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Class-XII-Maths Determinants
1 Practice more on Determinants www.embibe.com
CBSE NCERT Solutions for Class 12 Maths Chapter 04
Back of Chapter Questions
Exercise 𝟒. 𝟏
1. Evaluate the determinant |2 4
−5 −1|
Solution:
Given determinant is |2 4
−5 −1|
Expanding along 𝑅1, we get
= (2 × (−1)) − (4 × (−5))
= −2 + 20 = 18
Hence, |2 4
−5 −1| = 18
2. Evaluate the determinants
(i) |cos𝜃 − sin𝜃sin𝜃 cos 𝜃
|
(ii) |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1
|
Solution:
i) Given determinant is |cos 𝜃 −sin 𝜃sin 𝜃 cos 𝜃
|
Now, |cos 𝜃 −sin 𝜃sin 𝜃 cos 𝜃
|
= (cos 𝜃 × cos 𝜃) − (sin𝜃 × (− sin 𝜃))
= cos2 𝜃 + sin2 𝜃 = 1
Hence |cos 𝜃 − sin𝜃sin 𝜃 cos𝜃
| = 1
ii) Given determinant is |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1
|
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Now, |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1
|
= (𝑥2 − 𝑥 + 1) × (𝑥 + 1) − (𝑥 − 1) × (𝑥 + 1)
= 𝑥3 + 𝑥2 − 𝑥2 − 𝑥 + 𝑥 + 1 − (𝑥2 − 1)
= 𝑥3 − 𝑥2 + 2
Hence, |𝑥2 − 𝑥 + 1 𝑥 − 1𝑥 + 1 𝑥 + 1
| = 𝑥3 − 𝑥2 + 2
3. If 𝐴 = [1 24 2
], then show that |2𝐴| = 4|𝐴|
Solution:
Given that 𝐴 = [1 24 2
]
Now, 2𝐴 = [2 × 1 2 × 22 × 4 2 × 2
]
= [2 48 4
]
|2𝐴| = |2 48 4
|
Expanding along 𝑅1, we get
= 2 × 4 − 4 × 8 = 8 − 32 = −24 …(i)
And 4|𝐴| = 4 |1 24 2
|
Expanding along 𝑅1, we get
= 4(1 × 2 − 2 × 4) = 4(−6) = −24 …(ii)
From the equation (i) and (ii),
we get |2𝐴| = 4|𝐴|
Hence proved.
4. If 𝐴 = [1 0 10 1 20 0 4
], then show that |3𝐴| = 27|𝐴|
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Solution:
Given that 𝐴 = [1 0 10 1 20 0 4
]
⇒ 3𝐴 = [3 0 30 3 60 0 12
]
Now, |3𝐴| = |3 0 30 3 60 0 12
|
Expanding along 𝑅1, we get
= 3(36 − 0) − 0(0 − 0) + 3(0 − 0) = 108 …(i)
And 27|𝐴| = 27 |1 0 10 1 20 0 4
|
Expanding along 𝑅1, we get
= 27{1(4 − 0) − 0(0 − 0) + 1(0 − 0)} = 27(4) = 108 …(ii)
From the equation (i) and (ii),
we get |3𝐴| = 27|𝐴|
Hence proved.
5. Evaluate the determinants:
(i) |3 −1 −20 0 −13 −5 0
|
(ii) |3 −4 51 1 −22 3 1
|
(iii) |0 1 2
−1 0 −3−2 3 0
|
(iv) |2 −1 −20 2 −13 −5 0
|
Solution:
(i) Given determinant is |3 −1 −20 0 −13 −5 0
|
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Now, expanding the determinant along 𝑅1, we get
|3 −1 −20 0 −13 −5 0
| = 3(0 − 5) + 1(0 + 3) − 2(0 − 0) = −15 + 3 − 0 = −12
Hence |3 −1 −20 0 −13 −5 0
| = −12
Given determinant is |3 −4 51 1 −22 3 1
|
Now, expanding the determinant along 𝑅1, we get
|3 −4 51 1 −22 3 1
| = 3(1 + 6) + 4(1 + 4) + 5(3 − 2) = 21 + 20 + 5 = 46
Hence |3 −4 51 1 −22 3 1
| = 46
(iii) Given determinant is |0 1 2
−1 0 −3−2 3 0
|
Now, expanding the determinant along 𝑅1, we get
|0 1 2
−1 0 −3−2 3 0
| = 0(0 + 9) − 1(0 − 6) + 2(−3 − 0) = 0 + 6 − 6
So, the given determinant |0 1 2
−1 0 −3−2 3 0
| = 0
Given determinant is |2 −1 −20 2 −13 −5 0
|
Now, expanding the determinant along 𝑅1, we get
|2 −1 −20 2 −13 −5 0
| = 2(0 − 5) + 1(0 + 3) − 2(0 − 6) = −10 + 3 + 12 = 5
Hence, |2 −1 −20 2 −13 −5 0
| = 5
6. If 𝐴 = [1 1 −22 1 −35 4 −9
], find |𝐴|.
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Solution:
Given that 𝐴 = [1 1 −22 1 −35 4 −9
]
Now, |𝐴| = |1 1 −22 1 −35 4 −9
|
Expanding the determinant along 𝑅1, we get
= 1(−9 + 12) − 1(−18 + 15) − 2(8 − 5) = 3 + 3 − 6 = 0
Hence, |𝐴| = 0
7. Find values of 𝑥, if
(i) |2 45 1
| = |2𝑥 46 𝑥
|
(ii) |2 34 5
| = |𝑥 32𝑥 5
|
Solution:
(i) Given that |2 45 1
| = |2𝑥 46 𝑥
|
⇒ 2 − 20 = 2𝑥2 − 24
⇒ 𝑥2 = 3
⇒ 𝑥 = ±√3
Hence, the values of 𝑥 are ± √3
(ii) Given that |2 34 5
| = |𝑥 32𝑥 5
|
⇒ 10 − 12 = 5𝑥 − 6𝑥
⇒ −2 = −𝑥
⇒ 𝑥 = 2
Hence, the value of 𝑥 is 2
8. If |𝑥 218 𝑥
| = |6 218 6
|, then 𝑥 is equal to:
(A) 6
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(B) ±6
(C) −6
(D) 0
Solution:
(B)
Given that |𝑥 218 𝑥
| = |6 218 6
|
⇒ 𝑥2 − 36 = 36 − 36
⇒ 𝑥2 = 36
⇒ 𝑥 = ±6
Hence, the option (𝐵) is correct.
Exercise 𝟒. 𝟐
Using the property of determinants and without expanding, prove that:
1. |𝑥 𝑎 𝑥 + 𝑎𝑦 𝑏 𝑦 + 𝑏𝑧 𝑐 𝑧 + 𝑐
| = 0
Solution:
LHS = |𝑥 𝑎 𝑥 + 𝑎𝑦 𝑏 𝑦 + 𝑏𝑧 𝑐 𝑧 + 𝑐
|
= |𝑥 + 𝑎 𝑎 𝑥 + 𝑎𝑦 + 𝑏 𝑏 𝑦 + 𝑏𝑧 + 𝑐 𝑐 𝑧 + 𝑐
| [Applying C1 → 𝐶1 + 𝐶2]
= 0 = RHS [∵ 𝐶1 = 𝐶3]
Hence, LHS = RHS
2. |𝑎 − 𝑏 𝑏 − 𝑐 𝑐 − 𝑎𝑏 − 𝑐 𝑐 − 𝑎 𝑎 − 𝑏𝑐 − 𝑎 𝑎 − 𝑏 𝑏 − 𝑐
| = 0
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Solution:
LHS = |𝑎 − 𝑏 𝑏 − 𝑐 𝑐 − 𝑎𝑏 − 𝑐 𝑐 − 𝑎 𝑎 − 𝑏𝑐 − 𝑎 𝑎 − 𝑏 𝑏 − 𝑐
|
= |0 𝑏 − 𝑐 𝑐 − 𝑎0 𝑐 − 𝑎 𝑎 − 𝑏0 𝑎 − 𝑏 𝑏 − 𝑐
| [Applying 𝐶1 → 𝐶1 + 𝐶2 + 𝐶3]
= 0 = RHS [∵ In column 𝐶1 every element is zero. ]
Hence, LHS = RHS
3. |2 7 653 8 755 9 86
| = 0
Solution:
LHS = |2 7 653 8 755 9 86
|
= |2 7 633 8 725 9 81
| [Applying C3 → 𝐶3 − 𝐶1]
= 9 [2 7 73 8 85 9 9
] [Taking common 9 from C3]
= 0 = RHS [∵ 𝐶2 = 𝐶3]
Hence, LHS = RHS
4. |1 𝑏𝑐 𝑎(𝑏 + 𝑐)
1 𝑐𝑎 𝑏(𝑐 + 𝑎)
1 𝑎𝑏 𝑐(𝑎 + 𝑏)| = 0
Solution:
𝐿𝐻𝑆 = |
1 𝑏𝑐 𝑎(𝑏 + 𝑐)
1 𝑐𝑎 𝑏(𝑐 + 𝑎)
1 𝑎𝑏 𝑐(𝑎 + 𝑏)|
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= |1 𝑏𝑐 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎1 𝑐𝑎 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎1 𝑎𝑏 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎
| [Applying C3 → 𝐶3 + 𝐶2]
= (𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) |1 𝑏𝑐 11 𝑐𝑎 11 𝑎𝑏 1
| [Taking 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎 as common from 𝐶3]
= 0 = 𝑅𝐻𝑆 [∵ 𝐶1 = 𝐶3]
Hence, LHS = RHS
5. |𝑏 + 𝑐 𝑞 + 𝑟 𝑦 + 𝑧𝑐 + 𝑎 𝑟 + 𝑝 𝑧 + 𝑥𝑎 + 𝑏 𝑝 + 𝑞 𝑥 + 𝑦
| = 2 |𝑎 𝑝 𝑥𝑏 𝑞 𝑦𝑐 𝑟 𝑧
|
Solution:
𝐿𝐻𝑆 = |
𝑏 + 𝑐 𝑞 + 𝑟 𝑦 + 𝑧𝑐 + 𝑎 𝑟 + 𝑝 𝑧 + 𝑥𝑎 + 𝑏 𝑝 + 𝑞 𝑥 + 𝑦
|
= |
2𝑐 2𝑟 2𝑧𝑐 + 𝑎 𝑟 + 𝑝 𝑧 + 𝑥𝑎 + 𝑏 𝑝 + 𝑞 𝑥 + 𝑦
| [Applying R1 → 𝑅1 + 𝑅2 − 𝑅3]
= 2 |
𝑐 𝑟 𝑧𝑐 + 𝑎 𝑟 + 𝑝 𝑧 + 𝑥𝑎 + 𝑏 𝑝 + 𝑞 𝑥 + 𝑦
| [Taking 2 as common from R1]
= 2 |
𝑐 𝑟 𝑧𝑎 𝑝 𝑥
𝑎 + 𝑏 𝑝 + 𝑞 𝑥 + 𝑦| [Applying 𝑅2 → 𝑅2 − 𝑅1]
= 2 |
𝑐 𝑟 𝑧𝑎 𝑝 𝑥𝑏 𝑞 𝑦
| [Applying 𝑅3 → 𝑅3 − 𝑅2]
= −2 |
𝑎 𝑝 𝑥𝑐 𝑟 𝑧𝑏 𝑞 𝑦
| [Interchanging 𝑅1 ⟷ 𝑅2]
= 2 |𝑎 𝑝 𝑥𝑏 𝑞 𝑦𝑐 𝑟 𝑧
| = 𝑅𝐻𝑆 [Interchanging 𝑅2 ⟷ 𝑅3]
Hence, LHS = RHS
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6. |0 𝑎 −𝑏
−𝑎 0 −𝑐𝑏 𝑐 0
| = 0
Solution:
𝐿𝐻𝑆 = |0 𝑎 −𝑏
−𝑎 0 −𝑐𝑏 𝑐 0
|
= |0 𝑎 −𝑏
−𝑎𝑏 0 −𝑏𝑐𝑎𝑏 𝑎𝑐 0
| [Applying 𝑅2 → 𝑏𝑅2 and R3 → 𝑎𝑅3]
= |0 𝑎 −𝑏0 𝑎𝑐 −𝑏𝑐𝑎𝑏 𝑎𝑐 0
| [Applying 𝑅2 → 𝑅2 + 𝑅3]
= 𝑎𝑏(−𝑎𝑏𝑐 + 𝑎𝑏𝑐) [Expanding along 𝐶1]
= 𝑎𝑏(0) = 0 = 𝑅𝐻𝑆
Hence, LHS = RHS
7. |−𝑎2 𝑎𝑏 𝑎𝑐𝑏𝑎 −𝑏2 𝑏𝑐𝑐𝑎 𝑐𝑏 −𝑐2
| = 4𝑎2𝑏2𝑐2
Solution:
𝐿𝐻𝑆 = |−𝑎2 𝑎𝑏 𝑎𝑐𝑏𝑎 −𝑏2 𝑏𝑐𝑐𝑎 𝑐𝑏 −𝑐2
|
= 𝑎𝑏𝑐 |−𝑎 𝑎 𝑎𝑏 −𝑏 𝑏𝑐 𝑐 −𝑐
| [Taking a, b, c as common from C1, 𝐶2, 𝐶3 respectively]
= 𝑎2𝑏2𝑐2 |−1 1 11 −1 11 1 −1
| [Taking a, b, c as common from R1, 𝑅2, 𝑅3 respectively]
= 𝑎2𝑏2𝑐2 |0 1 10 −1 12 1 −1
| [Applying 𝐶1 → 𝐶1 + 𝐶2]
= 𝑎2𝑏2𝑐2{2(1 + 1)} [Expanding along 𝐶1]
= 4𝑎2𝑏2𝑐2 = 𝑅𝐻𝑆
Hence, LHS = RHS
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8. Using properties of determinants, show that
(i) |1 𝑎 𝑎2
1 𝑏 𝑏2
1 𝑐 𝑐2
| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)
(ii) |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3
| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐)
Solution:
(i) 𝐿𝐻𝑆 = |1 𝑎 𝑎2
1 𝑏 𝑏2
1 𝑐 𝑐2
|
= |0 𝑎 − 𝑏 𝑎2 − 𝑏2
0 𝑏 − 𝑐 𝑏2 − 𝑐2
1 𝑐 𝑐2
| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (𝑎 − 𝑏)(𝑏 − 𝑐) |0 1 𝑎 + 𝑏0 1 𝑏 + 𝑐1 𝑐 𝑐2
| [Taking common a − b from R1and b − c from R2] ]
= (𝑎 − 𝑏)(𝑏 − 𝑐){1(𝑏 + 𝑐 − 𝑎 − 𝑏)} [Expanding along 𝐶1]
= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎) = 𝑅𝐻𝑆
Hence, LHS = RHS
(ii) |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3
| = (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐)
𝐿𝐻𝑆 = |1 1 1𝑎 𝑏 𝑐𝑎3 𝑏3 𝑐3
|
= |0 0 1
𝑎 − 𝑏 𝑏 − 𝑐 𝑐𝑎3 − 𝑏3 𝑏3 − 𝑐3 𝑐3
| [By C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]
