t ⎦ + 3 ⎡ 1 –1 ⎤
⎣ 0 2 ⎦ = 3 ⎡ 3 5 ⎤
⎣ 4 6 ⎦

Solution:

Step 1: Expand each term

2 ×
⎡ x z ⎤ = ⎡ 2x 2z ⎤
⎣ y t ⎦ ⎣ 2y 2t ⎦

3 ×
⎡ 1 –1 ⎤ = ⎡ 3 –3 ⎤
⎣ 0 2 ⎦ ⎣ 0 6 ⎦

3 ×
⎡ 3 5 ⎤ = ⎡ 9 15 ⎤
⎣ 4 6 ⎦ ⎣ 12 18 ⎦

Step 2: Substitute back

⎡ 2x 2z ⎤ + ⎡ 3 –3 ⎤ = ⎡ 9 15 ⎤
⎣ 2y 2t ⎦ ⎣ 0 6 ⎦ ⎣ 12 18 ⎦

Step 3: Add the matrices on the left-hand side

⎡ 2x + 3 2z – 3 ⎤
⎣ 2y + 0 2t + 6 ⎦

⎡ 9 15 ⎤
⎣ 12 18 ⎦

Step 4: Equate corresponding elements

2x + 3 = 9
2z – 3 = 15
2y = 12
2t + 6 = 18

Step 5: Solve for each variable

2x = 6 → x = 3
2z = 18 → z = 9
2y = 12 → y = 6
2t = 12 → t = 6

Final Answers:

x = 3
y = 6
z = 9
t = 6

_____________________________________________________________

Q11: If x[2 3] + y[–1 1] = [10 5], find the values of x and y.

Solution:
We have
x[2 3] + y[–1 1] = [10 5]

⇒ [2x – y, 3x + y] = [10, 5]

Equating the corresponding elements,
2x – y = 10 …(1)
3x + y = 5 …(2)

Add (1) and (2):
(2x – y) + (3x + y) = 10 + 5
5x = 15
x = 3

Substitute x = 3 in (1):
2(3) – y = 10
6 – y = 10
y = –4

Final Answer:
x = 3, y = –4

_____________________________________________________________

Q12:

Given
3 ⎡ x y ⎤ = ⎡ x 6 ⎤ + ⎡ 4 x + y ⎤
⎣ z w ⎦ ⎣ –1 2w ⎦ ⎣ z + w 3 ⎦,

find the values of x, y, z and w.

Solution:

Left-hand side:

3 ⎡ x y ⎤
⎣ z w ⎦

⎡ 3x 3y ⎤
⎣ 3z 3w ⎦

Right-hand side:
⎡ x 6 ⎤ + ⎡ 4 x + y ⎤ = ⎡ x + 4 x + y + 6 ⎤
⎣ –1 2w ⎦ ⎣ z + w 3 ⎦ ⎡ z + w – 1 2w + 3 ⎦

Now equate both sides:

⎡ 3x 3y ⎤ = ⎡ x + 4 x + y + 6 ⎤
⎣ 3z 3w ⎦ ⎣ z + w – 1 2w + 3 ⎦

Comparing corresponding elements:
3x = x + 4
3y = x + y + 6
3z = z + w – 1
3w = 2w + 3

From 3x = x + 4 → 2x = 4 → x = 2
From 3w = 2w + 3 → w = 3
From 3y = x + y + 6 → 2y = 2 + 6 → y = 4
From 3z = z + w – 1 → 2z = 3 – 1 → z = 1

Final Answer:
x = 2, y = 4, z = 1, w = 3

_____________________________________________________________

Q13:

If

⎡ cos x −sin x 0 ⎤

⎢ sin x cos x 0 ⎥

⎣ 0 0 1 ⎦

show that F(x) F(y) = F(x + y).

Solution

F(x) F(y) =

⎡ cos x −sin x 0 ⎤ ⎡ cos y −sin y 0 ⎤

⎢ sin x cos x 0 ⎥ × ⎢ sin y cos y 0 ⎥

⎣ 0 0 1 ⎦ ⎣ 0 0 1 ⎦

Now multiply (row × column):

First row, first column:

(cos x)(cos y) + (−sin x)(sin y) + 0·0 = cos x cos y − sin x sin y = cos(x + y)

First row, second column:

(cos x)(−sin y) + (−sin x)(cos y) + 0·0 = −(cos x sin y + sin x cos y) = −sin(x + y)

First row, third column:

(cos x)(0) + (−sin x)(0) + 0·1 = 0

Second row, first column:

(sin x)(cos y) + (cos x)(sin y) + 0·0 = sin x cos y + cos x sin y = sin(x + y)

Second row, second column:

(sin x)(−sin y) + (cos x)(cos y) + 0·0 = −sin x sin y + cos x cos y = cos(x + y)

Second row, third column:

(sin x)(0) + (cos x)(0) + 0·1 = 0

Third row:

(0)(cos y) + (0)(sin y) + (1)(0) = 0

(0)(−sin y) + (0)(cos y) + (1)(0) = 0

(0)(0) + (0)(0) + (1)(1) = 1

Therefore,

F(x) F(y) =

⎡ cos(x + y) −sin(x + y) 0 ⎤

⎢ sin(x + y) cos(x + y) 0 ⎥

⎣ 0 0 1 ⎦

= F(x + y).

