DETERMINANTS

Exercise 4.1

Q1: Evaluate the determinants

1. | 2 4 |
| -5 -1 |

Solution:
Determinant = (2)(-1) − (4)(-5)
= −2 + 20
= 18.

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Q2: (i)

| cos θ − sin θ |
| sin θ cos θ |

Solution:
Determinant = (cos θ)(cos θ) − (− sin θ)(sin θ)
= cos² θ + sin² θ
= 1.

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Q2: (ii)

| x² − x + 1 x − 1 |
| x + 1 x + 1 |

Solution:
Determinant = (x² − x + 1)(x + 1) − (x − 1)(x + 1)
= (x + 1)[(x² − x + 1) − (x − 1)]
= (x + 1)[x² − x + 1 − x + 1]
= (x + 1)(x² − 2x + 2).

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Q3: If A = | 1 2 |
     | 4 2 | , then show that | 2A | = 4 | A |

Solution:

A = | 1 2 |
  | 4 2 |

First find | A |:

| A | = (1 × 2) − (2 × 4)
| A | = 2 − 8
| A | = −6

Now find 2A:

2A = | 2 4 |
   | 8 4 |

Now find | 2A |:

| 2A | = (2 × 4) − (4 × 8)
| 2A | = 8 − 32
| 2A | = −24

Now compare:

4 | A | = 4 × (−6) = −24

Therefore,

| 2A | = 4 | A |

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Q4: If A = | 1 0 1 |
     | 0 1 2 |
     | 0 0 4 | , then show that | 3A | = 27 | A |

Solution:

Since A is an upper triangular matrix, its determinant is the product of diagonal elements.

| A | = 1 × 1 × 4 = 4

Now find 3A:

3A = | 3 0 3 |
   | 0 3 6 |
   | 0 0 12 |

Now find | 3A |:

| 3A | = 3 × 3 × 12
| 3A | = 108

Now compare:

27 | A | = 27 × 4 = 108

Therefore,

| 3A | = 27 | A |

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Q5: Evaluate the following determinants:

(i) | 3 −1 −2 |
  | 0 0 −1 |
  | 3 −5 0 |

Solution:

Expand along the second row (because it has two zeros):

| A | = 0 × C₂₁ + 0 × C₂₂ + (−1) × C₂₃

Now find the cofactor C₂₃:

C₂₃ = (−1)²⁺³ × | 3 −1 |
         | 3 −5 |

C₂₃ = − (3 × (−5) − (−1 × 3))
C₂₃ = − (−15 + 3)
C₂₃ = − (−12)
C₂₃ = 12

Now,

| A | = (−1) × 12
| A | = −12

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(ii) | 3 −4 5 |
   | 1 1 −2 |
   | 2 3 1 |

Solution:

Expand along the first row:

| A | = 3 × | 1 −2 | − (−4) × | 1 −2 | + 5 × | 1 1 |
     | 3 1 |     | 2 1 |    | 2 3 |

Now evaluate each 2 × 2 determinant:

First minor:

| 1 −2 | = (1 × 1) − (−2 × 3) = 1 + 6 = 7
| 3 1 |

Second minor:

| 1 −2 | = (1 × 1) − (−2 × 2) = 1 + 4 = 5
| 2 1 |

Third minor:

| 1 1 | = (1 × 3) − (1 × 2) = 3 − 2 = 1
| 2 3 |

Now substitute:

| A | = 3(7) + 4(5) + 5(1)
| A | = 21 + 20 + 5
| A | = 46

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(iii)

| 0 1 2 |
|−1 0 −3 |
|−2 3 0 |

Solution (expand along the first row):

Determinant = a11·C11 − a12·C12 + a13·C13
= 0 · (…) − 1 · det([−1 −3; −2 0]) + 2 · det([−1 0; −2 3])

Compute the 2×2 minors:

det([−1 −3; −2 0]) = (−1)(0) − (−3)(−2) = 0 − 6 = −6.
det([−1 0; −2 3]) = (−1)(3) − (0)(−2) = −3.

Now substitute:

Determinant = 0 − 1·(−6) + 2·(−3)
= 0 + 6 − 6
= 0.

Therefore |A| = 0.