= (𝑎 − 𝑏)(𝑏 − 𝑐) |0 0 11 1 𝑐
𝑎2 + 𝑎𝑏 + 𝑏2 𝑏2 + 𝑏𝑐 + 𝑐2 𝑐3| [Taking common a − b from C1and b − c from C2]
= (𝑎 − 𝑏)(𝑏 − 𝑐){1(𝑏2 + 𝑏𝑐 + 𝑐2) − (𝑎2 + 𝑎𝑏 + 𝑏2)} [Expanding along 𝑅1]
= (𝑎 − 𝑏)(𝑏 − 𝑐){𝑐2 − 𝑎2 + 𝑏𝑐 − 𝑎𝑏}
= (𝑎 − 𝑏)(𝑏 − 𝑐){(𝑐 − 𝑎)(𝑐 + 𝑎) + 𝑏(𝑐 − 𝑎)}
= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎){𝑐 + 𝑎 + 𝑏}
= (𝑎 − 𝑏)(𝑏 − 𝑐)(𝑐 − 𝑎)(𝑎 + 𝑏 + 𝑐) = 𝑅𝐻𝑆
Hence, LHS = RHS
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9. Using properties of determinants, show that
|
𝑥 𝑥2 𝑦𝑧
𝑦 𝑦2 𝑧𝑥
𝑧 𝑧2 𝑥𝑦
| = (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥)
Solution:
𝐿𝐻𝑆 = |
𝑥 𝑥2 𝑦𝑧
𝑦 𝑦2 𝑧𝑥
𝑧 𝑧2 𝑥𝑦
|
= |
𝑥2 𝑥3 𝑥𝑦𝑧
𝑦2 𝑦3 𝑥𝑦𝑧
𝑧2 𝑧3 𝑥𝑦𝑧
| [Applying 𝑅1 → 𝑥𝑅1, 𝑅2 → 𝑦𝑅2, 𝑅3 → 𝑧𝑅3]
= 𝑥𝑦𝑧 |𝑥2 𝑥3 1𝑦2 𝑦3 1
𝑧2 𝑧3 1
| [Taking xyz as common from C3]
= 𝑥𝑦𝑧 |𝑥2 − 𝑦2 𝑥3 − 𝑦3 0
𝑦2 − 𝑧2 𝑦3 − 𝑧3 0
𝑧2 𝑧3 1
| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧) |𝑥 + 𝑦 𝑥2 + 𝑥𝑦 + 𝑦2 0
𝑦 + 𝑧 𝑦2 + 𝑦𝑧 + 𝑧2 0
𝑧2 𝑧3 1
| [Taking 𝑥 − 𝑦 as common from R1and y − z from R2]
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){(𝑥 + 𝑦)(𝑦2 + 𝑦𝑧 + 𝑧2) − (𝑦 + 𝑧)(𝑥2 + 𝑥𝑦 + 𝑦2)} [Expanding along C3]
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑦2 + 𝑥𝑦𝑧 + 𝑥𝑧2 + 𝑦3 + 𝑦2𝑧 + 𝑦𝑧2 − (𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 + 𝑥2𝑧 +
𝑥𝑦𝑧 + 𝑦2𝑧)}
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧2 + 𝑦𝑧2 − 𝑥2𝑦 − 𝑥2𝑧}
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧2 − 𝑥2𝑧 + 𝑦𝑧2 − 𝑥2𝑦}
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧){𝑥𝑧(𝑧 − 𝑥) + 𝑦(𝑧2 − 𝑥2)}
= 𝑥𝑦𝑧(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥){𝑥𝑧 + 𝑦(𝑧 + 𝑥)}
= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)(𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 𝑅𝐻𝑆
Hence, LHS = RHS
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10. Using properties of determinants, show that
(i) |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4
| = (5𝑥 + 4)(4 − 𝑥)2
(ii) |
𝑦 + 𝑘 𝑦 𝑦𝑦 𝑦 + 𝑘 𝑦𝑦 𝑦 𝑦 + 𝑘
| = 𝑘2(3𝑦 + 𝑘)
Solution:
(i) |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4
| = (5𝑥 + 4)(4 − 𝑥)2
𝐿𝐻𝑆 = |𝑥 + 4 2𝑥 2𝑥2𝑥 𝑥 + 4 2𝑥2𝑥 2𝑥 𝑥 + 4
|
= |5𝑥 + 4 2𝑥 2𝑥5𝑥 + 4 𝑥 + 4 2𝑥5𝑥 + 4 2𝑥 𝑥 + 4
| [Applying 𝐶1 → 𝐶1 + 𝐶2 + 𝐶3]
= (5𝑥 + 4) |1 2𝑥 2𝑥1 𝑥 + 4 2𝑥1 2𝑥 𝑥 + 4
| [Taking 5x + 4 as common from C1]
= (5𝑥 + 4) |0 𝑥 − 4 00 4 − 𝑥 𝑥 − 41 2𝑥 𝑥 + 4
| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (5𝑥 + 4){(𝑥 − 4)(𝑥 − 4) − (4 − 𝑥)0} [Expanding along C1]
= (5𝑥 + 4)(4 − 𝑥)2 = 𝑅𝐻𝑆
Hence, LHS = RHS
(iii) |
𝑦 + 𝑘 𝑦 𝑦𝑦 𝑦 + 𝑘 𝑦𝑦 𝑦 𝑦 + 𝑘
| = 𝑘2(3𝑦 + 𝑘)
𝐿𝐻𝑆 = |
𝑦 + 𝑘 𝑦 𝑦𝑦 𝑦 + 𝑘 𝑦𝑦 𝑦 𝑦 + 𝑘
|
= |
3𝑦 + 𝑘 𝑦 𝑦3𝑦 + 𝑘 𝑦 + 𝑘 𝑦3𝑦 + 𝑘 𝑦 𝑦 + 𝑘
| [Applying C1 → 𝐶1 + 𝐶2 + 𝐶3]
= (3𝑦 + 𝑘) |
1 𝑦 𝑦1 𝑦 + 𝑘 𝑦1 𝑦 𝑦 + 𝑘
| [Taking 3y + k as common from C1]
= (3𝑦 + 𝑘) |0 −𝑘 00 𝑘 −𝑘1 𝑦 𝑦 + 𝑘
| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
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= (3𝑦 + 𝑘){(−𝑘)(−𝑘) − (𝑘)0} [Expanding along C1]
= (3𝑦 + 𝑘)𝑘2 = 𝑅𝐻𝑆
Hence, LHS = RHS
11. Using properties of determinants, show that
(i) |𝑎 − 𝑏 − 𝑐 2𝑎 2𝑎
2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏
| = (𝑎 + 𝑏 + 𝑐)3
(ii) |
𝑥 + 𝑦 + 2𝑧 𝑥 𝑦𝑧 𝑦 + 𝑧 + 2𝑥 𝑦𝑧 𝑥 𝑧 + 𝑥 + 2𝑦
| = 2(𝑥 + 𝑦 + 𝑧)3
Solution:
(i) 𝐿𝐻𝑆 = |𝑎 − 𝑏 − 𝑐 2𝑎 2𝑎
2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏
|
= |𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐
2𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏
| [Applying R1 → 𝑅1 + 𝑅2 + 𝑅3]
= (𝑎 + 𝑏 + 𝑐) |1 1 12𝑏 𝑏 − 𝑐 − 𝑎 2𝑏2𝑐 2𝑐 𝑐 − 𝑎 − 𝑏
| [Taking a + b + c as commong from R1]
= (𝑎 + 𝑏 + 𝑐) |0 0 1
𝑎 + 𝑏 + 𝑐 −𝑎 − 𝑏 − 𝑐 2𝑏0 𝑎 + 𝑏 + 𝑐 𝑐 − 𝑎 − 𝑏
| [By C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]
= (𝑎 + 𝑏 + 𝑐){(𝑎 + 𝑏 + 𝑐)2 − 0} [Expanding along R1]
= (𝑎 + 𝑏 + 𝑐)3 = 𝑅𝐻𝑆
(𝑖𝑖) |
𝑥 + 𝑦 + 2𝑧 𝑥 𝑦𝑧 𝑦 + 𝑧 + 2𝑥 𝑦𝑧 𝑥 𝑧 + 𝑥 + 2𝑦
| = 2(𝑥 + 𝑦 + 𝑧)3
𝐿𝐻𝑆 = |
𝑥 + 𝑦 + 2𝑧 𝑥 𝑦𝑧 𝑦 + 𝑧 + 2𝑥 𝑦𝑧 𝑥 𝑧 + 𝑥 + 2𝑦
|
= |
2(𝑥 + 𝑦 + 𝑧) 𝑥 𝑦
2(𝑥 + 𝑦 + 𝑧) 𝑦 + 𝑧 + 2𝑥 𝑦
2(𝑥 + 𝑦 + 𝑧) 𝑥 𝑧 + 𝑥 + 2𝑦
| [Applying C1 → C1 + C2 + C3]
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= 2(𝑥 + 𝑦 + 𝑧) |
1 𝑥 𝑦1 𝑦 + 𝑧 + 2𝑥 𝑦1 𝑥 𝑧 + 𝑥 + 2𝑦
| [Taking 2(𝑥 + 𝑦 + 𝑧) common from C1]
= 2(𝑥 + 𝑦 + 𝑧) |
0 −(𝑥 + 𝑦 + 𝑧) 0
0 𝑥 + 𝑦 + 𝑧 −(𝑥 + 𝑦 + 𝑧)1 𝑥 𝑧 + 𝑥 + 2𝑦
| [By R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= 2(𝑥 + 𝑦 + 𝑧){(𝑥 + 𝑦 + 𝑧)2 − 0} [Expanding along C1]
= 2(𝑥 + 𝑦 + 𝑧)3 = 𝑅𝐻𝑆
Hence, LHS = RHS
12. Using properties of determinants, show that
|1 𝑥 𝑥2
𝑥2 1 𝑥𝑥 𝑥2 1
| = (1 − 𝑥3)2
Solution:
𝐿𝐻𝑆 = |1 𝑥 𝑥2
𝑥2 1 𝑥𝑥 𝑥2 1
|
= |1 + 𝑥 + 𝑥2 𝑥 𝑥2
1 + 𝑥 + 𝑥2 1 𝑥1 + 𝑥 + 𝑥2 𝑥2 1
| [Applying C1 → 𝐶1 + 𝐶2 + 𝐶3]
= (1 + 𝑥 + 𝑥2) |1 𝑥 𝑥2
1 1 𝑥1 𝑥2 1
| [Taking 1 + 𝑥 + 𝑥2 as common from C1]
= (1 + 𝑥 + 𝑥2) |0 𝑥 − 1 𝑥2 − 𝑥0 1 − 𝑥2 𝑥 − 11 𝑥2 1
| [Applying R1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (1 + 𝑥 + 𝑥2)(1 − 𝑥)2 |0 −1 −𝑥0 1 − 𝑥 −11 𝑥2 1
| [Taking 1 − 𝑥 as common from R1 and R2]
= (1 + 𝑥 + 𝑥2)(1 − 𝑥)2{1 + 𝑥(1 + 𝑥)} [Expanding along C1]
= (1 + 𝑥 + 𝑥2)(1 − 𝑥2)(1 + 𝑥 + 𝑥2) = (1 − 𝑥3)2 = 𝑅𝐻𝑆
Hence, LHS = RHS
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13. Using properties of determinants, show that
|1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑏
2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2𝑏 −2𝑎 1 − 𝑎2 − 𝑏2
| = (1 + 𝑎2 + 𝑏2)3
Solution:
𝐿𝐻𝑆 = |1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑏
2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2𝑏 −2𝑎 1 − 𝑎2 − 𝑏2
|
=1
𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 −2𝑎𝑏
2𝑎𝑏 1 − 𝑎2 + 𝑏2 2𝑎2
2𝑏 −2𝑎 𝑎 − 𝑎3 − 𝑎𝑏2
| [Applying C3 → 𝑎𝐶3]
=1
𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 0
2𝑎𝑏 1 − 𝑎2 + 𝑏2 1 + 𝑎2 + 𝑏2
2𝑏 −2𝑎 −𝑎 − 𝑎3 − 𝑎𝑏2
| [Applying C3 → 𝐶3 + 𝐶2]
=1 + 𝑎2 + 𝑏2
𝑎|1 + 𝑎2 − 𝑏2 2𝑎𝑏 0
2𝑎𝑏 1 − 𝑎2 + 𝑏2 12𝑏 −2𝑎 −𝑎
| [Taking 1 + a2 + 𝑏2 as common from 𝐶3]
=1+𝑎2+𝑏2
𝑎2 |1 + 𝑎2 − 𝑏2 2𝑎𝑏 0
2𝑎2𝑏 𝑎 − 𝑎3 + 𝑎𝑏2 𝑎2𝑏 −2𝑎 −𝑎
| [Applying 𝑅2 → 𝑎𝑅2]
=1+𝑎2+𝑏2
𝑎2 |1 + 𝑎2 − 𝑏2 2𝑎𝑏 02𝑎2𝑏 + 2𝑏 −𝑎 − 𝑎3 + 𝑎𝑏2 0
2𝑏 −2𝑎 −𝑎
| [Applying R2 → R2 + 𝑅3]
= (1 + 𝑎2 + 𝑏2) |1 + 𝑎2 − 𝑏2 2𝑏 02𝑎2𝑏 + 2𝑏 −1 − 𝑎2 + 𝑏2 0
2𝑏 −2 −1
| [Taking 𝑎 as common from C2 and C3]
= (1 + 𝑎2 + 𝑏2)(−1){(1 + 𝑎2 − 𝑏2)(−1 − 𝑎2 + 𝑏2) − 2𝑏(2𝑎2𝑏 + 2𝑏)} [Expanding along C3]
= −(1 + 𝑎2 + 𝑏2){−1 − 𝑎2 + 𝑏2 − 𝑎2 − 𝑎4 + 𝑎2𝑏2 + 𝑏2 + 𝑎2𝑏2 − 𝑏4 − 4𝑎2𝑏2 − 4𝑏2}
= (1 + 𝑎2 + 𝑏2){1 + 𝑎4 + 4 + 2𝑎2 + 2𝑎2𝑏2 + 2𝑏2}
= (1 + 𝑎2 + 𝑏2)(1 + 𝑎2 + 𝑏2)2 = (1 + 𝑎2 + 𝑏2)3 = 𝑅𝐻𝑆
Hence, LHS = RHS
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14. Using properties of determinants, show that
|𝑎2 + 1 𝑎𝑏 𝑎𝑐
𝑎𝑏 𝑏2 + 1 𝑏𝑐𝑐𝑎 𝑐𝑏 𝑐2 + 1
| = 1 + 𝑎2 + 𝑏2 + 𝑐2
Solution:
𝐿𝐻𝑆 = |𝑎2 + 1 𝑎𝑏 𝑎𝑐
𝑎𝑏 𝑏2 + 1 𝑏𝑐𝑐𝑎 𝑐𝑏 𝑐2 + 1
|
=1
𝑎𝑏𝑐|𝑎2 + 𝑎 𝑎2𝑏 𝑎2𝑐𝑎𝑏2 𝑏3 + 𝑏 𝑏2𝑐𝑐2𝑎 𝑐2𝑏 𝑐3 + 𝑐
| [Applying R1 → 𝑎𝑅1, 𝑅3 → 𝑏𝑅3, 𝑅3 → 𝑐𝑅3]
=𝑎𝑏𝑐
𝑎𝑏𝑐|𝑎2 + 1 𝑎2 𝑎2
𝑏2 𝑏2 + 1 𝑏2
𝑐2 𝑐2 𝑐2 + 1
| [Taking a as common from C1, 𝑏 from C2 and c from C3]
= |1 + 𝑎2 + 𝑏2 + 𝑐2 1 + 𝑎2 + 𝑏2 + 𝑐2 1 + 𝑎2 + 𝑏2 + 𝑐2
𝑏2 𝑏2 + 1 𝑏2
𝑐2 𝑐2 𝑐2 + 1
| [By R1 → 𝑅1 + 𝑅2 + 𝑅3]
= (1 + 𝑎2 + 𝑏2 + 𝑐2) |1 1 1𝑏2 𝑏2 + 1 𝑏2
𝑐2 𝑐2 𝑐2 + 1| [Taking 1 + a2 + 𝑏2 + 𝑐2 as common from R1]
= (1 + 𝑎2 + 𝑏2 + 𝑐2) |0 0 1
−1 1 𝑏2
0 −1 𝑐2 + 1| [Applying C1 → 𝐶1 − 𝐶2, 𝐶2 → 𝐶2 − 𝐶3]
= (1 + 𝑎2 + 𝑏2 + 𝑐2){1 − 0} [Expanding along R1]
= 1 + 𝑎2 + 𝑏2 + 𝑐2 = 𝑅𝐻𝑆
Hence, LHS = RHS
15. Let 𝐴 be a square matrix of order 3 × 3, then |𝑘𝐴| is equal to:
(A) 𝑘|𝐴|
(B) 𝑘2|𝐴|
(C) 𝑘3|𝐴|
(D) 3𝑘|𝐴|
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Solution:
If 𝐵 be a square matrix of order 𝑛 × 𝑛, then |𝑘𝐵| = 𝑘𝑛−1|𝐵|
Therefore, |𝑘𝐴| = 𝑘3−1|𝐴| = 𝑘2|𝐴|
Hence, the option (𝐵) is correct.