Hence proved that F(x) F(y) = F(x + y).

_____________________________________________________________

Q14: Show that

(i)
[ 5 −1 ] [ 2 1 ] ≠ [ 2 1 ] [ 5 −1 ]
[ 6 7 ] [ 3 4 ] [ 3 4 ] [ 6 7 ]

Solution:

[ 5 −1 ] [ 2 1 ]
[ 6 7 ] [ 3 4 ]

= [ (5×2 + (−1)×3) (5×1 + (−1)×4) ]
[ (6×2 + 7×3) (6×1 + 7×4) ]

= [ 7 1 ]
[ 33 34 ]

Now,

[ 2 1 ] [ 5 −1 ]
[ 3 4 ] [ 6 7 ]

= [ (2×5 + 1×6) (2×(−1) + 1×7) ]
[ (3×5 + 4×6) (3×(−1) + 4×7) ]

= [16 5 ]
[39 25 ]

Hence,

[ 7 1 ] ≠ [ 16 5 ]
[ 33 34 ] [39 25 ]

Therefore,

[ 5 −1 ] [ 2 1 ] ≠ [ 2 1 ] [ 5 −1 ]
[ 6 7 ] [ 3 4 ] [ 3 4 ] [ 6 7 ]

_____________________________________________________________

(ii)
[ 1 2 3 ] [ −1 1 0 ] ≠ [ −1 1 0 ] [ 1 2 3 ]
[ 0 1 0 ] [ 0 −1 1 ] [ 0 −1 1 ] [ 0 1 0 ]
[ 1 1 0 ] [ 2 3 4 ] [ 2 3 4 ] [ 1 1 0 ]

Solution:

[ 1 2 3 ] [ −1 1 0 ]
[ 0 1 0 ] [ 0 −1 1 ]
[ 1 1 0 ] [ 2 3 4 ]

= [ (1×(−1)+2×0+3×2) (1×1+2×(−1)+3×3) (1×0+2×1+3×4) ]
[ (0×(−1)+1×0+0×2) (0×1+1×(−1)+0×3) (0×0+1×1+0×4) ]
[ (1×(−1)+1×0+0×2) (1×1+1×(−1)+0×3) (1×0+1×1+0×4) ]

= [ 5 8 14 ]
[ 0 −1 1 ]
[ −1 0 1 ]

Now,

[−1 1 0 ] [ 1 2 3 ]
[ 0 −1 1 ] [ 0 1 0 ]
[ 2 3 4 ] [ 1 1 0 ]

= [ ((−1)×1 + 1×0 + 0×1) ((−1)×2 + 1×1 + 0×1) ((−1)×3 + 1×0 + 0×0) ]
[ (0×1 + (−1)×0 + 1×1) (0×2 + (−1)×1 + 1×1) (0×3 + (−1)×0 + 1×0) ]
[ (2×1 + 3×0 + 4×1) (2×2 + 3×1 + 4×1) (2×3 + 3×0 + 4×0) ]

= [ −1 −1 −3 ]
[ 1 0 0 ]
[ 6 9 6 ]

Hence, the two results are not equal.

Therefore,

[ 1 2 3 ] [ −1 1 0 ] ≠ [ −1 1 0 ] [1 2 3 ]
[ 0 1 0 ] [ 0 −1 1 ] [ 0 −1 1 ] [ 0 1 0 ]
[ 1 1 0 ] [ 2 3 4 ] [ 2 3 4 ] [ 1 1 0 ]

_____________________________________________________________

Q15:

Find A² − 5A + 6I, if

A =

[ 2 0 1 ]

[ 2 1 3 ]

[ 1 −1 0 ]

Solution:

First,

A² = A × A

=

[ 2 0 1 ] [ 2 0 1 ]

[ 2 1 3 ] [ 2 1 3 ]

[ 1 −1 0 ] [ 1 −1 0 ]

=

[ (2×2 + 0×2 + 1×1) (2×0 + 0×1 + 1×(−1)) (2×1 + 0×3 + 1×0) ]