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(iv)

| 2 −1 −2 |
| 0 2 −1 |
| 3 −5 0 |

Solution (expand along the first row):

Determinant = a11·M11 − a12·M12 + a13·M13
= 2·det([ 2 −1; −5 0 ]) − (−1)·det([ 0 −1; 3 0 ]) + (−2)·det([ 0 2; 3 −5 ])

Compute the 2×2 minors:

M11 = det([ 2 −1; −5 0 ]) = (2)(0) − (−1)(−5) = 0 − 5 = −5.
M12 = det([ 0 −1; 3 0 ]) = (0)(0) − (−1)(3) = 3.
M13 = det([ 0 2; 3 −5 ]) = (0)(−5) − (2)(3) = −6.

Now substitute:

Determinant = 2·(−5) − (−1)·(3) + (−2)·(−6)
= −10 + 3 + 12
= 5.

Therefore |A| = 5.

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Q6: If A = | 1 1 −2 |
     | 2 1 −3 |
     | 5 4 −9 | , find | A |

Solution:

Expand along the first row:

| A | = 1 × | 1 −3 | − 1 × | 2 −3 | + (−2) × | 2 1 |
     | 4 −9 |   | 5 −9 |     | 5 4 |

Now evaluate each 2 × 2 determinant:

First minor:
| 1 −3 | = (1)(−9) − (−3)(4) = −9 + 12 = 3
| 4 −9 |

Second minor:
| 2 −3 | = (2)(−9) − (−3)(5) = −18 + 15 = −3
| 5 −9 |

Third minor:
| 2 1 | = (2)(4) − (1)(5) = 8 − 5 = 3
| 5 4 |

Now substitute:

| A | = 1(3) − 1(−3) + (−2)(3)
| A | = 3 + 3 − 6
| A | = 0

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Q7: Find values of x, if

(i) | 2 4 | = | 2x 4 |
  | 5 1 | | 6 x |

Solution:

Left side determinant:

|2 4| = (2)(1) − (4)(5) = 2 − 20 = −18
|5 1|

Right side determinant:

| 2x 4 | = (2x)(x) − (4)(6) = 2x² − 24
| 6 x |

Since both determinants are equal:

2x² − 24 = −18
2x² = 6
x² = 3
x = ±√3

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(ii) | 2 3 | = | x 3 |
  | 4 5 | | 2x 5 |

Solution:

Left side determinant:

| 2 3 | = (2)(5) − (3)(4) = 10 − 12 = −2
| 4 5 |

Right side determinant:

| x 3 | = (x)(5) − (3)(2x) = 5x − 6x = −x
| 2x 5 |

Since both determinants are equal:

−x = −2
x = 2

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Q8: If | x 2 | = | 6 2 | , then x is equal to
   |18 x | |18 6 |

(A) 6
(B) ± 6
(C) − 6
(D) 0

Solution:

Find the determinant of the left side:

| x 2 | = (x)(x) − (2)(18)
| 18 x | = x² − 36

Find the determinant of the right side:

| 6 2 | = (6)(6) − (2)(18)
| 18 6 | = 36 − 36
= 0

Since both determinants are equal:

x² − 36 = 0
x² = 36
x = ± 6

Correct Answer: (B) ± 6

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Exercise 4.2

Q1: Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3)

(ii) (2, 7), (1, 1), (10, 8)

(iii) (−2, −3), (3, 2), (−1, −8)

Formula used:

Area of triangle
= 1/2 | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |

(i) Vertices: (1, 0), (6, 0), (4, 3)

Area
= 1/2 | 1(0 − 3) + 6(3 − 0) + 4(0 − 0) |
= 1/2 | −3 + 18 + 0 |
= 1/2 (15)

Area = 15/2 square units

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(ii) Vertices: (2, 7), (1, 1), (10, 8)

Area
= 1/2 | 2(1 − 8) + 1(8 − 7) + 10(7 − 1) |
= 1/2 | 2(−7) + 1 + 10(6) |
= 1/2 | −14 + 1 + 60 |
= 1/2 (47)

Area = 47/2 square units

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(iii) Vertices: (−2, −3), (3, 2), (−1, −8)

Area
= 1/2 | −2(2 − (−8)) + 3((−8) − (−3)) + (−1)((−3) − 2) |
= 1/2 | −2(10) + 3(−5) + (−1)(−5) |
= 1/2 | −20 − 15 + 5 |
= 1/2 (30)

Area = 15 square units

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Q2: Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

Solution:

Three points are collinear if the area of the triangle formed by them is zero.