16. Which of the following is correct
(A) Determinant is a square matrix
(B) Determinant is a number associated to a matrix
(C) Determinant is a number associated to a square matrix
(D) None of these
Solution:
Determinant is a number associated to a square matrix
Hence, the option (𝐶) is correct.
Exercise 𝟒. 𝟑
1. 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)
Solution:
i) Given vertices of the triangle are (1, 0), (6, 0), (4, 3)
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Area of triangle =1
2|1 0 16 0 14 3 1
|
=1
2[1(0 − 3) − 0(6 − 4) + 1(18 − 0)] =
1
2(15) = 7.5 square units
Given vertices of the triangle are (2, 7), (1, 1), (10, 8)
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We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Area of triangle =1
2|2 7 11 1 110 8 1
|
=1
2[2(1 − 8) − 7(1 − 10) + 1(8 − 10)] =
1
2(47) = 25.5 square units
Given vertices of the triangle are (−2,−3), (3, 2), (−1,−8)
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Area of triangle =1
2|−2 −3 13 2 1
−1 −8 1|
=1
2[−2(2 + 8) + 3(3 + 1) + 1(−24 + 2)] =
1
2(−30) = 15
Area of triangle = 15 square units
2. Show that points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎), 𝐶(𝑐, 𝑎 + 𝑏) are collinear.
Solution:
If the points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎) and 𝐶(𝑐, 𝑎 + 𝑏) are collinear, then the area of triangle
𝐴𝐵𝐶 will be zero.
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Area of triangle 𝐴𝐵𝐶 =1
2|𝑎 𝑏 + 𝑐 1𝑏 𝑐 + 𝑎 1𝑐 𝑎 + 𝑏 1
|
=1
2|𝑎 𝑎 + 𝑏 + 𝑐 1𝑏 𝑎 + 𝑏 + 𝑐 1𝑐 𝑎 + 𝑏 + 𝑐 1
| [Applying C2 → 𝐶1 + 𝐶2]
=1
2(𝑎 + 𝑏 + 𝑐) |
𝑎 1 1𝑏 1 1𝑐 1 1
| [Taking 𝑎 + 𝑏 + 𝑐 as common from C2]
= 0 [∵ 𝐶1 = 𝐶3]
Hence, the points 𝐴(𝑎, 𝑏 + 𝑐), 𝐵(𝑏, 𝑐 + 𝑎) and 𝐶(𝑐, 𝑎 + 𝑏) are collinear.
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3. Find values of 𝑘 if area of triangle is 4 square units and vertices are
(i) (𝑘, 0), (4, 0), (0, 2)
(ii) (−2, 0), (0, 4), (0, 𝑘)
Solution:
(i)Given vertices of the triangle are (𝑘, 0), (4, 0), (0, 2)
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Area of triangle =1
2|𝑘 0 14 0 10 2 1
|
=1
2[𝑘(0 − 2) − 0(4 − 0) + 1(8 − 0)] =
1
2(−2𝑘 + 8) = −𝑘 + 4
According to question, area of triangle 𝐴𝐵𝐶 = 4 square units
Therefore, |−𝑘 + 4| = 4 ⇒ −𝑘 + 4 = ±4
⇒ −𝑘 + 4 = 4 or −𝑘 + 4 = −4
⇒ 𝑘 = 0 or 𝑘 = 8
Hence, the value of 𝑘 are 0 and 8.
ii) Given vertices of the triangle are (−2, 0), (0, 4), (0, 𝑘)
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Now, area of triangle =1
2|−2 0 10 4 10 𝑘 1
|
=1
2[−2(4 − 𝑘) − 0(0 − 0) + 1(0 − 0)] =
1
2(−8 + 2𝑘) = −4 + 𝑘
According to question, Area of triangle 𝐴𝐵𝐶 = 4 square units
Therefore, |−4 + 𝑘| = 4 ⇒ −4 + 𝑘 = ±4
⇒ −4 + 𝑘 = 4 or −4 + 𝑘 = −4
⇒ 𝑘 = 8 or 𝑘 = 0
Hence, the value of 𝑘 are 0 and 8.
4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
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(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
Solution:
(i) Let, 𝑃(𝑥, 𝑦) be any point lie on the line joining 𝐴(1, 2) and 𝐵(3, 6). Hence, the points 𝐴, 𝐵
and 𝑃 will be collinear and area of triangle 𝐴𝐵𝐶 will be zero.
Therefore, area of triangle 𝐴𝐵𝑃 =1
2|1 2 13 6 1𝑥 𝑦 1
| = 0
⇒1
2[1(6 − 𝑦) − 2(3 − 𝑥) + 1(3𝑦 − 6𝑥)] = 0
⇒ 6 − 𝑦 − 6 + 2𝑥 + 3𝑦 − 6𝑥 = 0
⇒ −4𝑥 + 2𝑦 = 0
⇒ 2𝑥 = 𝑦
Hence the equation of line joining the points (1, 2) and (3, 6) is 2𝑥 − 𝑦 = 0
(ii) Let, 𝑃(𝑥, 𝑦) be any point lie on the line joining 𝐴(3, 1) and 𝐵(9, 3). Hence, the points 𝐴, 𝐵
and 𝑃 will be collinear and area of triangle 𝐴𝐵𝐶 will be zero.
Therefore, area of triangle 𝐴𝐵𝑃 =1
2|3 1 19 3 1𝑥 𝑦 1
| = 0
⇒1
2[3(3 − 𝑦) − 1(9 − 𝑥) + 1(9𝑦 − 3𝑥)] = 0
⇒ 9 − 3𝑦 − 9 + 𝑥 + 9𝑦 − 3𝑥 = 0
⇒ −2𝑥 + 6𝑦 = 0
⇒ 𝑥 = 3𝑦
Hence the equation of line joining the points (3, 1) and (9, 3) is 𝑥 − 3𝑦 = 0
5. If area of triangle is 35 sq. units with vertices (2, −6), (5, 4) and (𝑘, 4). Then 𝑘 is
(A) 12
(B) −2
(C) −12,−2
(D) 12,−2
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Solution:
Given vertices of the triangle are (2, −6), (5, 4), (𝑘, 4)
We know, area of triangle =1
2|
𝑥1 𝑦1 1𝑥2 𝑦2 1𝑥3 𝑦3 1
|
Hence, area of triangle =1
2|2 −6 15 4 1𝑘 4 1
|
=1
2[2(4 − 4) + 6(5 − 𝑘) + 1(20 − 4𝑘)] =
1
2(30 − 6𝑘 + 20 − 4𝑘) = 25 − 5𝑘
According to question, area of triangle = 35 square units
Therefore, |25 − 5𝑘| = 35
⇒ 25 − 5𝑘 = ±35
⇒ 25 − 5𝑘 = 35 or 25 − 5𝑘 = −35
⇒ 𝑘 =−10
5= −2 or 𝑘 =
60
5= 12
Hence, the option (𝐷) is correct.
Exercise 𝟒. 𝟒
1. Write Minors and Cofactors of the elements of following determinants:
(i) |2 −40 3
|
(ii) |𝑎 𝑐𝑏 𝑑
|
Solution:
(i)
Given determinant is |2 −40 3
|
We know, the minor of element 𝑎𝑖𝑗 is 𝑀𝑖𝑗 and the cofactor is 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗, therefore,
The minor of element 𝑎11 is 𝑀11 = 3 and the cofactor is 𝐴11 = (−1)1+1𝑀11 = 3
The minor of element 𝑎12 is 𝑀12 = 0 and the cofactor is 𝐴12 = (−1)1+2𝑀12 = 0
The minor of element 𝑎21 is 𝑀21 = −4 and the cofactor is 𝐴21 = (−1)2+1𝑀21 = 4
The minor of element 𝑎22 is 𝑀22 = 2 and the cofactor is 𝐴22 = (−1)2+2𝑀22 = 2
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(ii)
Given determinant is |𝑎 𝑐𝑏 𝑑
|
We know, the minor of element 𝑎𝑖𝑗 is 𝑀𝑖𝑗 and the cofactor is 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗, therefore,
The minor of element 𝑎11 is 𝑀11 = 𝑑 and the cofactor is 𝐴11 = (−1)1+1𝑀11 = 𝑑
The minor of element 𝑎12 is 𝑀12 = 𝑏 and the cofactor is 𝐴12 = (−1)1+2𝑀12 = −𝑏
The minor of element 𝑎21 is 𝑀21 = 𝑐 and the cofactor is 𝐴21 = (−1)2+1𝑀21 = −𝑐
The minor of element 𝑎22 is 𝑀22 = 𝑎 and the cofactor is 𝐴22 = (−1)2+2𝑀22 = 𝑎
2. Write Minors and Cofactors of the elements of following determinants:
(i) |1 0 00 1 00 0 1
|
(ii) |1 0 43 5 −10 1 2
|
Solution:
(i)
Given determinant is |1 0 00 1 00 0 1
|
We know, minor of an element 𝑎𝑖𝑗 of a determinant is the determinant obtained by
deleting its 𝑖𝑡ℎ row and 𝑗𝑡ℎ column in which element 𝑎𝑖𝑗 lies. Minor of an element 𝑎𝑖𝑗 is
denoted by 𝑀𝑖𝑗.
Hence,
𝑀11 = |1 00 1
| = 1 − 0 = 1
𝑀12 = |0 00 1
| = 0 − 0 = 0
𝑀13 = |0 10 0
| = 0 − 0 = 0
𝑀21 = |0 00 1
| = 0 − 0 = 0
𝑀22 = |1 00 1
| = 1 − 0 = 1
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𝑀23 = |1 00 0
| = 0 − 0 = 0
𝑀31 = |0 01 0
| = 0 − 0 = 0
𝑀32 = |1 00 0
| = 0 − 0 = 0
𝑀33 = |1 00 1
| = 1 − 0 = 1
And we know, cofactor of an element 𝑎𝑖𝑗, is denoted by 𝐴𝑖𝑗 is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗,
therefore
𝐴11 = (−1)1+1𝑀11 = 1 𝐴12 = (−1)1+2𝑀12 = 0 𝐴13 = (−1)1+3𝑀13 = 0
𝐴21 = (−1)2+1𝑀21 = 0 𝐴22 = (−1)2+2𝑀22 = 1 𝐴23 = (−1)2+3𝑀233 = 0
𝐴31 = (−1)3+1𝑀31 = 0 𝐴32 = (−1)3+2𝑀32 = 0 𝐴33 = (−1)3+3𝑀33 = 1
(ii)
Given determinant is |1 0 43 5 −10 1 2
|
We know, minor of an element 𝑎𝑖𝑗 of a determinant is the determinant obtained by
deleting its 𝑖𝑡ℎ row and 𝑗𝑡ℎ column in which element 𝑎𝑖𝑗 lies. Minor of an element 𝑎𝑖𝑗 is
denoted by 𝑀𝑖𝑗.
Hence,
𝑀11 = |5 −11 2
| = 10 + 1 = 11
𝑀12 = |3 −10 2
| = 6 − 0 = 6
𝑀13 = |3 50 1
| = 3 − 0 = 3
𝑀21 = |0 41 2
| = 0 − 4 = −4
𝑀22 = |1 40 2
| = 2 − 0 = 2
𝑀23 = |1 00 1
| = 1 − 0 = 1
𝑀31 = |0 45 −1
| = 0 − 20 = −20
𝑀32 = |1 43 −1
| = −1 − 12 = −13
𝑀33 = |1 03 5
| = 5 − 0 = 5
And we know, cofactor of an element 𝑎𝑖𝑗, is denoted by 𝐴𝑖𝑗 is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗𝑀𝑖𝑗,
therefore
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𝐴11 = (−1)1+1𝑀11 = 11 𝐴12 = (−1)1+2𝑀12 = −6 𝐴13 = (−1)1+3𝑀13 = 3
𝐴21 = (−1)2+1𝑀21 = 4 𝐴22 = (−1)2+2𝑀22 = 2 𝐴23 = (−1)2+3𝑀233 = −1
𝐴31 = (−1)3+1𝑀31 = −20 𝐴32 = (−1)3+2𝑀32 = 13 𝐴33 = (−1)3+3𝑀33 = 5
3. Using cofactors of elements of second row, evaluate ∆= |5 3 82 0 11 2 3
|.
Solution:
We know, ∆ = |5 3 82 0 11 2 3
| = 𝑎21𝐴21 + 𝑎22𝐴22 + 𝑎23𝐴23
Here, 𝑎21 = 2, 𝑎22 = 0, 𝑎23 = 1 and
𝐴21 = (−1)2+1 |3 82 3
| = −(9 − 16) = 7
𝐴22 = (−1)2+2 |5 81 3
| = 15 − 8 = 7
𝐴23 = (−1)2+3 |5 31 2
| = −(10 − 3) = −7
Therefore, ∆= |5 3 82 0 11 2 3
| = 2(7) + 0(7) + 1(−7) = 7
4. Using Cofactors of elements of third column, evaluate ∆= |
1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦
|.