[ (2×2 + 1×2 + 3×1) (2×0 + 1×1 + 3×(−1)) (2×1 + 1×3 + 3×0) ]

[ (1×2 + (−1)×2 + 0×1) (1×0 + (−1)×1 + 0×(−1)) (1×1 + (−1)×3 + 0×0) ]

=

[ 5 −1 2 ]

[ 9 −2 5 ]

[ 0 −1 −2 ]

Now,

5A =

[ 10 0 5 ]

[ 10 5 15 ]

[ 5 −5 0 ]

and

6I =

[ 6 0 0 ]

[ 0 6 0 ]

[ 0 0 6 ]

∴ A² − 5A + 6I =

[ 5 −1 2 ] [ 10 0 5 ] + [ 6 0 0 ]

[ 9 −2 5 ] − [ 10 5 15 ] [ 0 6 0 ]

[ 0 −1 −2 ] [ 5 −5 0 ] [ 0 0 6 ]

=

[ (5−10+6) (−1−0+0) (2−5+0) ]

[ (9−10+0) (−2−5+6) (5−15+0) ]

[ (0−5+0) (−1−(−5)+0) (−2−0+6) ]

=

[ 1 −1 −3 ]

[ −1 −1 −10 ]

[ −5 4 4 ]

Answer: A² − 5A + 6I =

[ 1 −1 −3 ]

[ −1 −1 −10 ]

[ −5 4 4 ]

_____________________________________________________________

Q16: If

A =

[ 1 0 2 ]

[ 0 2 1 ]

[ 2 0 3 ],

prove that A³ − 6A² + 7A + 2I = 0.

Solution:

First, find A²:

A² = A × A

=

[ 1 0 2 ] [ 1 0 2 ]

[ 0 2 1 ] [ 0 2 1 ]

[ 2 0 3 ] [ 2 0 3 ]

=

[ (1×1 + 0×0 + 2×2) (1×0 + 0×2 + 2×0) (1×2 + 0×1 + 2×3) ]

[ (0×1 + 2×0 + 1×2) (0×0 + 2×2 + 1×0) (0×2 + 2×1 + 1×3) ]

[ (2×1 + 0×0 + 3×2) (2×0 + 0×2 + 3×0) (2×2 + 0×1 + 3×3) ]

=

[ 5 0 8 ]

[ 2 4 5 ]

[ 8 0 13 ]

Now, find A³ = A × A²

A³ =

[ 1 0 2 ] [ 5 0 8 ]

[ 0 2 1 ] [ 2 4 5 ]

[ 2 0 3 ] [ 8 0 13 ]

=

[(1×5 + 0×2 + 2×8) (1×0 + 0×4 + 2×0) (1×8 + 0×5 + 2×13) ]

[ (0×5 + 2×2 + 1×8) (0×0 + 2×4 + 1×0) (0×8 + 2×5 + 1×13) ]

[ (2×5 + 0×2 + 3×8) (2×0 + 0×4 + 3×0) (2×8 + 0×5 + 3×13) ]

=

[ 21 0 34 ]

[ 12 8 23 ]

[ 34 0 55 ]

Now compute each term in A³ − 6A² + 7A + 2I:

6A² =

[ 30 0 48 ]

[ 12 24 30 ]

[ 48 0 78 ]

7A =

[ 7 0 14 ]

[ 0 14 7 ]

[ 14 0 21 ]

2I =

[ 2 0 0 ]

[ 0 2 0 ]

[ 0 0 2 ]

∴ A³ − 6A² + 7A + 2I =

[ 21 0 34 ] [ 30 0 48 ] [ 7 0 14 ] [ 2 0 0 ]

[ 12 8 23 ] − [ 12 24 30 ] + [ 0 14 7 ] + [ 0 2 0 ]

[ 34 0 55 ] [ 48 0 78 ] [ 14 0 21 ] [ 0 0 2 ]

=

[ (21−30+7+2) (0−0+0+0) (34−48+14+0) ]

[ (12−12+0+0) (8−24+14+2) (23−30+7+0) ]

[ (34−48+14+0) (0−0+0+0) (55−78+21+2) ]

=

[ 0 0 0 ]

[ 0 0 0 ]

[ 0 0 0 ]

∴ A³ − 6A² + 7A + 2I = 0, proved.

_____________________________________________________________

Q17: If

A = [ 3 −2 ]

  [ 4 −2 ] and I = [ 1 0 ]

          [ 0 1 ],

find k so that A² = kA − 2I.