Area = 1/2 |
a b + c 1
b c + a 1
c a + b 1 |

Evaluate the determinant:

= a[(c + a) − (a + b)] − (b + c)[b − c] + 1[b(a + b) − c(c + a)]

= a(c − b) − (b + c)(b − c) + [ab + b² − c² − ac]

= a(c − b) − (b² − c²) + a(b − c) + (b² − c²)

= a(c − b + b − c)

= 0

Since the area is zero, the given points are collinear.

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Q3: Find values of k if area of triangle is 4 sq. units and vertices are:

(i) (k, 0), (4, 0), (0, 2)

Solution:

Area
= 1/2 | k(0 − 2) + 4(2 − 0) + 0(0 − 0) |
= 1/2 | −2k + 8 |

Given area = 4,

1/2 | −2k + 8 | = 4
| −2k + 8 | = 8

Case 1: −2k + 8 = 8
−2k = 0
k = 0

Case 2: −2k + 8 = −8
−2k = −16
k = 8

Therefore,
k = 0 or k = 8

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(ii) (−2, 0), (0, 4), (0, k)

Solution:

Area
= 1/2 | −2(4 − k) + 0(k − 0) + 0(0 − 4) |
= 1/2 | −8 + 2k |

Given area = 4,

1/2 | −8 + 2k | = 4
| −8 + 2k | = 8

Case 1: −8 + 2k = 8
2k = 16
k = 8

Case 2: −8 + 2k = −8
2k = 0
k = 0

Therefore,
k = 0 or k = 8

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Q4:

(i) Find equation of line joining (1, 2) and (3, 6) using determinants.

Solution:

The equation of a line through points (x₁, y₁) and (x₂, y₂) is:

| x y 1 |
| x₁ y₁ 1 | = 0
| x₂ y₂ 1 |

Substituting (1, 2) and (3, 6):

| x y 1 |
| 1 2 1 | = 0
| 3 6 1 |

Expanding:

x(2 − 6) − y(1 − 3) + 1(6 − 6) = 0
x(−4) − y(−2) + 0 = 0
−4x + 2y = 0

Equation of the line:

2y = 4x
y = 2x

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(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Solution:

Using determinant form:

| x y 1 |
| 3 1 1 | = 0
| 9 3 1 |

Expanding:

x(1 − 3) − y(3 − 9) + 1(9 − 9) = 0
x(−2) − y(−6) + 0 = 0
−2x + 6y = 0

Equation of the line:

6y = 2x
y = x / 3

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Q5: If area of triangle is 35 sq. units with vertices (2, −6), (5, 4) and (k, 4). Then k is

Solution:

Notice the points (5, 4) and (k, 4) have the same y-coordinate, so the base of the triangle along the line y = 4 has length |k − 5|.
The vertical distance (height) from (2, −6) up to y = 4 is 4 − (−6) = 10.

Area = 1/2 × base × height
35 = 1/2 × |k − 5| × 10
35 = 5 · |k − 5|
|k − 5| = 7

So k − 5 = 7 or k − 5 = −7
Hence k = 12 or k = −2.

Therefore the required values are k = 12 and k = −2.

Answer: D (12, -2)

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Exercise 4.3

Q1: Write Minors and Cofactors of the elements of following determinants:

1. (i)

| 2 –4 |
| 0 3 |

Solution:

M11 = 3, M12 = 0, M21 = –4, M22 = 2,
A11 = 3, A12 = 0, A21 = 4, A22 = 2

(ii)
| a c |
| b d |

Solution:

M11 = d, M12 = b, M21 = c, M22 = a

A11 = d, A12 = –b, A21 = –c, A22 = a

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2. (i)
| 1 0 0 |
| 0 1 0 |
| 0 0 1 |

Solution:

M11 = 1, M12 = 0, M13 = 0,
M21 = 0, M22 = 1, M23 = 0,
M31 = 0, M32 = 0, M33 = 1

A11 = 1, A12 = 0, A13 = 0,
A21 = 0, A22 = 1, A23 = 0,
A31 = 0, A32 = 0, A33 = 1

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2. (ii)
  
| 1 0 4 |
| 3 5 –1 |
| 0 1 2 |

M11 = 11, M12 = 6, M13 = 3,
M21 = –4, M22 = 2, M23 = 1,
M31 = –20, M32 = –13, M33 = 5

A11 = 11, A12 = –6, A13 = 3,
A21 = 4, A22 = 2, A23 = –1,
A31 = –20, A32 = 13, A33 = 5

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Q3: Using cofactors of elements of second row, evaluate Δ =

| 5 3 8 |
| 2 0 1 |
| 1 2 3 |

Solution:

Second row elements: 2, 0, 1

Step 1: Cofactors of the second row

C21 = (–1)^(2+1) × determinant of
| 3 8 |
| 2 3 |
= –1 × (3×3 – 8×2)
= –1 × (9 – 16)
= 7

C22 = (–1)^(2+2) × determinant of
| 5 8 |
| 1 3 |
= 1 × (5×3 – 8×1)
= 15 – 8
= 7

C23 = (–1)^(2+3) × determinant of
| 5 3 |
| 1 2 |
= –1 × (5×2 – 3×1)
= –1 × (10 – 3)
= –7

Step 2: Expansion using second row

Δ = 2×C21 + 0×C22 + 1×C23
Δ = 2×7 + 0 + (–7)
Δ = 14 – 7
Δ = 7

Final Answer: Δ = 7

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Q4: Using cofactors of elements of third column, evaluate Δ where

| 1 x yz |
| 1 y zx |
| 1 z xy |

Third column elements: yz, zx, xy

Step 1 — Minors and cofactors of third column

Minor for element at (1,3):
M13 = determinant of the 2×2 matrix formed by removing row 1 and column 3, i.e.
| 1 y |
| 1 z |
M13 = 1·z − y·1 = z − y
Sign factor = (−1)^(1+3) = (−1)^4 = 1
C13 = 1 × M13 = z − y

Minor for element at (2,3):
M23 = determinant of
| 1 x |
| 1 z |
M23 = 1·z − x·1 = z − x
Sign factor = (−1)^(2+3) = (−1)^5 = −1
C23 = −1 × M23 = −(z − x) = x − z

Minor for element at (3,3):
M33 = determinant of
| 1 x |
| 1 y |
M33 = 1·y − x·1 = y − x
Sign factor = (−1)^(3+3) = (−1)^6 = 1
C33 = 1 × M33 = y − x

Step 2 — Expand Δ along the third column

Δ = (element 1,3)·C13 + (element 2,3)·C23 + (element 3,3)·C33

Substitute elements and cofactors:

Δ = yz·(z − y) + zx·(x − z) + xy·(y − x)

Step 3 — Simplify and factorise

Expand (or group) and factor:

yz(z − y) + zx(x − z) + xy(y − x)
= y z z − y^2 z + z x x − z^2 x + x y y − x^2 y
= z^2 y − y^2 z + x^2 z − x z^2 + x y^2 − x^2 y

This expression factorises to the product of three simple factors:

Δ = (x − y) (y − z) (z − x)

Final Answer: (x − y) (y − z) (z − x)

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Q5: We have a 3×3 determinant:

Δ =
| a11 a12 a13 |
| a21 a22 a23 |
| a31 a32 a33 |

Aij = cofactor of aij.

Key fact of determinants

A determinant can always be expanded along any column.

For expansion along the first column, the formula is:

Δ = a11·A11 + a21·A21 + a31·A31

This is a basic property:

Element × its cofactor, added down the chosen column.

Now check the options.

Option D is:

a11 A11 + a21 A21 + a31 A31

This matches exactly the expansion of Δ along the first column.

Therefore, the correct answer is: (D)

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Exercise 4.4

Q1: Find adjoint of each of the matrices:

1. [ 1 2 ]
[ 3 4 ]

Adjoint of a 2×2 matrix [a b; c d] is [d −b; −c a].

For [ 1 2; 3 4 ]:

Adjoint =

[ 4 −2
−3 1 ]

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2.