Solution:
We know, ∆= |
1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦
| = 𝑎13𝐴13 + 𝑎23𝐴23 + 𝑎33𝐴33
Here, 𝑎13 = 𝑦𝑧, 𝑎23 = 𝑧𝑥, 𝑎33 = 𝑥𝑦 and
𝐴13 = (−1)1+3 |1 𝑦1 𝑧
| = 𝑧 − 𝑦
𝐴23 = (−1)2+3 |1 𝑥1 𝑧
| = −(𝑧 − 𝑥) = 𝑥 − 𝑧
𝐴33 = (−1)3+3 |1 𝑥1 𝑦
| = 𝑦 − 𝑥
Therefore, ∆= |
1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦
| = 𝑦𝑧(𝑧 − 𝑦) + 𝑧𝑥(𝑥 − 𝑧) + 𝑥𝑦(𝑦 − 𝑥)
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= 𝑦𝑧2 − 𝑦2𝑧 + 𝑧𝑥2 − 𝑥𝑧2 + 𝑥𝑦2 − 𝑥2𝑦
= 𝑧𝑥2 − 𝑥2𝑦 − 𝑥𝑧2 + 𝑥𝑦2 + 𝑦𝑧2 − 𝑦2𝑧
= 𝑥2(𝑧 − 𝑦) − 𝑥(𝑧2 − 𝑦2) + 𝑦𝑧(𝑧 − 𝑦)
= (𝑧 − 𝑦)[𝑥2 − 𝑥(𝑧 + 𝑦) + 𝑦𝑧]
= (𝑧 − 𝑦)[𝑥2 − 𝑥𝑧 − 𝑥𝑦 + 𝑦𝑧]
= (𝑧 − 𝑦)[𝑥(𝑥 − 𝑧) − 𝑦(𝑥 − 𝑧)]
= (𝑥 − 𝑧)(𝑧 − 𝑦)(𝑥 − 𝑦)
= (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)
Hence, |
1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦
| = (𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥)
5. If ∆= |
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
| 𝐴𝑖𝑗 is Cofactors of 𝑎𝑖𝑗 , then value of ∆ is given by
(A) 𝑎11𝐴11 + 𝑎12𝐴32 + 𝑎13𝐴33
(B) 𝑎11𝐴11 + 𝑎12𝐴21 + 𝑎13𝐴31
(C) 𝑎21𝐴11 + 𝑎22𝐴12 + 𝑎23𝐴13
(D) 𝑎11𝐴11 + 𝑎21𝐴21 + 𝑎31𝐴31
Solution:
The value of |
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
| is given by: 𝑎11𝐴11 + 𝑎21𝐴21 + 𝑎31𝐴31
Hence, the option (𝐷) is correct.
Exercise 𝟒. 𝟓
1. Find adjoint of the matrix [1 23 4
]
Solution:
Given matrix is [1 23 4
]
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We know, the adjoint of a square matrix 𝐴 = [𝑎𝑖𝑗]𝑛×𝑛 is defined as the transpose of the
matrix [𝐴𝑖𝑗]𝑛×𝑛, where 𝐴𝑖𝑗 is the cofactor of the element 𝑎𝑖𝑗. Adjoint of the matrix 𝐴 is
denoted by 𝑎𝑑𝑗 𝐴.
Let 𝐴 = [1 23 4
]
therefore, 𝐴11 = 4, 𝐴12 = −3, 𝐴21 = −2, 𝐴22 = 1
Hence, adjoint of matrix 𝐴 = [𝐴11 𝐴21
𝐴12 𝐴22] = [
4 −2−3 1
]
2. Find adjoint of the matrix [1 −1 22 3 5
−2 0 1]
Solution:
Given matrix is [1 −1 22 3 5
−2 0 1]
We know, the adjoint of a square matrix 𝐴 = [𝑎𝑖𝑗]𝑛×𝑛 is defined as the transpose of the
matrix [𝐴𝑖𝑗]𝑛×𝑛, where 𝐴𝑖𝑗 is the cofactor of the element 𝑎𝑖𝑗. Adjoint of the matrix 𝐴 is
denoted by 𝑎𝑑𝑗 𝐴.
Let 𝐴 = [1 −1 22 3 5
−2 0 1], therefore
𝐴11 = 3 𝐴12 = −12 𝐴13 = 6𝐴21 = 1 𝐴22 = 5 𝐴23 = 2
𝐴31 = −11 𝐴32 = −1 𝐴33 = 5
Hence, adjoint of matrix 𝐴 = [
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
] = [3 1 −11
−12 5 −16 2 5
]
3. Verify 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 for the matrix [2 3
−4 −6]
Solution:
Given matrix is [2 3
−4 −6]
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Let 𝐴 = [2 3
−4 −6],
therefore, 𝐴11 = −6 𝐴12 = 4 𝐴21 = −3 𝐴22 = 2
|𝐴| = −12 + 12 = 0
𝑎𝑑𝑗 𝐴 = [𝐴11 𝐴21
𝐴12 𝐴22] = [
−6 −34 2
]
𝐴(𝑎𝑑𝑗 𝐴) = [2 3
−4 −6] [
−6 −34 2
] = [−12 + 12 −6 + 624 − 24 12 − 12
] = [0 00 0
]
(𝑎𝑑𝑗 𝐴). 𝐴 = [−6 −34 2
] [2 3
−4 −6] = [
−12 + 12 −18 + 188 − 8 12 − 12
] = [0 00 0
]
|𝐴|. 𝐼 = 0. [1 00 1
] = [0 00 0
]
Hence, 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 = [0 00 0
]
4. Verify 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 for the matrix [1 −1 23 0 −21 0 3
]
Solution:
Given matrix is [1 −1 23 0 −21 0 3
]
Here, 𝐴 = [1 −1 23 0 −21 0 3
], therefore, |𝐴| = 1(0 − 0) + 1(9 + 2) + 2(0 − 0) = 11
𝐴11 = 0 𝐴12 = −11 𝐴13 = 0𝐴21 = 3 𝐴22 = 1 𝐴23 = −1𝐴31 = 2 𝐴32 = 8 𝐴33 = 3
Adjoint of matrix 𝐴 = [
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
] = [0 3 2
−11 1 80 −1 3
]
𝐴(𝑎𝑑𝑗 𝐴) = [1 −1 23 0 −21 0 3
] [0 3 2
−11 1 80 −1 3
]
= [0 + 11 + 0 3 − 1 − 2 2 − 8 + 60 + 0 + 0 9 + 0 + 2 6 + 0 − 60 + 0 + 0 3 + 0 − 3 2 + 0 + 9
] = [11 0 00 11 00 0 11
]
Now, (𝑎𝑑𝑗 𝐴). 𝐴 = [0 3 2
−11 1 80 −1 3
] [1 −1 23 0 −21 0 3
]
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= [0 + 9 + 2 0 + 0 + 0 0 − 6 + 6
−11 + 3 + 8 11 + 0 + 0 −22 − 2 + 240 − 3 + 3 0 + 0 + 0 0 + 2 + 9
] = [11 0 00 11 00 0 11
]
|𝐴|. 𝐼 = 11. [1 0 00 1 00 0 1
] = [11 0 00 11 00 0 11
]
Hence, 𝐴(𝑎𝑑𝑗 𝐴) = (𝑎𝑑𝑗 𝐴). 𝐴 = |𝐴|. 𝐼 = [11 0 00 11 00 0 11
]
5. Find the inverse of the matrix (if it exists)
[2 −24 3
]
Solution:
Given matrix is [2 −24 3
]
Let 𝐴 = [2 −24 3
]
Therefore, 𝐴11 = 3 𝐴12 = −4 𝐴21 = 2 𝐴22 = 2
And |𝐴| = 6 + 8 = 14 ≠ 0 ⇒ 𝐴−1 exists.
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴 =
⇒ 𝐴−1 =1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22]
=> 1
14[
3 2−4 2
]
Hence, the inverse of the matrix [2 −24 3
] is 1
14[
3 2−4 2
]
6. Find the inverse of the matrix (if it exists)
[−1 5−3 2
]
Solution:
Given matrix is [−1 5−3 2
]
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Let 𝐴 = [−1 5−3 2
]
Therefore, 𝐴11 = 2 𝐴12 = 3 𝐴21 = −5 𝐴22 = −1
|𝐴| = −2 + 15 = 13 ≠ 0 ⇒ 𝐴−1 exists.
We know 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22] =
1
13[2 −53 −1
]
Hence, the inverse of the matrix [−1 5−3 2
] is 1
13[2 −53 −1
]
7. Find the inverse of the matrix (if it exists)
[1 2 30 2 40 0 5
]
Solution:
Given matrix is [1 2 30 2 40 0 5
]
Let 𝐴 = [1 2 30 2 40 0 5
]
Therefore, |𝐴| = 1(10 − 0) − 2(0 − 0) + 3(0 − 0) = 10 ≠ 0 ⇒ 𝐴−1 exists.
𝐴11 = 10 𝐴12 = 0 𝐴13 = 0𝐴21 = −10 𝐴22 = 5 𝐴23 = 0𝐴31 = 2 𝐴32 = −4 𝐴33 = 2
We know 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
]
=1
10[10 −10 20 5 −40 0 2
]
Hence, the inverse of the matrix [1 2 30 2 40 0 5
] is 1
10[10 −10 20 5 −40 0 2
]
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8. Find the inverse of the matrix (if it exists)
[1 0 03 3 05 2 −1
]
Solution:
Given matrix is [1 0 03 3 05 2 −1
]
Let 𝐴 = [1 0 03 3 05 2 −1
],
Therefore, |𝐴| = 1(−3 − 0) − 0(−3 − 0) + 0(6 − 15) = −3 ≠ 0 ⇒ 𝐴−1 exists.
𝐴11 = −3 𝐴12 = 3 𝐴13 = −9𝐴21 = 0 𝐴22 = −1 𝐴23 = −2𝐴31 = 0 𝐴32 = 0 𝐴33 = 3
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
]
=1
3[−3 0 03 −1 0
−9 −2 3]
Hence, the inverse of the matrix [1 0 03 3 05 2 −1
] is 1
3[−3 0 03 −1 0
−9 −2 3]
9. Find the inverse of the matrix (if it exists)
[2 1 34 −1 0
−7 2 1]
Solution:
Given matrix is [2 1 34 −1 0
−7 2 1]
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Let 𝐴 = [2 1 34 −1 0
−7 2 1]
Therefore, |𝐴| = 2(−1 − 0) − 1(4 − 0) + 3(8 − 7) = −3 ≠ 0 ⇒ 𝐴−1 exists.
𝐴11 = −1 𝐴12 = −4 𝐴13 = 1𝐴21 = 5 𝐴22 = 23 𝐴23 = −11𝐴31 = 3 𝐴32 = 12 𝐴33 = −6
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
]
=1
3[−1 5 3−4 23 121 −11 −6
]
Hence, the inverse of the matrix [2 1 34 −1 0
−7 2 1] is
1
3[−1 5 3−4 23 121 −11 −6
]
10. Find the inverse of the matrix (if it exists)
[1 −1 20 2 −33 −2 4
]
Solution:
Given matrix is [1 −1 20 2 −33 −2 4
]
Let 𝐴 = [1 −1 20 2 −33 −2 4
]
Therefore, [𝐴] = 1(8 − 6) + 1(0 + 9) + 2(0 − 6) = −1 ≠ 0 ⇒ 𝐴−1 exists.
𝐴11 = 2 𝐴12 = −9 𝐴13 = −6𝐴21 = 0 𝐴22 = −2 𝐴23 = −1
𝐴31 = −1 𝐴32 = 3 𝐴33 = 2
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
]
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=1
−1[
2 0 −1−9 −2 3−6 −1 2
] = [−2 0 19 2 −36 1 −2
]
Hence, the inverse of the matrix [1 −1 20 2 −33 −2 4
] is [−2 0 19 2 −36 1 −2
]
11. Find the inverse of the matrix (if it exists)
[1 0 00 cos𝛼 sin 𝛼0 sin𝛼 − cos𝛼
]
Solution:
Given matrix is [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼
]
Let 𝐴 = [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼
], therefore
Now, |𝐴| = 1(− cos2 𝛼 − sin2 𝛼) + 0(0 − 0) + 0(0 − 0) = −1 ≠ 0
⇒ 𝐴−1 exists. Therefore
𝐴11 = 1 𝐴12 = 0 𝐴13 = 0𝐴21 = 0 𝐴22 = −cos𝛼 𝐴23 = −sin𝛼𝐴31 = 0 𝐴32 = −sin𝛼 𝐴33 = cos𝛼
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
]
=1
−1[1 0 00 − cos𝛼 − sin𝛼0 −sin𝛼 cos𝛼
]
⇒ 𝐴−1 = [−1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼
]
Hence, the inverse of the matrix [1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼
] is [−1 0 00 cos𝛼 sin𝛼0 sin𝛼 − cos𝛼
]
12. Let 𝐴 = [3 72 5
] and 𝐵 = [6 87 9
]. Verify that (𝐴𝐵)−1 = 𝐵−1𝐴−1.
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Solution:
Given matrices are 𝐴 = [3 72 5
] and 𝐵 = [6 87 9
]
Here, 𝐴 = [3 72 5
], therefore, 𝐴11 = 5 𝐴12 = −2 𝐴21 = −7 𝐴22 = 3
Now, |𝐴| = 15 − 14 = 1 ≠ 0 ⇒ 𝐴−1 exists.
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22]
=1
1[
5 −7−2 3
]
= [5 −7
−2 3]
and 𝐵 = [6 87 9
], therefore, 𝐵11 = 9 𝐵12 = −7 𝐵21 = −8 𝐵22 = 6
|𝐵| = 54 − 56 = −2 ≠ 0 ⇒ 𝐵−1 exists.
𝐵−1 =1
|𝐵|𝑎𝑑𝑗 𝐵
=1
|𝐵|[𝐵11 𝐵21
𝐵12 𝐵22]
=1
−2[
9 −8−7 6
]
= [−
9
24
7
2−3
]
Now, 𝐵−1𝐴−1 = [−9 2⁄ 47 2⁄ −3
] [5 −7
−2 3]
= [−
45
2− 8
63
2+ 12
35
2+ 6 −
49
2− 9
]
= [−
61
2
87
247
2−
67
2
] …(i)
And 𝐴𝐵 = [3 72 5
] [6 87 9
] = [18 + 49 24 + 6312 + 35 16 + 45
] = [67 8747 61
]
|𝐴𝐵| = 67 × 61 − 87 × 47 = 4087 − 4089 = −2 ≠ 0 ⇒ (𝐴𝐵)−1 exists.