Solution:

A = [ 3 −2 ]

  [ 4 −2 ]

Now,

A² = A × A

= [ 3 −2 ] [ 3 −2 ]

 [ 4 −2 ] [ 4 −2 ]

= [ (3×3 + (−2)×4) (3×(−2) + (−2)×(−2)) ]

[ (4×3 + (−2)×4) (4×(−2) + (−2)×(−2)) ]

= [ (9 − 8)  (−6 + 4) ]

 [ (12 − 8) (−8 + 4) ]

= [ 1 −2 ]

[ 4 −4 ]

We are given that A² = kA − 2I.

So,

[ 1 −2 ] = k [ 3 −2 ] −  2 [ 1 0 ]

[ 4 −4 ]    [ 4 −2 ]    [ 0 1 ]

⇒ [ 1 −2 ] = [ 3k−2  −2k ]

      [ 4k  −2k−2 ]

Now equate the corresponding elements:

From (1,1): 1 = 3k − 2 ⇒ 3k = 3 ⇒ k = 1

Check with others:

(1,2): −2 = −2k ⇒ k = 1

(2,1): 4 = 4k ⇒ k = 1

(2,2): −4 = −2k − 2 ⇒ −4 = −2k − 2 ⇒ −2 = −2k ⇒ k = 1

∴ k = 1

_____________________________________________________________

Q18:

If A = [ 0 -tan(α/2)

tan(α/2) 0 ] and I is the 2×2 identity matrix, show that

I + A = (I − A) [ cos α -sin α

sin α cos α ]

Solution:

Let t = tan(α/2). Then

A = [ 0 -t

t 0 ].

So

I + A = [ 1 -t

t 1 ]

and

I − A = [ 1 t

-t 1 ].

Denote the rotation matrix by

R = [ cos α -sin α

sin α cos α ].

Compute (I − A) R by multiplying the matrices:

(I − A) R = [ 1 t

-t 1 ] × [ cos α -sin α

sin α cos α ]

= [ 1·cos α + t·sin α 1·(−sin α) + t·cos α

(−t)·cos α + 1·sin α (−t)·(−sin α) + 1·cos α ]

= [ cos α + t sin α −sin α + t cos α

−t cos α + sin α t sin α + cos α ].

We must show this equals I + A = [ 1 −t

t 1 ].

Use the half-angle formulas (in terms of t = tan(α/2)):

cos α = (1 − t^2) / (1 + t^2),

sin α = 2t / (1 + t^2).

Now evaluate the entries:

1. cos α + t sin α

= (1 − t^2)/(1 + t^2) + t·(2t/(1 + t^2))

= (1 − t^2 + 2t^2)/(1 + t^2)

= (1 + t^2)/(1 + t^2)

= 1.

2. −sin α + t cos α

= −(2t/(1 + t^2)) + t·((1 − t^2)/(1 + t^2))

= (−2t + t − t^3)/(1 + t^2)

= −t(1 + t^2)/(1 + t^2)

= −t.

Similarly, the lower-left entry:

−t cos α + sin α = t (we get the same simplification), and

the lower-right entry: t sin α + cos α = 1.

Therefore (I − A) R = [ 1 −t

t 1 ] = I + A.

Hence proved:

I + A = (I − A) [ cos α −sin α

sin α cos α ].

_____________________________________________________________

Q19: A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) ₹1800
(b) ₹2000

Solution:

A trust fund has ₹30,000 to be invested in two types of bonds.

Let

x = amount invested at 5%

y = amount invested at 7%

Total amount condition:

x + y = 30,000 … (1)

Total interest condition:

0.05x + 0.07y = required interest … (2)

We solve for each case.

(a) Required annual interest = ₹1800

From (2):

0.05x + 0.07y = 1800

Use (1): y = 30000 − x

Substitute:

0.05x + 0.07(30000 − x) = 1800

0.05x + 2100 − 0.07x = 1800

−0.02x = −300

x = 15000

Then

y = 30000 − 15000 = 15000

Answer for (a):

Invest ₹15,000 in the 5% bond and ₹15,000 in the 7% bond.

(b) Required annual interest = ₹2000

Equation:

0.05x + 0.07(30000 − x) = 2000

0.05x + 2100 − 0.07x = 2000

−0.02x = −100

x = 5000

Then

y = 30000 − 5000 = 25000

Answer for (b):

Invest ₹5,000 in the 5% bond and ₹25,000 in the 7% bond.