[ 1 −1 2 ]
[ 2 3 5 ]
[ −2 0 1 ]

Step 1 — Compute cofactors (Cij):

C11 = 3, C12 = −12, C13 = 6
C21 = 1, C22 = 5, C23 = 2
C31 = −11, C32 = −1, C33 = 5

Step 2 — Adjoint = transpose of cofactor matrix

Cofactor matrix =
[ 3 −12 6 ]
[ 1 5 2 ]
[−11 −1 5 ]

Adjoint = transpose =
[ 3 1 −11 ]
[−12 5 −1 ]
[ 6 2 5 ]

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Verify A (adj A) = (adj A) A = |A| I

3. A =[ 2 3 ]
[ −4 −6 ]

Step 1 — Determinant

|A| = 2×(−6) − 3×(−4) = −12 + 12 = 0

Step 2 — Adjoint (classical adjoint = transpose of cofactor matrix)

For a 2×2 [a b; c d], adj = [d −b; −c a].

So adj A = [ −6 −3
4 2 ]

Step 3 — Multiply

A × adj A = [2 3; −4 −6] × [−6 −3; 4 2] = [0 0; 0 0]

adj A × A = [−6 −3; 4 2] × [2 3; −4 −6] = [0 0; 0 0]

det(A) I = 0 × I2 = [0 0; 0 0]

Thus A(adj A) = (adj A)A = |A| I (all equal to the zero 2×2 matrix).

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4. A =
[ 1 −1 2 ]
[ 3 0 −2 ]
[ 1 0 3 ]

Step 1 — Determinant

Compute |A| = 11

Step 2 — Adjoint (transpose of cofactor matrix)

adj A =
[ 0 3 2 ]
[−11 1 8 ]
[ 0 −1 3 ]

Step 3 — Multiply

A × adj A =
[11 0 0 ]
[0 11 0 ]
[0 0 11 ]

adj A × A =
[ 11 0 0 ]
[ 0 11 0 ]
[ 0 0 11 ]

det(A) I = 11 × I3 =
[ 11 0 0 ]
[ 0 11 0 ]
[ 0 0 11 ]

Thus A(adj A) = (adj A)A = |A| I (all equal to 11 times the 3×3 identity).

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Find the inverse of each of the matrices (if it exists)

5. [ 2 −2 ]
[ 4 3 ]

Determinant = 2×3 − (−2)×4 = 6 + 8 = 14 (nonzero → inverse exists)

Inverse (2×2 formula 1/det [d −b; −c a]):

A5^−1 = (1/14) [ 3 2
−4 2 ]

So

A5^−1 = [ 3/14 1/7
−2/7 1/7 ]

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6. [ −1 5 ]
[ −3 2 ]

Determinant = (−1)×2 − 5×(−3) = −2 + 15 = 13 (nonzero → inverse exists)

Inverse:

A6^−1 = (1/13) [ 2 −5
3 −1 ]

So

A6^−1 = [ 2/13 −5/13
3/13 −1/13 ]

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7. [ 1 2 3 ]
[ 0 2 4 ]
[ 0 0 5 ]

This is upper-triangular; determinant = product of diagonal = 1×2×5 = 10 (nonzero → inverse exists).

Compute inverse (solve for upper-triangular inverse or use formula):

A7^−1 =
[ 1 −1 1/5 ]
[ 0 1/2 −2/5 ]
[ 0 0 1/5 ]

(You can check A7 × A7^−1 = I3.)

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8. [ 1 0 0 ]
[ 3 3 0 ]
[ 5 2 −1 ]

Determinant = −3 (nonzero → inverse exists)

Inverse:

A8^−1 =
[ 1 0 0 ]
[ −1 1/3 0 ]
[ 3 2/3 −1 ]

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9. [ 2 1 3 ]
[ 4 −1 0 ]
[ −7 2 1 ]

Determinant = −3 (nonzero → inverse exists)

Inverse:

A9^−1 = (1/3) ×
[ 1 −5 −3 ]
[ 4 −23 −12 ]
[ −1 11 6 ]

Or written entrywise:

A9^−1 =
[ 1/3 −5/3 −1 ]
[ 4/3 −23/3 −4 ]
[ −1/3 11/3 2 ]

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10. [1 −1 2 ]
[0 2 −3 ]
[3 −2 4 ]

Determinant = −1 (nonzero → inverse exists)

Inverse:

A10^−1 =
[ −2 0 1 ]
[ 9 2 −3 ]
[ 6 1 −2 ]

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11. Find the inverse of the matrix

[ 1 0 0 ]
[ 0 cos α sin α ]
[ 0 sin α −cos α ]

First, find determinant.

|A| = 1 × determinant of
[ cos α sin α
sin α −cos α ]

= 1 × ( −cos²α − sin²α )
= −(cos²α + sin²α)
= −1

So inverse exists.