𝐴𝐵11 = 61 𝐴𝐵12 = −47 𝐴𝐵21 = −87 𝐴𝐵22 = 67
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(𝐴𝐵)−1 =1
|𝐴𝐵|𝑎𝑑𝑗 𝐴𝐵 =
1
|𝐴𝐵|[𝐴𝐵11 𝐴𝐵21
𝐴𝐵12 𝐴𝐵22] =
1
−2[
61 −87−47 67
] = [−
61
2
87
247
2−
67
2
] …(ii)
From (i) and (ii)
(𝐴𝐵)−1 = 𝐵−1𝐴−1
Hence verified.
13. If 𝐴 = [3 1
−1 2] show that 𝐴2 − 5𝐴 + 7𝐼 = 0. Hence, find 𝐴−1.
Solution:
Given matrix is 𝐴 = [3 1
−1 2]
𝐿𝐻𝑆 = 𝐴2 − 5𝐴 + 7𝐼 = 𝐴𝐴 − 5𝐴 + 7𝐼
= [3 1
−1 2] [
3 1−1 2
] − 5 [3 1
−1 2] + 7 [
1 00 1
]
= [9 − 1 3 + 2
−3 − 2 −1 + 4] − [
15 5−5 10
] + [7 00 7
]
= [8 − 15 + 7 5 − 5 + 0−5 + 5 + 0 3 − 10 + 7
] = [0 00 0
] = 𝑂 = 𝑅𝐻𝑆
⇒ 𝐴2 − 5𝐴 + 7𝐼 = 𝑂
⇒ 𝐴2 − 5𝐴 = −7𝐼
Multiplying by 𝐴−1(because |𝐴| ≠ 0)
⇒ 𝐴𝐴𝐴−1 − 5𝐴𝐴−1 = −7𝐼𝐴−1
⇒ 𝐴𝐼 − 5𝐼 = −7𝐴−1 [∵ 𝐴𝐴−1 = 𝐼]
⇒ 7𝐴−1 = 5𝐼 − 𝐴 = 5 [1 00 1
] − [3 1
−1 2] = [
5 00 5
] − [3 1
−1 2] = [
2 −11 3
]
⇒ 𝐴−1 =1
7[2 −11 3
]
14. For the matrix 𝐴 = [3 21 1
], find the numbers 𝑎 and 𝑏 such that 𝐴2 + 𝑎𝐴 + 𝑏𝐼 = 𝑂.
Solution:
Given matrix is 𝐴 = [3 21 1
]
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And 𝐴2 + 𝑎𝐴 + 𝑏𝐼 = 𝑂
⇒ [3 21 1
] [3 21 1
] + 𝑎 [3 21 1
] + 𝑏 [1 00 1
] = [0 00 0
]
⇒ [9 + 2 6 + 23 + 1 2 + 1
] + [3𝑎 2𝑎𝑎 𝑎
] + [𝑏 00 𝑏
] = [0 00 0
]
⇒ [11 + 3𝑎 + 𝑏 8 + 2𝑎 + 04 + 𝑎 + 0 3 + 𝑎 + 𝑏
] = [0 00 0
]
⇒ 4 + 𝑎 = 0
⇒ 𝑎 = −4
and 3 + 𝑎 + 𝑏 = 0
⇒ 𝑏 = −3 − 𝑎 = −3 + 4
= 1
Hence, 𝑎 = −4, 𝑏 = 1
15. For the matrix 𝐴 = [1 1 11 2 −32 −1 3
]
show that 𝐴3 − 6𝐴2 + 5𝐴 + 11𝐼 = 0. Hence, find 𝐴−1.
Solution:
Given matrix is 𝐴 = [1 1 11 2 −32 −1 3
]
𝐴2 = 𝐴. 𝐴 = [1 1 11 2 −32 −1 3
] [1 1 11 2 −32 −1 3
]
= [1 + 1 + 2 1 + 2 − 1 1 − 3 + 31 + 2 − 6 1 + 4 + 3 1 − 6 − 92 − 1 + 6 2 − 2 − 3 2 + 3 + 9
] = [4 2 1
−3 8 −147 −3 14
]
𝐴3 = 𝐴2. 𝐴 = [4 2 1
−3 8 −147 −3 14
] [1 1 11 2 −32 −1 3
]
= [4 + 2 + 2 4 + 4 − 1 4 − 6 + 3
−3 + 8 − 28 −3 + 16 + 14 −3 − 24 − 427 − 3 + 28 7 − 6 − 14 7 + 9 + 42
] = [8 7 1
−23 27 −6932 −13 58
]
LHS = 𝐴3 − 6𝐴2 + 5𝐴 + 11𝐼
= [8 7 1
−23 27 −6932 −13 58
] − 6 [4 2 1
−3 8 −147 −3 14
] + 5 [1 1 11 2 −32 −1 3
] + 11 [1 0 00 1 00 0 1
]
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36 Practice more on Determinants www.embibe.com
= [8 7 1
−23 27 −6932 −13 58
] − [24 12 6
−18 48 −8442 −18 84
] + [5 5 55 10 −1510 −5 15
] + [11 0 00 11 00 0 11
]
= [8 − 24 + 5 + 11 7 − 12 + 5 + 0 1 − 6 + 5 + 0
−23 + 18 + 5 + 0 27 − 48 + 10 + 11 −69 + 84 − 15 + 032 − 42 + 10 + 0 −13 + 18 − 5 + 0 58 − 84 + 15 + 11
]
= [0 0 00 0 00 0 0
] = 𝑂 = 𝑅𝐻𝑆
Now, 𝐴3 − 6𝐴2 + 5𝐴 + 11 𝐼 = 𝑂
⇒ 𝐴3 − 6𝐴2 + 5𝐴 = −11 𝐼
Multiplying by 𝐴−1 (because |𝐴| ≠ 0)
⇒ 𝐴2𝐴𝐴−1 − 6𝐴𝐴𝐴−1 + 5𝐴𝐴−1 = −11𝐴−1
⇒ 𝐴2𝐼 − 6𝐴𝐼 + 5𝐼 = −11𝐴−1 [∵ 𝐴𝐴−1 = 𝐼]
⇒ 11𝐴−1 = −𝐴2 + 6𝐴 − 5𝐼
⇒ 11𝐴−1 = −[4 2 1
−3 8 −147 −3 14
] + 6 [1 1 11 2 −32 −1 3
] − 5 [1 0 00 1 00 0 1
]
⇒ 11𝐴−1 = [−4 −2 −13 −8 14
−7 3 −14] + [
6 6 66 12 −1812 −6 18
] − [5 0 00 5 00 0 5
]
⇒ 11𝐴−1 = [−4 + 6 − 5 −2 + 6 + 0 −1 + 6 + 03 + 6 − 0 −8 + 12 − 5 14 − 18 + 0
−7 + 12 + 0 3 − 6 + 0 −14 + 18 − 5] = [
−3 4 59 −1 −45 −3 −1
]
⇒ 𝐴−1 =1
11[−3 4 59 −1 −45 −3 −1
]
Hence, 𝐴−1 =1
11[−3 4 59 −1 −45 −3 −1
]
16. If 𝐴 = [2 −1 1
−1 2 −11 −1 2
]
verify that 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼 = 𝑂 and hence find 𝐴−1.
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Solution:
𝐴2 = 𝐴. 𝐴 = [2 −1 1
−1 2 −11 −1 2
] [2 −1 1
−1 2 −11 −1 2
]
= [4 + 1 + 1 −2 − 2 − 1 2 + 1 + 2
−2 − 2 − 1 1 + 4 + 1 −1 − 2 − 22 + 1 + 2 −1 − 2 − 2 1 + 1 + 4
] = [6 −5 5
−5 6 −55 −5 6
]
𝐴3 = 𝐴2. 𝐴 = [6 −5 5
−5 6 −55 −5 6
] [2 −1 1
−1 2 −11 −1 2
]
= [12 + 5 + 5 −6 − 10 − 5 6 + 5 + 10
−10 − 6 − 5 5 + 12 + 5 −5 − 6 − 1010 + 5 + 6 −5 − 10 − 6 5 + 5 + 12
] = [22 −21 21
−21 22 −2121 −21 22
]
LHS = 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼
= [22 −21 21
−21 22 −2121 −21 22
] − 6 [6 −5 5
−5 6 −55 −5 6
] + 9 [2 −1 1
−1 2 −11 −1 2
] − 4 [1 0 00 1 00 0 1
]
= [22 −21 21
−21 22 −2121 −21 22
] − [36 −30 30
−30 36 −3030 −30 36
] + [18 −9 9−9 18 −99 −9 18
] − [4 0 00 4 00 0 4
]
= [22 − 36 + 18 − 4 −21 + 30 − 9 + 0 21 − 30 + 9 + 0−21 + 30 − 9 − 0 22 − 36 + 18 − 4 −21 + 30 − 9 + 021 − 30 + 9 − 0 −21 + 30 − 9 − 0 22 − 36 + 18 − 4
]
= [0 0 00 0 00 0 0
] = 𝑂 = 𝑅𝐻𝑆
⇒ 𝐴3 − 6𝐴2 + 9𝐴 − 4𝐼 = 𝑂 ⇒ 𝐴3 − 6𝐴2 + 9𝐴 = 4𝐼
Post multiplying by 𝐴−1(because |𝐴| ≠ 0)
𝐴2𝐴𝐴−1 − 6𝐴𝐴𝐴−1 + 9𝐴𝐴−1 = 4𝐼𝐴−1
⇒ 𝐴2𝐼 − 6𝐴𝐼 + 9𝐼 = 4𝐴−1 [Because 𝐴𝐴−1 = 𝐼]
⇒ 4𝐴−1 = 𝐴2 − 6𝐴 + 9𝐼
⇒ 4𝐴−1 = [6 −5 5
−5 6 −55 −5 6
] − 6 [2 −1 1
−1 2 −11 −1 2
] + 9 [1 0 00 1 00 0 1
]
⇒ 4𝐴−1 = [6 −5 5
−5 6 −55 −5 6
] − [12 −6 6−6 12 −66 −6 12
] + [9 0 00 9 00 0 9
]
⇒ 4𝐴−1 = [6 − 12 + 9 −5 + 6 + 0 5 − 6 + 0−5 + 6 + 0 6 − 12 + 9 −5 + 6 + 05 − 6 + 0 −5 + 6 + 0 6 − 12 + 9
] = [3 1 −11 3 1
−1 1 3]
⇒ 𝐴−1 =1
4[
3 1 −11 3 1
−1 1 3]
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Hence, 𝐴−1 =1
4[
3 1 −11 3 1
−1 1 3]
17. Let 𝐴 be a non-singular square matrix of order 3 × 3. Then |𝑎𝑑𝑗 𝐴| is equal to:
(A) |𝐴|
(B) |𝐴|2
(C) |𝐴|3
(D) 3|𝐴|
Solution:
Given that 𝐴 be a non-singular square matrix of order 3 × 3.
We know that 𝑎𝑑𝑗 𝐴 = |𝐴|𝐼
⇒ (𝑎𝑑𝑗 𝐴)𝐴 = |𝐴| [1 0 00 1 00 0 1
]
⇒ |(𝑎𝑑𝑗 𝐴)𝐴| = |𝐴| |1 0 00 1 00 0 1
| = |
|𝐴| 0 00 |𝐴| 00 0 |𝐴|
| = |𝐴|3
⇒ |𝑎𝑑𝑗 𝐴| = |𝐴|2,
Hence, the option (𝐵) is correct.
18. If 𝐴 is an invertible matrix of order 2, then det(𝐴−1) is equal to:
(A) det(𝐴)
(B) 1
det(𝐴)
(C) 1
(D) 0
Solution:
Given that the matrix 𝐴 is invertible, hence, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
The order of matrix is 2, so, let 𝐴 = |𝑎 𝑏𝑐 𝑑
|
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Therefore, |𝐴| = 𝑎𝑑 − 𝑏𝑐 and 𝑎𝑑𝑗 𝐴 = [𝑑 −𝑏−𝑐 𝑎
]
𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴 = 𝐴−1 =
1
|𝐴|[𝑑 −𝑏−𝑐 𝑎
] = [
𝑑
|𝐴|−
𝑏
|𝐴|
−𝑐
|𝐴|
𝑎
|𝐴|
]
det(𝐴−1) = |𝐴−1| = |
𝑑
|𝐴|−
𝑏
|𝐴|
−𝑐
|𝐴|
𝑎
|𝐴|
|
=1
|𝐴|2|𝑑 −𝑏−𝑐 𝑎
| =1
|𝐴|2(𝑎𝑑 − 𝑏𝑐) =
1
|𝐴|2|𝐴| =
1
|𝐴|
Hence, the option (𝐵) is correct.
Exercise 𝟒. 𝟔
1. Examine the consistency of the system of equations
𝑥 + 2𝑦 = 2
2𝑥 + 3𝑦 = 3
Solution:
The given system of equations are 𝑥 + 2𝑦 = 2 and 2𝑥 + 3𝑦 = 3
This system of equations can be written as 𝐴𝑋 = 𝐵.
Where, 𝐴 = [1 22 3
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
23]
Now, |𝐴| = 3 − 4 = −1 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
2. Examine the consistency of the system of equations
2𝑥 − 𝑦 = 5
𝑥 + 𝑦 = 4
Solution:
The given system of equations are 2𝑥 − 𝑦 = 5 and 𝑥 + 𝑦 = 4
This system of equations can be written as 𝐴𝑋 = 𝐵.
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Where, 𝐴 = [2 −11 1
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
54]
Now, |𝐴| = 2 + 1 = 3 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
3. Examine the consistency of the system of equations
𝑥 + 3𝑦 = 5
2𝑥 + 6𝑦 = 8
Solution:
The given system of equations: 𝑥 + 3𝑦 = 5 and 2𝑥 + 6𝑦 = 8
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [1 32 6
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
58]
Now, |𝐴| = 6 − 6 = 0 ⇒ 𝐴 is a singular matrix and so 𝐴−1 does not exist.
Now, 𝑎𝑑𝑗 𝐴 = [6 −3
−2 1]
(𝑎𝑑𝑗 𝐴) 𝐵 = [6 −3
−2 1] [
58] = [
30 − 24−10 + 8
] = [6
−2] ≠ 𝑂
So, there is no solutions of the given system of equations.
Hence, the system of equations are inconsistent.