Final Answers

(a) ₹15,000 at 5% and ₹15,000 at 7%

(b) ₹5,000 at 5% and ₹25,000 at 7%

_____________________________________________________________

Q20: The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Solution:

Step 1 — convert dozens to numbers of books:

10 dozen = 10 × 12 = 120 chemistry books

8 dozen = 8 × 12 = 96 physics books

10 dozen = 10 × 12 = 120 economics books

Write quantity vector Q and price row vector P:

Q = [120

96

120] (a 3 × 1 column)

P = [80 60 40] (a 1 × 3 row)

Total revenue = P × Q (1×3 times 3×1 gives scalar)

Compute:

P × Q = 80×120 + 60×96 + 40×120

= 9600 + 5760 + 4800

= 20,160

Answer:

The bookshop will receive ₹20,160 in total from selling all the books.

_____________________________________________________________

Q21: The restriction on n, k and p so that PY + WY will be defined are:
(A) k = 3, p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3

Solution:

X is 2 × n

Y is 3 × k

Z is 2 × p

W is n × 3

P is p × k

We need PY + WY to be defined.

That means:

1. PY must be defined

P is p × k

Y is 3 × k

For multiplication PY, the number of columns of P must equal number of rows of Y:

k = 3

Then PY becomes p × k × 3 × k → p × k.

2. WY must be defined

W is n × 3

Y is 3 × k

3 = 3 → multiplication is valid.

Result is n × k.

3. For addition PY + WY, both results must have the same order:

PY is p × k

WY is n × k

So p = n.

Final restrictions:

k = 3 and p = n

Correct Answer: (A) k = 3, p = n

_____________________________________________________________

Q22: If n = p, then the order of the matrix 7X – 5Z is:

(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n

Solution:

From the earlier given matrix orders:

X is of order 2 × n

Z is of order 2 × p

Given n = p, Z becomes 2 × n.

So:

7X has order 2 × n

5Z has order 2 × n

Since subtraction is possible only when orders match,

7X – 5Z has order 2 × n.

Correct Answer: (B) 2 × n

_____________________________________________________________

EXERCISE 3.3

Q1: Find the transpose of each of the following matrices:

(i) [ 5

1/2

-1 ]

(ii) [ 1 -1

2 3 ]

(iii) [ -1 5 6

√3 5 6

2 3 -1 ]

Answers:

(i) The transpose of the 3×1 column [ 5 ; 1/2 ; −1 ] is the 1×3 row:

[ 5 1/2 −1 ]

(ii) The transpose of the 2×2 matrix [ 1 −1 ; 2 3 ] is:

[ 1 2

−1 3 ]

(iii) The transpose of the 3×3 matrix

[ −1 5 6

√3 5 6

2 3 −1 ]

is

[ −1 √3 2

5 5 3

6 6 −1 ]

_____________________________________________________________

Q2: If A =

[ -1 2 3

5 7 9

-2 1 1 ]

and B =

[ -4 1 -5

1 2 0

1 3 1 ], then verify that

(i) (A + B)' = A' + B'

(ii) (A – B)' = A' – B'

Solution

First find A + B

A + B =

[ (-1 + -4) (2 + 1) (3 + -5)

(5 + 1) (7 + 2) (9 + 0)

(-2 + 1) (1 + 3) (1 + 1) ]

A + B =

[ -5 3 -2

6 9 9

-1 4 2 ]

Now find (A + B)'

(A + B)' =

[ -5 6 -1

3 9 4

-2 9 2 ]

Now find A' and B'

A' =

[ -1 5 -2

2 7 1

3 9 1 ]

B' =

[ -4 1 1

1 2 3

-5 0 1 ]

Now find A' + B'

A' + B' =

[ (-1 + -4) (5 + 1) (-2 + 1)

(2 + 1) (7 + 2) (1 + 3)

(3 + -5) (9 + 0) (1 + 1) ]

A' + B' =

[ -5 6 -1

3 9 4

-2 9 2 ]

Thus (A + B)' = A' + B' verified.

Part (ii)

First find A – B

A – B =

[ (-1 - -4) (2 - 1) (3 - -5)

(5 - 1) (7 - 2) (9 - 0)

(-2 - 1) (1 - 3) (1 - 1) ]

A – B =

[ 3 1 8

4 5 9

-3 -2 0 ]

Now find (A – B)'

(A – B)' =

[ 3 4 -3

1 5 -2

8 9 0 ]

Now find A' – B'

A' – B' =

[ (-1 - -4) (5 - 1) (-2 - 1)

(2 - 1) (7 - 2) (1 - 3)

(3 - -5) (9 - 0) (1 - 1) ]

A' – B' =

[ 3 4 -3

1 5 -2

8 9 0 ]

Thus (A – B)' = A' – B' verified.