Now find adjoint.

Cofactors:

C11 = −1
C12 = 0
C13 = 0

C21 = 0
C22 = −cos α
C23 = −sin α

C31 = 0
C32 = −sin α
C33 = cos α

Adjoint (transpose of cofactor matrix):

adj A =
[ −1 0 0 ]
[ 0 −cos α −sin α ]
[ 0 −sin α cos α ]

Inverse:

A⁻¹ = (1/|A|) adj A = −1 × adj A

So,

A⁻¹ =
[ 1 0 0 ]
[ 0 cos α sin α ]
[ 0 sin α −cos α ]

Hence, A⁻¹ = A

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12.

A = [ 3 7 ]
[ 2 5 ]

B = [ 6 8 ]
[ 7 9 ]

Verify that (AB)⁻¹ = B⁻¹ A⁻¹

Step 1: Find AB

AB =
[ 3 7 ] [ 6 8 ]
[ 2 5 ] [ 7 9 ]

AB =
[ 67 87 ]
[ 47 61 ]

Step 2: Find (AB)⁻¹

|AB| = 67×61 − 87×47
= 4087 − 4089
= −2

adj(AB) =
[ 61 −87 ]
[ −47 67 ]

(AB)⁻¹ = (1/−2) adj(AB)

(AB)⁻¹ =
[ −61/2 87/2 ]
[ 47/2 −67/2 ]

Step 3: Find A⁻¹

|A| = 3×5 − 7×2 = 15 − 14 = 1

A⁻¹ =
[ 5 −7 ]
[ −2 3 ]

Step 4: Find B⁻¹

|B| = 6×9 − 8×7 = 54 − 56 = −2

B⁻¹ = (1/−2)
[ 9 −8 ]
[ −7 6 ]

B⁻¹ =
[ −9/2 4 ]
[ 7/2 −3 ]

Step 5: Find B⁻¹ A⁻¹

B⁻¹ A⁻¹ =
[ −9/2 4 ] [ 5 −7 ]
[ 7/2 −3 ] [ −2 3 ]

Multiplying,

B⁻¹ A⁻¹ =
[ −61/2 87/2 ]
[ 47/2 −67/2 ]

Conclusion:

(AB)⁻¹ =
[ −61/2 87/2 ]
[ 47/2 −67/2 ]

B⁻¹ A⁻¹ =
[ −61/2 87/2 ]
[ 47/2 −67/2 ]

Hence,

(AB)⁻¹ = B⁻¹ A⁻¹

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13: If A =
[ 3 1 ]
[ −1 2 ]

Show that A² − 5A + 7I = O. Hence find A⁻¹.

Step 1: Find A²

A² =
[ 3 1 ] [ 3 1 ]
[ −1 2 ] [ −1 2 ]

A² =
[ 8 5 ]
[ −5 3 ]

Step 2: Find −5A

−5A =
[ −15 −5 ]
[ 5 −10 ]

Step 3: Find 7I

7I =
[ 7 0 ]
[ 0 7 ]

Step 4: Compute A² − 5A + 7I

A² − 5A =
[ 8 5 ] [ −15 −5 ]
[ −5 3 ] + [ 5 −10 ]

=
[ −7 0 ]
[ 0 −7 ]

Now add 7I:

[ −7 0 ] + [ 7 0 ] = [ 0 0 ]
[ 0 −7 ] [ 0 7 ] [ 0 0 ]

Hence,

A² − 5A + 7I = O

Hence find A⁻¹

From
A² − 5A + 7I = O

Multiply both sides by A⁻¹:

A − 5I + 7A⁻¹ = O

So,

7A⁻¹ = 5I − A

A⁻¹ = (1/7)(5I − A)

5I − A =
[ 5 0 ] − [ 3 1 ] = [ 2 −1 ]
[ 0 5 ] [ −1 2 ] [ 1 3 ]

Therefore,

A⁻¹ = (1/7)
[ 2 −1 ]
[ 1 3 ]

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14. For the matrix A =
[ 3 2 ]
[ 1 1 ]

Find the numbers a and b such that A² + aA + bI = O.