4. Examine the consistency of the system of equations
𝑥 + 𝑦 + 𝑧 = 1
2𝑥 + 3𝑦 + 2𝑧 = 2
𝑎𝑥 + 𝑎𝑦 + 2𝑎𝑧 = 4
Solution:
The given system of equations are 𝑥 + 𝑦 + 𝑧 = 1, 2𝑥 + 3𝑦 + 2𝑧 = 2 and 𝑎𝑥 + 𝑎𝑦 + 2𝑎𝑧 =4
This system of equations can be written as 𝐴𝑋 = 𝐵, where
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𝐴 = [1 1 12 3 2𝑎 𝑎 2𝑎
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
123]
Now, |𝐴| = 1(6𝑎 − 2𝑎) − 1(4𝑎 − 2𝑎) + 1(2𝑎 − 3𝑎) = 𝑎 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
5. Examine the consistency of the system of equations
3𝑥 − 𝑦 − 2𝑧 = 2
2𝑦 − 𝑧 = −1
3𝑥 − 5𝑦 = 3
Solution:
The given system of equations are 3𝑥 − 𝑦 − 2𝑧 = 2, 2𝑦 − 𝑧 = −1 and 3𝑥 − 5𝑦 = 3
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [3 −1 −20 2 −13 −5 0
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
2−13
]
|𝐴| = 3(0 − 5) + 1(0 + 3) − 2(0 − 6) = −15 + 3 + 12 = 0
⇒ 𝐴 is a singular matrix and so 𝐴−1 does not exists. Now,
𝐴11 = −5 𝐴12 = −3 𝐴13 = −6𝐴21 = 10 𝐴22 = 6 𝐴23 = 12𝐴31 = 5 𝐴32 = 3 𝐴33 = 6
Now, 𝑎𝑑𝑗 𝐴 = [−5 10 5−3 6 3−6 12 6
]
Also (𝑎𝑑𝑗 𝐴)𝐵 = [−5 10 5−3 6 3−6 12 6
] [2
−13
] = [−10 − 10 + 15
−6 − 6 + 9−12 − 12 + 16
] = [−5−3−6
] ≠ 𝑂
So, there is no solutions of the given system of equations.
Hence, the system of equations are inconsistent.
6. Examine the consistency of the system of equations
5𝑥 − 𝑦 + 4𝑧 = 5
2𝑥 + 3𝑦 + 5𝑧 = 2
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5𝑥 − 2𝑦 + 6𝑧 = −1
Solution:
The given system of equations are 5𝑥 − 𝑦 + 4𝑧 = 5, 2𝑥 + 3𝑦 + 5𝑧 = 2 and 5𝑥 − 2𝑦 +
6𝑧 = −1
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [5 −1 42 3 55 −2 6
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
52
−1]
Now, |𝐴| = 5(18 + 10) + 1(12 − 25) + 4(−4 − 15) = 140 − 13 − 76 = 51 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
7. Solve system of linear equations, using matrix method
5𝑥 + 2𝑦 = 4
7𝑥 + 3𝑦 = 5
Solution:
The given system of equations are 5𝑥 + 2𝑦 = 4 and 7𝑥 + 3𝑦 = 5
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [5 27 3
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
45]
|𝐴| = 15 − 14 = 1 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
Now, 𝐴11 = 3 𝐴12 = −7 𝐴21 = −2 𝐴22 = 5
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
⇒ 𝐴−1 =1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22] =
1
1[
3 −2−7 5
]
Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] = [
3 −2−7 5
] [45]
⇒ [𝑥𝑦] = [
12 − 10−28 + 25
] ⇒ [𝑥𝑦] = [
2−3
] ⇒ 𝑥 = 2, 𝑦 = −3
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8. Solve system of linear equations, using matrix method
2𝑥 − 𝑦 = −2
3𝑥 + 4𝑦 = 3
Solution:
The given system of equations are 2𝑥 − 𝑦 = −2 and 3𝑥 + 4𝑦 = 3
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [2 −13 4
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
−23
]
|𝐴| = 8 + 3 = 11 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
Hence, the system of equations are consistent.
Now, 𝐴11 = 4 𝐴12 = −3 𝐴21 = 1 𝐴22 = 2
𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22] =
1
11[
4 1−3 2
]
𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] =
1
11[
4 1−3 2
] [−23
]
⇒ [𝑥𝑦] =
1
11[−8 + 36 + 6
] ⇒ [𝑥𝑦] = [
−5
1112
11
] ⇒ 𝑥 = −5
11, 𝑦 =
12
11
9. Solve system of linear equations, using matrix method
4𝑥 − 3𝑦 = 3
3𝑥 − 5𝑦 = 7
Solution:
The given system of equations are 4𝑥 − 3𝑦 = 3 and 3𝑥 − 5𝑦 = 7
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [4 −33 −5
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
37]
|𝐴| = −20 + 9 = −11 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
Now, 𝐴11 = −5 𝐴12 = −3 𝐴21 = 3 𝐴22 = 4
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𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22] =
1
−11[−5 3−3 4
]
Now, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] = −
1
11[−5 3−3 4
] [37]
⇒ [𝑥𝑦] = −
1
11[−15 + 21−9 + 28
] ⇒ [𝑥𝑦] = [
−6
11
−19
11
] ⇒ 𝑥 = −6
11, 𝑦 = −
19
11
10. Solve system of linear equations, using matrix method
5𝑥 + 2𝑦 = 3
3𝑥 + 2𝑦 = 5
Solution:
The given system of equations are 5𝑥 + 2𝑦 = 3 and 3𝑥 + 2𝑦 = 5
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [5 23 2
] , 𝑋 = [𝑥𝑦] and 𝐵 = [
35]
|𝐴| = 10 − 6 = 4 ≠ 0 ⇒ 𝐴 is non-singular and so 𝐴−1 exists.
Hence, the system of equations are consistent.
Now, 𝐴11 = 2 𝐴12 = −3 𝐴21 = −2 𝐴22 = 5
𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
|𝐴|[𝐴11 𝐴21
𝐴12 𝐴22] =
1
4[
2 −2−3 5
]
𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦] =
1
4[
2 −2−3 5
] [35]
⇒ [𝑥𝑦] =
1
4[
6 − 10−9 + 25
] ⇒ [𝑥𝑦] = [
−4
416
4
] ⇒ 𝑥 = −1, 𝑦 = 4
11. Solve system of linear equations, using matrix method
2𝑥 + 𝑦 + 𝑧 = 1
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𝑥 − 2𝑦 − 𝑧 =3
2
3𝑦 − 5𝑧 = 9
Solution:
The given system of equations are 2𝑥 + 𝑦 + 𝑧 = 1, 𝑥 − 2𝑦 − 𝑧 =3
2 and 3𝑦 − 5𝑧 = 9
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [2 1 11 −2 −10 3 −5
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
13
2
9
]
|𝐴| = 2(10 + 3) − 1(−5 − 0) + 1(3 − 0) = 26 + 5 + 3 = 34 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
𝐴11 = 13 𝐴12 = 5 𝐴13 = 3𝐴21 = 8 𝐴22 = −10 𝐴23 = −6𝐴31 = 1 𝐴32 = 3 𝐴33 = −5
𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
34[13 8 15 −10 33 −6 −5
]
Now, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =
1
34[13 8 15 −10 33 −6 −5
] [
13
2
9
]
⇒ [𝑥𝑦𝑧] =
1
34[13 + 12 + 95 − 15 + 273 − 9 − 45
] ⇒ [𝑥𝑦𝑧] =
1
34[
3417
−51] = [
11
2
−3
2
] ⇒ 𝑥 = 1, 𝑦 =1
2, 𝑧 = −
3
2
12. Solve system of linear equations, using matrix method
𝑥 − 𝑦 + 𝑧 = 4
2𝑥 + 𝑦 − 3𝑧 = 0
𝑥 + 𝑦 + 𝑧 = 2
Solution:
The given system of equations are 𝑥 − 𝑦 + 𝑧 = 4, 2𝑥 + 𝑦 − 3𝑧 = 0 and
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𝑥 + 𝑦 + 𝑧 = 2
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [1 −1 12 1 −31 1 1
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
402]
|𝐴| = 1(1 + 3) + 1(2 + 3) + 1(2 − 1) = 4 + 5 + 1 = 10 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
𝐴11 = 4 𝐴12 = −5 𝐴13 = 1𝐴21 = 2 𝐴22 = 0 𝐴23 = −2𝐴31 = 2 𝐴32 = 5 𝐴33 = 3
Now, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
10[
4 2 2−5 0 51 −2 3
]
Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =
1
10[
4 2 2−5 0 51 −2 3
] [402]
⇒ [𝑥𝑦𝑧] =
1
10[
16 + 0 + 4−20 + 0 + 10
4 + 0 + 6] ⇒ [
𝑥𝑦𝑧] =
1
10[
20−1010
] = [2
−11
]
⇒ 𝑥 = 2, 𝑦 = −1, 𝑧 = 1
13. Solve system of linear equations, using matrix method
2𝑥 + 3𝑦 + 3𝑧 = 5
𝑥 − 2𝑦 + 𝑧 = −4
3𝑥 − 𝑦 − 2𝑧 = 3
Solution:
The given system of equations are 2𝑥 + 3𝑦 + 3𝑧 = 5, 𝑥 − 2𝑦 + 𝑧 = −4 and 3𝑥 − 𝑦 − 2𝑧 =
3
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [2 3 31 −2 13 −1 −2
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
5−43
]
|𝐴| = 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6) = 10 + 15 + 15 = 40 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
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𝐴11 = 5 𝐴12 = 5 𝐴13 = 5𝐴21 = 3 𝐴22 = −13 𝐴23 = 11𝐴31 = 9 𝐴32 = 1 𝐴33 = −7
Now, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
40[5 3 95 −13 15 11 −7
]
Also, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =
1
40[5 3 95 −13 15 11 −7
] [5
−43
]
⇒ [𝑥𝑦𝑧] =
1
40[25 − 12 + 2725 + 52 + 325 − 44 − 21
] ⇒ [𝑥𝑦𝑧] =
1
40[
4080
−40] = [
12
−1]
⇒ 𝑥 = 1, 𝑦 = 2, 𝑧 = −1
14. Solve system of linear equations, using matrix method
𝑥 − 𝑦 + 2𝑧 = 7
3𝑥 + 4𝑦 − 5𝑧 = −5
2𝑥 − 𝑦 + 3𝑧 = 12
Solution:
The given system of equations are 𝑥 − 𝑦 + 2𝑧 = 7, 3𝑥 + 4𝑦 − 5𝑧 = −5 and 2𝑥 − 𝑦 + 3𝑧 =
12
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [1 −1 23 4 −52 −1 3
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
7−512
]
|𝐴| = 1(12 − 5) + 1(9 + 10) + 2(−3 − 8) = 7 + 19 − 22 = 4 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
𝐴11 = 7 𝐴12 = −19 𝐴13 = −11𝐴21 = 1 𝐴22 = −1 𝐴23 = −1
𝐴31 = −3 𝐴32 = 11 𝐴33 = 7
Now, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
4[
7 1 −3−19 −1 11−11 −1 7
]
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Also, 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =
1
4[
7 1 −3−19 −1 11−11 −1 7
] [7
−512
]
⇒ [𝑥𝑦𝑧] =
1
4[
49 − 5 − 36−133 + 5 + 132−77 + 5 + 84
]
⇒ [𝑥𝑦𝑧] =
1
4[4412
] = [113]
⇒ 𝑥 = 1, 𝑦 = 1, 𝑧 = 3
15. If 𝐴 = [2 −3 53 2 −41 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:
𝐴 = [2 −3 53 2 −41 1 −2
]
|𝐴| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 0 − 6 + 5 = −1 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
𝐴11 = 0 𝐴12 = 2 𝐴13 = 1𝐴21 = −1 𝐴22 = −9 𝐴23 = −5𝐴31 = 2 𝐴32 = 23 𝐴33 = 13
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
−1[0 −1 22 −9 231 −5 13
] = [0 1 −2
−2 9 −23−1 5 −13
]
The given system of equations:
2𝑥 − 3𝑦 + 5𝑧 = 11
3𝑥 + 2𝑦 − 4𝑧 = −5
𝑥 + 𝑦 − 2𝑧 = −3
This system of equations can be written as 𝐴𝑋 = 𝐵, where
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𝐴 = [2 −3 53 2 −41 1 −2
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
11−5−3
]
𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] = [
0 1 −2−2 9 −23−1 5 −13
] [11−5−3
]
⇒ [𝑥𝑦𝑧] = [
0 − 5 + 6−22 − 45 + 69−11 − 25 + 39
] ⇒ [𝑥𝑦𝑧] = [
123]
⇒ 𝑥 = 1, 𝑦 = 2, 𝑧 = 3
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.
Solution:
Let the cost of 1 kg of onion = ₹𝑥,
Let the cost of 1 kg of wheat = ₹𝑦 and
Let the cost of 1 kg rice = ₹𝑧
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. So 4𝑥 + 3𝑦 + 2𝑧 = 60
The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. So 2𝑥 + 4𝑦 + 6𝑧 = 90 and
The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. So 6𝑥 + 2𝑦 + 3𝑧 = 70
Therefore the system of equations are:
4𝑥 + 3𝑦 + 2𝑧 = 60
2𝑥 + 4𝑦 + 6𝑧 = 90
6𝑥 + 2𝑦 + 3𝑧 = 70
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [4 3 22 4 66 2 3
] , 𝑋 = [𝑥𝑦𝑧] and 𝐵 = [
609070
]
|𝐴| = 4(12 − 12) − 3(6 − 36) + 2(4 − 24) = 0 + 90 − 40 = 50 ≠ 0
⇒ 𝐴 is non-singular and so 𝐴−1 exists. Now,
𝐴11 = 0 𝐴12 = 30 𝐴13 = −20𝐴21 = −5 𝐴22 = 0 𝐴23 = 10𝐴31 = 10 𝐴32 = −20 𝐴33 = 10
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Now, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴 =
1
50[
0 −5 1030 0 −20
−20 10 10]
Also 𝑋 = 𝐴−1𝐵 ⇒ [𝑥𝑦𝑧] =
1
50[
0 −5 1030 0 −20
−20 10 10] [
609070
]
⇒ [𝑥𝑦𝑧] =
1
50[
0 − 450 + 7001800 + 0 − 1400
−1200 + 900 + 700]
⇒ [𝑥𝑦𝑧] =
1
50[250400400
] = [588]
⇒ 𝑥 = 5, 𝑦 = 8, 𝑧 = 8
Hence,
the cost of 1 kg of onion is ₹7, the cost of 1 kg of wheat is ₹8 and the cost of 1 kg rice is ₹8.
Miscellaneous Exercise on 𝟒
1. Prove that the determinant |𝑥 sin𝜃 cos 𝜃
− sin𝜃 −𝑥 1cos 𝜃 1 𝑥
| is independent of 𝜃.
Solution:
Let the given determinant be ∆= |𝑥 sin 𝜃 cos 𝜃
− sin𝜃 −𝑥 1cos 𝜃 1 𝑥
|
= 𝑥(−𝑥2 − 1) − sin 𝜃(−𝑥 sin 𝜃 − cos 𝜃) + cos 𝜃(− sin𝜃 + 𝑥 cos𝜃)
= −𝑥3 − 𝑥 + 𝑥 sin2 𝜃 + sin 𝜃 cos 𝜃 − cos𝜃 sin𝜃 + 𝑥 cos2 𝜃
= −𝑥3 − 𝑥 + 𝑥(sin2 𝜃 + cos2 𝜃)
= −𝑥3 − 𝑥 + 𝑥 = −𝑥3, which is independent of 𝜃.