_____________________________________________________________

Q3: If A' = [ 3 4

-1 2

0 1 ]

and B = [ -1 2 1

1 2 3 ], then verify that

(i) (A + B)' = A' + B'

(ii) (A – B)' = A' – B'

First convert A' to A (since A' is given)

A' =

[ 3 4

-1 2

0 1 ]

So A is the transpose of A':

A =

[ 3 -1 0

4 2 1 ]

B is already given as:

B =

[ -1 2 1

1 2 3 ]

(i) VERIFY (A + B)' = A' + B'

Step 1: Find A + B

A + B =

[ (3 + -1) (-1 + 2) (0 + 1)

(4 + 1) (2 + 2) (1 + 3) ]

A + B =

[ 2 1 1

5 4 4 ]

Step 2: Find (A + B)'

(A + B)' =

[ 2 5

1 4

1 4 ]

Step 3: Find B'

B' =

[ -1 1

2 2

1 3 ]

Step 4: Find A' + B'

A' + B' =

[ (3 + -1) (4 + 1)

(-1 + 2) (2 + 2)

(0 + 1) (1 + 3) ]

A' + B' =

[ 2 5

1 4

1 4 ]

Thus,

(A + B)' = A' + B' VERIFIED

(ii) VERIFY (A – B)' = A' – B'

Step 1: Find A – B

A – B =

[ (3 - -1) (-1 - 2) (0 - 1)

(4 - 1) (2 - 2) (1 - 3) ]

A – B =

[ 4 -3 -1

3 0 -2 ]

Step 2: Find (A – B)'

(A – B)' =

[ 4 3

-3 0

-1 -2 ]

Step 3: Find A' – B'

A' – B' =

[ (3 - -1) (4 - 1)

(-1 - 2) (2 - 2)

(0 - 1) (1 - 3) ]

A' – B' =

[ 4 3

-3 0

-1 -2 ]

Thus,

(A – B)' = A' – B' VERIFIED

_____________________________________________________________

Q4: If A' = [ -2 3

1 2 ]

and B = [ -1 0

1 2 ], then find (A + 2B)'

First get A by transposing A':

A' =

[ -2 3

1 2 ]

So A = (A')' =

[ -2 1

3 2 ]

Now compute 2B:

B =

[ -1 0

1 2 ]

2B =

[ -2 0

2 4 ]

Next compute A + 2B:

A + 2B =

[ (-2 + -2) (1 + 0)

(3 + 2) (2 + 4) ]

A + 2B =

[ -4 1

5 6 ]

Finally take the transpose:

(A + 2B)' =

[ -4 5

1 6 ]

Answer: (A + 2B)' = [ -4 5

1 6 ]

_____________________________________________________________

Q5: For the matrices A and B, verify that (AB)' = B'A', where

(i) A =

[ 1

-4

3 ], B = [ -1 2 1 ]

(ii) A =

[ 0

1

2 ], B = [ 1 5 7 ]

(i) VERIFY (AB)' = B'A'

Step 1: Compute AB

A is 3×1 and B is 1×3, so AB is 3×3.

A =

[ 1

-4

3 ]

B = [ -1 2 1 ]

Multiply:

AB =

[ 1 × -1 1 × 2 1 × 1

-4 × -1 -4 × 2 -4 × 1

3 × -1 3 × 2 3 × 1 ]

AB =

[ -1 2 1

4 -8 -4

-3 6 3]

Step 2: Find (AB)'

(AB)' =

[ -1 4 -3

2 -8 6

1 -4 3 ]

Step 3: Compute B'A'

First find B':

B = [ -1 2 1 ]

B' =

[ -1

2

1 ]

A' = [ 1 -4 3 ]

Now multiply B'A':

B'A' =

[ -1 × 1 -1 × -4 -1 × 3

2 × 1 2 × -4 2 × 3

1 × 1 1 × -4 1 × 3 ]

B'A' =

[ -1 4 -3

2 -8 6

1 -4 3 ]

Verified: (AB)' = B'A'

(ii) VERIFY (AB)' = B'A'

Step 1: Compute AB

A =

[ 0

1

2 ]

B = [ 1 5 7 ]

AB =

[ 0×1 0×5 0×7

1×1 1×5 1×7

2×1 2×5 2×7 ]

AB =

[ 0 0 0

1 5 7

2 10 14 ]

Step 2: Find (AB)'

(AB)' =

[ 0 1 2

0 5 10

0 7 14 ]

Step 3: Compute B'A'

B' =

[ 1

5

7 ]

A' = [ 0 1 2 ]

Now multiply:

B'A' =

[ 1×0 1×1 1×2

5×0 5×1 5×2

7×0 7×1 7×2 ]

B'A' =

[ 0 1 2

0 5 10

0 7 14 ]

Verified: (AB)' = B'A'

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Q6: If (i) A = [ cos α sin α

−sin α cos α ], then verify that A′A = I

Solution:

A′ denotes the transpose of A.