Step 1: Find A²

A² =
[ 3 2 ] [ 3 2 ]
[ 1 1 ] [ 1 1 ]

A² =
[ 11 8 ]
[ 4 3 ]

Step 2: Write A² + aA + bI = O

A² + aA + bI =
[ 11 8 ] + a[ 3 2 ] + b[ 1 0 ]
[ 4 3 ] [ 1 1 ] [ 0 1 ]

=
[ 11 + 3a + b 8 + 2a ]
[ 4 + a 3 + a + b ]

Step 3: Equate each entry to zero

11 + 3a + b = 0 …… (1)
8 + 2a = 0 …… (2)
4 + a = 0 …… (3)
3 + a + b = 0 …… (4)

From (3):
a = −4

Substitute a = −4 in (2):
8 + 2(−4) = 0

Substitute a = −4 in (1):
11 − 12 + b = 0
b = 1

Check in (4):
3 − 4 + 1 = 0

Final Answer:

a = −4
b = 1

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15. For the matrix A =

[ 1 1 1 ]
[ 1 2 −3 ]
[ 2 −1 3 ]

Show that A³ − 6A² + 5A + 11I = O. Hence find A⁻¹.

Given:
A³ − 6A² + 5A + 11I = O

Multiply both sides by A⁻¹:

A² − 6A + 5I + 11A⁻¹ = O

So,

11A⁻¹ = −(A² − 6A + 5I)

Now find A².

A² =
[ 4 2 1 ]
[−3 8 −14 ]
[ 7 −3 14 ]

Find A² − 6A + 5I:

A² − 6A + 5I =
[ 3 −4 −5 ]
[−9 1 4 ]
[ −5 3 1 ]

Hence,

A⁻¹ = −(1/11) ×
[ 3 −4 −5 ]
[ −9 1 4 ]
[ −5 3 1 ]

So,

A⁻¹ = (1/11)
[ −3 4 5 ]
[ 9 −1 −4 ]
[ 5 −3 −1 ]

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16. If A =

[ 2 −1 1 ]
[ −1 2 −1 ]
[ 1 −1 2 ]

Verify that A³ − 6A² + 9A − 4I = O and hence find A⁻¹.

Given:
A³ − 6A² + 9A − 4I = O

Multiply both sides by A⁻¹:

A² − 6A + 9I − 4A⁻¹ = O

So,

4A⁻¹ = A² − 6A + 9I

Find A².

A² =
[ 6 −5 5 ]
[−5 6 −5 ]
[ 5 −5 6 ]

Now compute A² − 6A + 9I:

A² − 6A + 9I =
[ 3 1 −1 ]
[ 1 3 1 ]
[ −1 1 3 ]

Hence,

A⁻¹ = (1/4)
[ 3 1 −1 ]
[ 1 3 1 ]
[−1 1 3 ]

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17. Let A be a nonsingular square matrix of order 3 × 3. Then | adj A | is equal to

Property used:
For an n × n matrix A,

| adj A | = |A|^(n−1)

Here, n = 3

So,

| adj A | = |A|^(3−1)
| adj A | = |A|²

Correct answer: (B) |A|²

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18. If A is an invertible matrix of order 2, then det(A⁻¹) is equal to

Property used:

det(A⁻¹) = 1 / det(A)

Correct answer: (B) 1 / det(A)

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Exercise 4.5

Examine the consistency of the system of equations

1.

x + 2y = 2
2x + 3y = 3

Solution:
Multiply the first equation by 2:
2x + 4y = 4

Subtract the second equation from this:
(2x + 4y) − (2x + 3y) = 4 − 3
y = 1

Substitute y = 1 in x + 2y = 2:
x + 2 = 2
x = 0

Conclusion:
The system is consistent and has a unique solution.
Solution: x = 0, y = 1

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2.

2x − y = 5
x + y = 4

Solution:
Add the two equations:
3x = 9
x = 3

Substitute x = 3 in x + y = 4:
3 + y = 4
y = 1

Conclusion:
The system is consistent and has a unique solution.
Solution: x = 3, y = 1

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3.

x + 3y = 5
2x + 6y = 8

Solution:
Multiply the first equation by 2:
2x + 6y = 10

But the second equation is:
2x + 6y = 8

Since 10 ≠ 8, the equations are contradictory.