2. Without expanding the determinant, prove that |𝑎 𝑎2 𝑏𝑐𝑏 𝑏2 𝑐𝑎𝑐 𝑐2 𝑎𝑏
| = |1 𝑎2 𝑎3
1 𝑏2 𝑏3
1 𝑐2 𝑐3
|
Solution:
LHS = |𝑎 𝑎2 𝑏𝑐𝑏 𝑏2 𝑐𝑎𝑐 𝑐2 𝑎𝑏
|
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=1
𝑎𝑏𝑐|𝑎2 𝑎3 𝑎𝑏𝑐𝑏2 𝑏3 𝑎𝑏𝑐𝑐2 𝑐3 𝑎𝑏𝑐
| [Applying 𝑅1 → 𝑎𝑅1, 𝑅2 → 𝑏𝑅2, 𝑅3 → 𝑐𝑅3]
=𝑎𝑏𝑐
𝑎𝑏𝑐|𝑎2 𝑎3 1𝑏2 𝑏3 1𝑐2 𝑐3 1
| [Taking 𝑎𝑏𝑐 as common from 𝐶3]
= (−1)1 |1 𝑎3 𝑎2
1 𝑏3 𝑏2
1 𝑐3 𝑐2
| [Interchanging 𝐶1 ⟷ 𝐶3]
= (−1)2 |1 𝑎2 𝑎3
1 𝑏2 𝑏3
1 𝑐2 𝑐3
| [Interchanging 𝐶2 ⟷ 𝐶3]
= |1 𝑎2 𝑎3
1 𝑏2 𝑏3
1 𝑐2 𝑐3
| = RHS
Hence proved.
3. Evaluate |
cos𝛼 cos𝛽 cos𝛼 sin𝛽 − sin𝛼− sin𝛽 cos𝛽 0
sin𝛼 cos𝛽 sin𝛼 sin𝛽 cos 𝛼|
Solution:
Given determinant is |
cos𝛼 cos𝛽 cos𝛼 sin𝛽 − sin𝛼− sin𝛽 cos𝛽 0
sin𝛼 cos𝛽 sin𝛼 sin𝛽 cos𝛼|
= −sin𝛼(− sin𝛼 sin2 𝛽 − sin𝛼 cos2 𝛽)
− 0(cos𝛼 cos𝛽 sin𝛼 sin𝛽 − cos𝛼 sin𝛽 sin𝛼 cos𝛽)
+cos𝛼(cos𝛼 cos2 𝛽 + cos𝛼 sin2 𝛽)
= sin2 𝛼(sin2 𝛽 + cos2 𝛽) + cos2 𝛼(cos2 𝛽 + sin2 𝛽) [Expanding along 𝐶3]
= sin2 𝛼(sin2 𝛽 + cos2 𝛽) + cos2 𝛼(cos2 𝛽 + sin2 𝛽)
= sin2 𝛼 + sin2 𝛼 = 1 [∵ sin2 𝜃 + cos2 𝜃 = 1]
4. If 𝑎, 𝑏 and 𝑐 are real numbers and
∆ = |𝑏 + 𝑐 𝑐 + 𝑎 𝑎 + 𝑏𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎
| = 0
show that either 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 = 𝑏 = 𝑐.
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Solution:
Given that ∆ = |𝑏 + 𝑐 𝑐 + 𝑎 𝑎 + 𝑏𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎
| = 0
⇒ |2(𝑎 + 𝑏 + 𝑐) 2(𝑎 + 𝑏 + 𝑐) 2(𝑎 + 𝑏 + 𝑐)
𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎
| = 0 [Applying 𝑅1 → 𝑅1 + 𝑅2 + 𝑅3]
⇒ 2(𝑎 + 𝑏 + 𝑐) |1 1 1
𝑐 + 𝑎 𝑎 + 𝑏 𝑏 + 𝑐𝑎 + 𝑏 𝑏 + 𝑐 𝑐 + 𝑎
| = 0 [Taking 2(𝑎 + 𝑏 + 𝑐) as common from 𝑅1][
⇒ 2(𝑎 + 𝑏 + 𝑐) |1 0 0
𝑐 + 𝑎 𝑏 − 𝑐 𝑏 − 𝑎𝑎 + 𝑏 𝑐 − 𝑎 𝑐 − 𝑏
| = 0 [Applying 𝐶2 → 𝐶2 − 𝐶1, 𝐶3 → 𝐶3 − 𝐶1]
⇒ 2(𝑎 + 𝑏 + 𝑐)1[(𝑏 − 𝑐)(𝑐 − 𝑏) − (𝑏 − 𝑎)(𝑐 − 𝑎)] = 0 [Expanding along 𝑅1]
⇒ 2(𝑎 + 𝑏 + 𝑐)[𝑏𝑐 − 𝑏2 − 𝑐2 + 𝑏𝑐 − (𝑏𝑐 − 𝑏𝑎 − 𝑎𝑐 + 𝑎2)] = 0
⇒ 2(𝑎 + 𝑏 + 𝑐)[𝑏𝑐 − 𝑏2 − 𝑐2 + 𝑎𝑏 + 𝑐𝑎 − 𝑎2] = 0
⇒ −(𝑎 + 𝑏 + 𝑐)[2𝑎2 + 2𝑏2 + 2𝑐2 − 2𝑎𝑏 − 2𝑏𝑐 − 2𝑐𝑎] = 0
⇒ −(𝑎 + 𝑏 + 𝑐)[(𝑎2 + 𝑏2 − 2𝑎𝑏) + (𝑏2 + 𝑐2 − 2𝑏𝑐) + (𝑐2 + 𝑎2 − 2𝑐𝑎)] = 0
⇒ −(𝑎 + 𝑏 + 𝑐)[(𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2] = 0
⇒ 𝑎 + 𝑏 + 𝑐 = 0 or (𝑎 − 𝑏)2 = 0, (𝑏 − 𝑐)2 = 0, (𝑐 − 𝑎)2 = 0
⇒ 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 − 𝑏 = 0, 𝑏 − 𝑐 = 0, 𝑐 − 𝑎 = 0
⇒ 𝑎 + 𝑏 + 𝑐 = 0 or a = b. 𝑏 = 𝑐, 𝑐 = 𝑎
Therefore, 𝑎 + 𝑏 + 𝑐 = 0 or 𝑎 = 𝑏 = 𝑐
5. Solve the equation |𝑥 + 𝑎 𝑥 𝑥
𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎
| = 0, 𝑎 ≠ 0
Solution:
Given that |𝑥 + 𝑎 𝑥 𝑥
𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎
| = 0
⇒ |3𝑥 + 𝑎 3𝑥 + 𝑎 3𝑥 + 𝑎
𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎
| = 0 [Applying 𝑅1 → 𝑅1 + 𝑅2 + 𝑅3]
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⇒ (3𝑥 + 𝑎) |1 1 1𝑥 𝑥 + 𝑎 𝑥𝑥 𝑥 𝑥 + 𝑎
| = 0 [Taking (3𝑥 + 𝑎) as common from 𝑅1]
⇒ (3𝑥 + 𝑎) |1 0 0𝑥 𝑎 0𝑥 0 𝑎
| = 0 [Applying 𝐶2 → 𝐶2 − 𝐶1, 𝐶3 → 𝐶3 + 𝐶1]
⇒ (3𝑥 + 𝑎)1[𝑎2 − 0] = 0 [Expanding along 𝑅1]
⇒ 𝑎2(3𝑥 + 𝑎) = 0
⇒ (3𝑥 + 𝑎) = 0 [∵ 𝑎 ≠ 0]
⇒ 𝑥 = −𝑎
3
Value of x is −𝑎
3
6. Prove that |𝑎2 𝑏𝑐 𝑎𝑐 + 𝑐2
𝑎2 + 𝑎𝑏 𝑏2 𝑎𝑐𝑎𝑏 𝑏2 + 𝑏𝑐 𝑐2
| = 4𝑎2𝑏2𝑐2
Solution:
LHS = |𝑎2 𝑏𝑐 𝑎𝑐 + 𝑐2
𝑎2 + 𝑎𝑏 𝑏2 𝑎𝑐𝑎𝑏 𝑏2 + 𝑏𝑐 𝑐2
|
= |2𝑎2 0 2𝑎𝑐
𝑎2 + 𝑎𝑏 𝑏2 𝑎𝑐𝑎𝑏 𝑏2 + 𝑏𝑐 𝑐2
| [Applying 𝑅1 → 𝑅1 + 𝑅2 − 𝑅3]
= 𝑎𝑏𝑐 |2𝑎 0 2𝑎
𝑎 + 𝑏 𝑏 𝑎𝑏 𝑏 + 𝑐 𝑐
| [Taking 𝑎, 𝑏, 𝑐 as common from 𝐶1, 𝐶2, 𝐶3]
= 𝑎𝑏𝑐 |2𝑎 0 0
𝑎 + 𝑏 𝑏 −𝑏𝑏 𝑏 + 𝑐 𝑐 − 𝑏
| [Applying 𝐶3 → 𝐶3 − 𝐶1]
= (𝑎𝑏𝑐)2𝑎[𝑏(𝑐 − 𝑏) − (−𝑏)(𝑏 + 𝑐)] [Expanding along 𝑅1]
= 2𝑎2𝑏𝑐[𝑏𝑐 − 𝑏2 + 𝑏2 + 𝑏𝑐] = 2𝑎2𝑏𝑐[2𝑏𝑐]
= 4𝑎2𝑏2𝑐2 = RHS
Hence proved
7. If 𝐴−1 = [3 −1 1
−15 6 −55 −2 2
] and 𝐵 = [1 2 −2
−1 3 00 −2 1
], find (𝐴𝐵)−1.
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Solution:
Given that 𝐴−1 = [3 −1 1
−15 6 −55 −2 2
] and 𝐵 = [1 2 −2
−1 3 00 −2 1
]
Here, 𝐵 = [1 2 −2
−1 3 00 −2 1
]
Therefore, |𝐵| = 1(3 − 0) − 2(−1 − 0) − 2(2 − 0) = 1 ≠ 0 ⇒ 𝐵−1 exists.
𝐵11 = 3 𝐵12 = 1 𝐵13 = 2𝐵21 = 2 𝐵22 = 1 𝐵23 = 2𝐵31 = 6 𝐵32 = 2 𝐵33 = 5
We know, 𝐵−1 =1
|𝐵|𝑎𝑑𝑗 𝐵
=1
1[
𝐵11 𝐵21 𝐵31
𝐵12 𝐵22 𝐵32
𝐵13 𝐵23 𝐵33
] = [3 2 61 1 22 2 5
]
Also (𝐴𝐵)−1 = 𝐵−1𝐴−1, therefore
(𝐴𝐵)−1 = 𝐵−1𝐴−1𝑧
= [3 2 61 1 22 2 5
] [3 −1 1
−15 6 −55 −2 2
]
= [9 − 30 + 30 −3 + 12 − 12 3 − 10 + 123 − 15 + 10 −1 + 6 − 4 1 − 5 + 46 − 30 + 25 −2 + 12 − 10 2 − 10 + 10
] = [9 −3 5
−2 1 01 0 2
]
Hence, (𝐴𝐵)−1 = [9 −3 5
−2 1 01 0 2
]
8. Let 𝐴 = [1 −2 1
−2 3 11 1 5
]. Verify that
(i) (𝑎𝑑𝑗 𝐴)−1 = 𝑎𝑑𝑗(𝐴−1)
(ii) (𝐴−1)−1 = 𝐴
Solution:
(i)
Given matrix is 𝐴 = [1 −2 1
−2 3 11 1 5
], therefore
|𝐴| = 1(15 − 1) + 2(−10 − 1) + 1(−2 − 3) = −13 ≠ 0 ⇒ 𝐴−1 exists.
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𝐴11 = 14 𝐴12 = 11 𝐴13 = −5𝐴21 = 11 𝐴22 = 4 𝐴23 = −3𝐴31 = −5 𝐴32 = −3 𝐴33 = −1
𝑎𝑑𝑗 𝐴 = [
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
] = [14 11 −511 4 −3−5 −3 −1
]
𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴 =
1
|𝐴|[
𝐴11 𝐴21 𝐴31
𝐴12 𝐴22 𝐴32
𝐴13 𝐴23 𝐴33
] =1
−13[14 11 −511 4 −3−5 −3 −1
] …(i)
Let, 𝐵 = 𝑎𝑑𝑗 𝐴, so, 𝐵 = [14 11 −511 4 −3−5 −3 −1
], therefore
|𝐵| = 14(−4 − 9) − 11(−11 − 15) − 5(−33 + 20) = −182 + 286 + 65 = 169 ≠ 0
⇒ 𝐵−1 exists.
𝐵11 = −13 𝐵12 = 26 𝐵13 = −13𝐵21 = 26 𝐵22 = −39 𝐵23 = −13
𝐵31 = −13 𝐵32 = −13 𝐵33 = −65
𝐵−1 =1
|𝐵|[
𝐵11 𝐵21 𝐵31
𝐵12 𝐵22 𝐵32
𝐵13 𝐵23 𝐵33
] =1
169[−13 26 −1326 −39 −13
−13 −13 −65] =
1
13[−1 2 −12 −3 −1
−1 −1 −5]
⇒ (𝑎𝑑𝑗 𝐴)−1 =1
13[−1 2 −12 −3 −1
−1 −1 −5] …(ii)
Let, 𝐶 = 𝐴−1, so, 𝐶 =1
−13[14 11 −511 4 −3−5 −3 −1
] =
[ −
14
13−
11
13
5
13
−11
13−
4
13
3
135
13
3
13
1
13]
, therefore
𝐶11 = −1
13𝐶12 =
2
13𝐶13 = −
1
13
𝐶21 =2
13𝐶22 = −
3
13𝐶23 = −
1
13
𝐶31 = −1
13𝐶32 = −
1
13𝐶33 = −
5
13
𝐴𝑑𝑗 𝐶 = [
𝐶11 𝐶21 𝐶31
𝐶12 𝐶22 𝐶32
𝐶13 𝐶23 𝐶33
] =
[ −
1
13
2
13−
1
132
13−
3
13−
1
13
−1
13−
1
13−
5
13]
=1
13[−1 2 −12 −3 −1
−1 −1 −5]
⇒ 𝐴𝑑𝑗 𝐶 = 𝑎𝑑𝑗 (𝐴−1) =1
13[−1 2 −12 −3 −1
−1 −1 −5] …(iii)
From the equations (ii) and (iii) we have, (𝑎𝑑𝑗 𝐴)−1 = 𝑎𝑑𝑗(𝐴−1)
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(ii) From the equation (i), we have,
𝐴−1 =1
−13[14 11 −511 4 −3−5 −3 −1
]
Let, 𝐷 = 𝐴−1, so, 𝐷 =1
−13[14 11 −511 4 −3−5 −3 −1
] =
[ −
14
13−
11
13
5
13
−11
13−
4
13
3
135
13
3
13
1
13]
, therefore
|𝐷| = −(1
13)3
[14(−4 − 9) − 11(−11 − 15) − 5(−33 + 20)]
= −(1
13)3(169) = −
1
13≠ 0 ⇒ 𝐷−1 exists.