So,

A′ = [ cos α −sin α

sin α cos α ]

Now multiply A′ and A:

A′A =

[ cos α −sin α ] [ cos α sin α ]

[ sin α cos α ] [ −sin α cos α ]

Compute each entry:

• (1,1) entry:

cos α · cos α + (−sin α) · (−sin α)

= cos²α + sin²α

= 1

• (1,2) entry:

cos α · sin α + (−sin α) · cos α

= cos α sin α − cos α sin α

= 0

• (2,1) entry:

sin α · cos α + cos α · (−sin α)

= sin α cos α − sin α cos α

= 0

• (2,2) entry:

sin α · sin α + cos α · cos α

= sin²α + cos²α

= 1

Therefore,

A′A = [ 1 0

0 1 ]

which is the identity matrix.

Hence verified.

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(ii) If A = [ sin α cos α

−cos α sin α ], then verify that A′A = I

Solution:

A′ = [ sin α −cos α

cos α sin α ]

Now multiply A′ and A:

A′A =

[ sin α −cos α ] [ sin α cos α ]

[ cos α sin α ] [ −cos α sin α ]

Compute entries:

• (1,1) entry:

sin α · sin α + (−cos α) · (−cos α)

= sin²α + cos²α

= 1

• (1,2) entry:

sin α · cos α + (−cos α) · sin α

= sin α cos α − sin α cos α

= 0

• (2,1) entry:

cos α · sin α + sin α · (−cos α)

= cos α sin α − cos α sin α

= 0

• (2,2) entry:

cos α · cos α + sin α · sin α

= cos²α + sin²α

= 1

Therefore,

A′A = [ 1 0

0 1 ]

which is the identity matrix.

Hence verified.

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Q7: (i) Show that the matrix

A =

[ 1 -1 5 ]

[ -1 2 1 ]

[ 5 1 3 ]

is a symmetric matrix.

Solution:

A matrix is symmetric if Aᵀ = A.

Write the transpose Aᵀ by interchanging rows and columns:

Aᵀ =

[ 1 -1 5 ]

[ -1 2 1 ]

[ 5 1 3 ]

We see that every element aᵢⱼ equals aⱼᵢ.

So Aᵀ = A.

Therefore, A is a symmetric matrix.

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(ii) Show that the matrix

A =

[0 1 -1 ]

[ -1 0 1 ]

[ 1 -1 0 ]

is a skew-symmetric matrix.

Solution:

A matrix is skew-symmetric if Aᵀ = -A.

Find Aᵀ by interchanging rows and columns:

Aᵀ =

[ 0 -1 1 ]

[ 1 0 -1 ]

[ -1 1 0 ]

Now find -A by changing the sign of every entry of A:

-A =

[ 0 -1 1 ]

[ 1 0 -1 ]

[ -1 1 0 ]

We see that Aᵀ = -A.

Therefore, A is a skew-symmetric matrix.

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Q8: For the matrix A =

[ 1 5 ]

[ 6 7 ]

(i) (A + Aᵀ) is a symmetric matrix

Solution:

First compute the transpose Aᵀ by interchanging rows and columns:

Aᵀ =

[ 1 6 ]

[ 5 7 ]

Now form A + Aᵀ:

A + Aᵀ =

[ 1+1 5+6 ]

[ 6+5 7+7 ]

So

A + Aᵀ =

[ 2 11 ]

[ 11 14 ]

Compare with its transpose:

(A + Aᵀ)ᵀ =

[ 2 11 ]

[ 11 14 ]

Since (A + Aᵀ)ᵀ = (A + Aᵀ), the matrix A + Aᵀ is symmetric.

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(ii) (A − Aᵀ) is a skew-symmetric matrix

Solution:

Compute A − Aᵀ:

A − Aᵀ =

[ 1−1 5−6 ]

[ 6−5 7−7 ]

So

A − Aᵀ =

[ 0 −1 ]

[ 1 0 ]

Find its transpose:

(A − Aᵀ)ᵀ =

[ 0 1 ]

[ −1 0 ]

Observe that (A − Aᵀ)ᵀ = −(A − Aᵀ). Therefore A − Aᵀ is skew-symmetric.

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Q9: Find 1/2 (A + Aᵀ) and 1/2 (A − Aᵀ), when A =

[ 0 a b ]

[ −a 0 c ]

[ −b −c 0 ]

Solution:

First compute the transpose Aᵀ (swap rows and columns):

Aᵀ =

[ 0 −a −b ]

[ a 0 −c ]

[ b c 0 ]

Now form the sum and difference.