Conclusion:
The system is inconsistent.
No solution exists.

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4.

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Solution:
First equation multiplied by 2:
2x + 2y + 2z = 2

Subtract from the second equation:
(2x + 3y + 2z) − (2x + 2y + 2z) = 2 − 2
y = 0

Substitute y = 0 in x + y + z = 1:
x + z = 1

Third equation becomes:
a(x + y + 2z) = 4
a(x + 2z) = 4

Using x + z = 1, we get:
x + 2z = 1 + z

So a(1 + z) = 4

This gives a unique solution only when a ≠ 0.

Conclusion:
The system is consistent for a ≠ 0.
For a = 0, the system is inconsistent.

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5.

3x − y − 2z = 2
2y − z = −1
3x − 5y = 3

Solution:
From the second equation:
z = 2y + 1

Substitute z in the first equation:
3x − y − 2(2y + 1) = 2
3x − y − 4y − 2 = 2
3x − 5y = 4

But the third equation is:
3x − 5y = 3

Since 4 ≠ 3, the equations contradict.

Conclusion:
The system is inconsistent.
No solution exists.

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6.

5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1

Solution:

Subtract (1) from (3):
y − 2z = 6 …(4)

Subtract (1) from (2):
−3x + 4y + z = −3 …(5)

From (4):
y = 6 + 2z

Substitute in (5):
−3x + 4(6 + 2z) + z = −3
−3x + 9z = −27
x = 9 + 3z

Substitute x and y in (1):
5(9 + 3z) − (6 + 2z) + 4z = 5
39 + 17z = 5
z = −2

Then,
y = 2, x = 3

Conclusion:
The system is consistent with unique solution
x = 3, y = 2, z = −2

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Solve system of linear equations, using matrix method, in Exercises 7 to 14.

7.

5x + 2y = 4
7x + 3y = 5

Answer:
x = 2, y = −3

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8.

2x − y = −2
3x + 4y = 3

Answer:
x = −5/11, y = 12/11

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9.

4x − 3y = 3
3x − 5y = 7

Answer:
x = −6/11, y = −19/11

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10.

5x + 2y = 3
3x + 2y = 5

Answer:
x = −1, y = 4

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11.

2x + y + z = 1
x − 2y − z = 3/2
3y − 5z = 9

Answer:
x = 1, y = 1/2, z = −3/2

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12.

x − y + z = 4
2x + y − 3z = 0
x + y + z = 2

Answer:
x = 2, y = −1, z = 1

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13.

2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3

Answer:
x = 1, y = 2, z = −1

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14.

x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Answer:
x = 2, y = 1, z = 3

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15.

If
A =
2 −3 5
3 2 −4
1 1 −2

find A⁻¹. Using A⁻¹ solve the system of equations

2x − 3y + 5z = 11
3x + 2y − 4z = −5
x + y − 2z = −3

Solution:

Determinant of A = −1

Hence A⁻¹ exists.

A⁻¹ =
0 1 −2
−2 9 −23
−1 5 −13

Using X = A⁻¹B, we obtain

x = 1
y = 2
z = 3

Verification:

2x − 3y + 5z = 2 − 6 + 15 = 11
3x + 2y − 4z = 3 + 4 – 12 = −5
x + y − 2z = 1 + 2 − 6 = −3

Answer:

x = 1, y = 2, z = 3

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16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Given:

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60.
The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90.
The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70.

Find the cost of each item per kg by matrix method.

Solution

Let
Cost of onion per kg = x
Cost of wheat per kg = y
Cost of rice per kg = z

Then the given information gives the following equations:

4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70

These can be written in matrix form as:

Coefficient matrix
4 3 2
2 4 6
6 2 3

Variable matrix
x
y
z

Constant matrix
60
90
70

Solving this system by matrix method, we obtain:

x = 5
y = 8
z = 8

Verification

4(5) + 3(8) + 2(8) = 20 + 24 + 16 = 60
2(5) + 4(8) + 6(8) = 10 + 32 + 48 = 90
6(5) + 2(8) + 3(8) = 30 + 16 + 24 = 70

Answer

Cost of onion per kg = ₹5
Cost of wheat per kg = ₹8
Cost of rice per kg = ₹8

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