A + Aᵀ =

[ 0+0 a+(−a) b+(−b) ]

[ (−a)+a 0+0 c+(−c) ]

[ (−b)+b (−c)+c 0+0 ]

So

A + Aᵀ =

[ 0 0 0 ]

[ 0 0 0 ]

[ 0 0 0 ]

Hence

1/2 (A + Aᵀ) =

[ 0 0 0 ]

[ 0 0 0 ]

[ 0 0 0 ] (the zero matrix)

Next,

A − Aᵀ =

[ 0−0 a−(−a) b−(−b) ]

[ (−a)−a 0−0 c−(−c) ]

[ (−b)−b (−c)−c 0−0 ]

So

A − Aᵀ =

[ 0 2a 2b ]

[ −2a 0 2c ]

[ −2b −2c 0 ]

Therefore

1/2 (A − Aᵀ) =

[ 0 a b ]

[ −a 0 c ]

[ −b −c 0 ]

which is exactly A. (This reflects that A is skew-symmetric, so its symmetric part is the zero matrix and its skew-symmetric part is itself.)

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Q10: Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

S = 1/2 (A + Aᵀ) is symmetric and K = 1/2 (A − Aᵀ) is skew-symmetric, and A = S + K.

(i) A = [ 3 5 ]

[ 1 −1 ]

Aᵀ = [ 3 1 ]

[ 5 −1 ]

S = 1/2 (A + Aᵀ) = 1/2 [ 3+3 5+1 ] = [ 3 3 ]

[ 1+5 −1+−1 ] [ 3 −1 ]

K = 1/2 (A − Aᵀ) = 1/2 [ 3−3 5−1 ] = [ 0 2 ]

[ 1−5 −1−(−1)] [ −2 0 ]

Check: S + K = [3 5; 1 −1] = A.

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(ii) A = [ 6 −2 2 ]

[ −2 3 −1 ]

[ 2 −1 3 ]

Since A = Aᵀ already (A is symmetric),

S = 1/2 (A + Aᵀ) = A = [ 6 −2 2 ]

[ −2 3 −1 ]

[ 2 −1 3 ]

K = 1/2 (A − Aᵀ) = 0 (the 3×3 zero matrix).

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(iii) A =

[ 3 3 −1 ]

[ −2 −2 1 ]

[ −4 −5 2 ]

Aᵀ =

[ 3 −2 −4 ]

[ 3 −2 −5 ]

[ −1 1 2 ]

S = 1/2 (A + Aᵀ) =

1/2 ·

[ 3+3 3+(−2) −1+(−4) ]

[ −2+3 −2+(−2) 1+(−5) ]

[ −4+(−1) −5+1 2+2 ]

= [ 3 1/2 −5/2 ]

[ 1/2 −2 −2 ]

[ −5/2 −2 2 ]

K = 1/2 (A − Aᵀ) =

1/2 ·

[ 3−3 3−(−2) −1−(−4) ]

[ −2−3 −2−(−2) 1−(−5) ]

[ −4−(−1) −5−1 2−2 ]

= [ 0 5/2 3/2 ]

[ −5/2 0 3 ]

[ −3/2 −3 0 ]

Check: S + K = A.

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(iv) A =

[ 1 5 ]

[ −1 2 ]

Aᵀ =

[ 1 −1 ]

[ 5 2 ]

S = 1/2 (A + Aᵀ) = 1/2 [ 1+1 5+(−1) ] = [ 1 2 ]

[ −1+5 2+2 ] [ 2 2 ]

K = 1/2 (A − Aᵀ) = 1/2 [ 1−1 5−(−1) ] = [ 0 3 ]

[ −1−5 2−2 ] [ −3 0 ]

Check: S + K = [1 5

−1 2] = A.

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Q11: If A, B are symmetric matrices of same order, then AB − BA is a

(A) Skew symmetric matrix

(B) Symmetric matrix

(C) Zero matrix

(D) Identity matrix

Solution (simple):

For symmetric matrices A and B,

Aᵀ = A and Bᵀ = B.

Now check (AB − BA):

(AB − BA)ᵀ = BᵀAᵀ − AᵀBᵀ

= BA − AB

= −(AB − BA)

Since the transpose gives the negative, AB − BA is skew symmetric.

Correct Answer: (A) Skew symmetric matrix

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Q12: If A = [ cos α −sin α

sin α cos α ] and A + A′ = I, then the value of α is
(A) π/6  (B) π/3  (C) π  (D) 3π/2

Solution (clean):

A + A′ = I gives:

2 cos α = 1

So, cos α = 1/2

Therefore,

α = π/3

Correct Answer: π/3

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