𝐷11 = −1
13𝐷12 =
2
13𝐷13 = −
1
13
𝐷21 =2
13𝐷22 = −
3
13𝐷23 = −
1
13
𝐷31 = −1
13𝐷32 = −
1
13𝐷33 = −
5
13
𝐷−1 =1
|𝐷|[
𝐷11 𝐷21 𝐷31
𝐷12 𝐷22 𝐷32
𝐷13 𝐷23 𝐷33
] =1
−1 13⁄
[ −
1
13
2
13−
1
132
13−
3
13−
1
13
−1
13−
1
13−
5
13]
= [1 −2 1
−2 3 11 1 5
]
⇒ 𝐷−1 = (𝐴−1)−1 = [1 −2 1
−2 3 11 1 5
] = 𝐴
9. Evaluate |
𝑥 𝑦 𝑥 + 𝑦𝑦 𝑥 + 𝑦 𝑥
𝑥 + 𝑦 𝑥 𝑦|.
Solution:
Given determinant is |
𝑥 𝑦 𝑥 + 𝑦𝑦 𝑥 + 𝑦 𝑥
𝑥 + 𝑦 𝑥 𝑦|
= |
2(𝑥 + 𝑦) 𝑦 𝑥 + 𝑦
2(𝑥 + 𝑦) 𝑥 + 𝑦 𝑥
2(𝑥 + 𝑦) 𝑥 𝑦| [Applying 𝐶1 → 𝐶1 + 𝐶2 + 𝐶3]
= 2(𝑥 + 𝑦) |
1 𝑦 𝑥 + 𝑦1 𝑥 + 𝑦 𝑥1 𝑥 𝑦
| [Taking 2(𝑥 + 𝑦) as common from 𝐶1]
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= 2(𝑥 + 𝑦) |
0 −𝑥 𝑦0 𝑦 𝑥 − 𝑦1 𝑦 𝑦 + 𝑘
| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= 2(𝑥 + 𝑦){(−𝑥)(𝑥 − 𝑦) − 𝑦. 𝑦} [Expanding along 𝐶1]
= 2(𝑥 + 𝑦)(−𝑥2 + 𝑥𝑦 − 𝑦2)
= −2(𝑥 + 𝑦)(𝑥2 − 𝑥𝑦 + 𝑦2) = −2(𝑥3 + 𝑦3)
Hence, |
𝑥 𝑦 𝑥 + 𝑦𝑦 𝑥 + 𝑦 𝑥
𝑥 + 𝑦 𝑥 𝑦| = −2(𝑥3 + 𝑦3)
10. Evaluate |
1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦
|
Solution:
Given determinant is |
1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦
|
= |
0 −𝑦 00 𝑦 −𝑥1 𝑥 𝑥 + 𝑦
| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= {(−𝑦)(−𝑥) − 𝑦. 0} [Expanding along 𝐶1]
= 𝑥𝑦
Hence, |
1 𝑥 𝑦1 𝑥 + 𝑦 𝑦1 𝑥 𝑥 + 𝑦
| = 𝑥𝑦
11. Using properties of determinants, prove that:
|
𝛼 𝛼2 𝛽 + 𝛾
𝛽 𝛽2 𝛾 + 𝛼
𝛾 𝛾2 𝛼 + 𝛽
| = (𝛽 − 𝛾)(𝛾 − 𝛼)(𝛼 − 𝛽)(𝛼 + 𝛽 + 𝛾)
Solution:
LHS = |
𝛼 𝛼2 𝛽 + 𝛾
𝛽 𝛽2 𝛾 + 𝛼
𝛾 𝛾2 𝛼 + 𝛽
|
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= |
𝛼 𝛼2 𝛼 + 𝛽 + 𝛾
𝛽 𝛽2 𝛼 + 𝛽 + 𝛾
𝛾 𝛾2 𝛼 + 𝛽 + 𝛾
| [Applying 𝐶3 → 𝐶1 + 𝐶3]
= (𝛼 + 𝛽 + 𝛾) |𝛼 𝛼2 1𝛽 𝛽2 1
𝛾 𝛾2 1
| [Taking α + β + γ as common from 𝐶3]
= (𝛼 + 𝛽 + 𝛾) |
𝛼 − 𝛽 𝛼2 − 𝛽2 0
𝛽 − 𝛾 𝛽2 − 𝛾2 0
𝛾 𝛾2 1
| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾) |
1 𝛼 + 𝛽 01 𝛽 + 𝛾 0
𝛾 𝛾2 1|
[Taking 𝛼 − 𝛽 as common from 𝑅1 and 𝛽 − 𝛾 as common from 𝑅2]
= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾){(𝛽 + 𝛾) − (𝛼 + 𝛽)} [Expanding along 𝐶3]
= (𝛼 + 𝛽 + 𝛾)(𝛼 − 𝛽)(𝛽 − 𝛾)(𝛾 − 𝛼) = RHS
Hence proved.
12. Using properties of determinants, prove that:
|
𝑥 𝑥2 1 + 𝑝𝑥3
𝑦 𝑦2 1 + 𝑝𝑦3
𝑧 𝑧2 1 + 𝑝𝑧3
| = (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥), where p is any scalar.
Solution:
LHS = |
𝑥 𝑥2 1 + 𝑝𝑥3
𝑦 𝑦2 1 + 𝑝𝑦3
𝑧 𝑧2 1 + 𝑝𝑧3
|
= |𝑥 𝑥2 1𝑦 𝑦2 1
𝑧 𝑧2 1
| + |
𝑥 𝑥2 𝑝𝑥3
𝑦 𝑦2 𝑝𝑦3
𝑧 𝑧2 𝑝𝑧3
|
= (−1)1 |1 𝑥2 𝑥1 𝑦2 𝑦
1 𝑧2 𝑧
| + 𝑝 |𝑥 𝑥2 𝑥3
𝑦 𝑦2 𝑦3
𝑧 𝑧2 𝑧3
| [Taking 𝑝 as common from 𝐶3]
= (−1)2 |1 𝑥 𝑥2
1 𝑦 𝑦2
1 𝑧 𝑧2
| + 𝑝𝑥𝑦𝑧 |1 𝑥 𝑥2
1 𝑦 𝑦2
1 𝑧 𝑧2
| [Taking 𝑥, 𝑦, 𝑧 as common from 𝑅1, 𝑅2, 𝑅3 respectively]
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= (1 + 𝑝𝑥𝑦𝑧) |1 𝑥 𝑥2
1 𝑦 𝑦2
1 𝑧 𝑧2
| [Taking |1 𝑥 𝑥2
1 𝑦 𝑦2
1 𝑧 𝑧2
| as common]
= (1 + 𝑝𝑥𝑦𝑧) |0 𝑥 − 𝑦 𝑥2 − 𝑦2
0 𝑦 − 𝑧 𝑦2 − 𝑧2
1 𝑧 𝑧2
| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧) |0 1 𝑥 + 𝑦0 1 𝑦 + 𝑧
1 𝑧 𝑧2
| [Taking 𝑥 − 𝑦 as common from 𝑅1 and 𝑦 − 𝑧 as common form 𝑅2]
= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧){(𝑦 + 𝑧) − (𝑥 + 𝑦)} [Expanding along 𝐶1]
= (1 + 𝑝𝑥𝑦𝑧)(𝑥 − 𝑦)(𝑦 − 𝑧)(𝑧 − 𝑥) = RHS
Hence proved.
13. Using properties of determinants, prove that:
|3𝑎 −𝑎 + 𝑏 −𝑎 + 𝑐
−𝑏 + 𝑎 3𝑏 −𝑏 + 𝑐−𝑐 + 𝑎 −𝑐 + 𝑏 3𝑐
| = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎)
Solution:
LHS = |3𝑎 −𝑎 + 𝑏 −𝑎 + 𝑐
−𝑏 + 𝑎 3𝑏 −𝑏 + 𝑐−𝑐 + 𝑎 −𝑐 + 𝑏 3𝑐
|
= |𝑎 + 𝑏 + 𝑐 −𝑎 + 𝑏 −𝑎 + 𝑐𝑎 + 𝑏 + 𝑐 3𝑏 −𝑏 + 𝑐𝑎 + 𝑏 + 𝑐 −𝑐 + 𝑏 3𝑐
| [Applying 𝐶1 → 𝐶1 + 𝐶2 + 𝐶3]
= (𝑎 + 𝑏 + 𝑐) |1 −𝑎 + 𝑏 −𝑎 + 𝑐1 3𝑏 −𝑏 + 𝑐1 −𝑐 + 𝑏 3𝑐
| [Taking (𝑎 + 𝑏 + 𝑐) as common from 𝐶1]
= (𝑎 + 𝑏 + 𝑐) |0 −𝑎 − 2𝑏 −𝑎 + 𝑏0 2𝑏 + 𝑐 −𝑏 − 2𝑐1 −𝑐 + 𝑏 3𝑐
| [Applying 𝑅1 → 𝑅1 − 𝑅2, 𝑅2 → 𝑅2 − 𝑅3]
= (𝑎 + 𝑏 + 𝑐){(−𝑎 − 2𝑏)(−𝑏 − 2𝑐) − (2𝑏 + 𝑐)(−𝑎 + 𝑏)} [Expanding along 𝐶1]
= (𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 2𝑎𝑐 + 2𝑏2 + 4𝑏𝑐 − (−2𝑎𝑏 + 2𝑏2 − 𝑎𝑐 + 𝑏𝑐))
= (𝑎 + 𝑏 + 𝑐)(3𝑎𝑏 + 3𝑏𝑐 + 3𝑐𝑎) = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎) = RHS
Hence proved.
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14. Using properties of determinants, prove that:
|
1 1 + 𝑝 1 + 𝑝 + 𝑞2 3 + 2𝑝 4 + 3𝑝 + 2𝑞3 6 + 3𝑝 10 + 6𝑝 + 3𝑞
| = 1
Solution:
LHS = |
1 1 + 𝑝 1 + 𝑝 + 𝑞2 3 + 2𝑝 4 + 3𝑝 + 2𝑞3 6 + 3𝑝 10 + 6𝑝 + 3𝑞
|
= |
1 1 + 𝑝 1 + 𝑝 + 𝑞0 1 2 + 𝑝0 3 7 + 3𝑝
| [Applying 𝑅2 → 𝑅2 − 2𝑅1, 𝑅3 → 𝑅3 − 3𝑅1]
= 1{1. (7 + 3𝑝) − (3)(2 + 𝑝)} [Expanding along 𝐶1]
= 7 + 3𝑝 − 6 − 3𝑝 = 1 = RHS
Hence proved.
15. Using properties of determinants, prove that:
|
sin 𝛼 cos𝛼 cos(𝛼 + 𝛿)
sin𝛽 cos𝛽 cos(𝛽 + 𝛿)
sin𝛾 cos 𝛾 cos(𝛾 + 𝛿)| = 0
Solution:
LHS = |
sin𝛼 cos 𝛼 cos(𝛼 + 𝛿)
sin𝛽 cos 𝛽 cos(𝛽 + 𝛿)
sin𝛾 cos 𝛾 cos(𝛾 + 𝛿)|
= |
sin𝛼 cos𝛼 cos𝛿 − sin𝛼 sin𝛿 cos(𝛼 + 𝛿)
sin𝛽 cos𝛽 cos𝛿 − sin𝛽 sin 𝛿 cos(𝛽 + 𝛿)
sin𝛾 cos 𝛾 cos 𝛿 − sin 𝛾 sin𝛿 cos(𝛾 + 𝛿)| [Applying 𝐶2 → cos𝛿𝐶2 − sin 𝛿 𝐶1]
= |
sin𝛼 cos(𝛼 + 𝛿) cos(𝛼 + 𝛿)
sin𝛽 cos(𝛽 + 𝛿) cos(𝛽 + 𝛿)
sin𝛾 cos(𝛾 + 𝛿) cos(𝛾 + 𝛿)|
= 0 = RHS [∵ 𝐶2 = 𝐶3]
Hence proved.
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16. Solve the system of equations:
2
𝑥+
3
𝑦+
10
𝑧= 4
4
𝑥−
6
𝑦+
5
𝑧= 1
6
𝑥+
9
𝑦−
20
𝑧= 2
Solution:
The given system of equations are
2
𝑥+
3
𝑦+
10
𝑧= 4
4
𝑥−
6
𝑦+
5
𝑧= 1
6
𝑥+
9
𝑦−
20
𝑧= 2
This system of equations can be written as 𝐴𝑋 = 𝐵, where
𝐴 = [2 3 104 −6 56 9 −20
] , 𝑋 = [
1 𝑥⁄
1 𝑦⁄
1 𝑧⁄] and 𝐵 = [
412]
Now, |𝐴| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 ≠ 0
⇒ 𝐴−1 exists.
Therefore
𝐴11 = 75 𝐴12 = 110 𝐴13 = 72𝐴21 = 150 𝐴22 = −100 𝐴23 = 0𝐴31 = 75 𝐴32 = 30 𝐴33 = −24
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴 =
1
1200[75 150 75110 −100 3072 0 −24
]
Also, 𝑋 = 𝐴−1𝐵
⇒ [
1 𝑥⁄
1 𝑦⁄
1 𝑧⁄] =
1
1200[75 150 75110 −100 3072 0 −24
] [412]
⇒
[ 1
𝑥1
𝑦1
𝑧]
=1
1200[300 + 150 + 150440 − 100 + 60288 + 0 − 48
]
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⇒
[ 1
𝑥1
𝑦
1
𝑧]
=1
1200[600400240
] =
[ 1
21
31
5]
⇒1
𝑥=
1
2,
1
𝑦=
1
3,
1
𝑧=
1
5⇒ 𝑥 = 2, 𝑦 = 3, 𝑧 = 5
17. Choose the correct answer.
If 𝑎, 𝑏, 𝑐 are in A.P., then the determinant |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎𝑥 + 3 𝑥 + 4 𝑥 + 2𝑏𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐
| is:
(A) 0
(B) 1
(C) 𝑥
(D) 2𝑥
Solution:
Given that 𝑎, 𝑏, 𝑐 are in A.P and determinant is |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎𝑥 + 3 𝑥 + 4 𝑥 + 2𝑏𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐
|
= |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎
0 0 2(2𝑏 − 𝑎 − 𝑐)𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐
| [Applying 𝑅2 → 2𝑅2 − (𝑅1 − 𝑅3)]
= |𝑥 + 2 𝑥 + 3 𝑥 + 2𝑎
0 0 0𝑥 + 4 𝑥 + 5 𝑥 + 2𝑐
| [∵ 𝑎, 𝑏, 𝑐 are in AP, therefore 2𝑏 = 𝑎 + 𝑐]
= 0 [∵ All the elements of 𝑅2 is zero]
Hence, the option (𝐴) is correct.
18. Choose the correct answer.
If 𝑥, 𝑦, 𝑧 are nonzero real numbers, then the inverse of matrix 𝐴 = [𝑥 0 00 𝑦 00 0 𝑧
] is:
(A) [𝑥−1 0 00 𝑦−1 0
0 0 𝑧−1
]
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(B) 𝑥𝑦𝑧 [𝑥−1 0 00 𝑦−1 0
0 0 𝑧−1
]
(C) 1
𝑥𝑦𝑧[𝑥 0 00 𝑦 00 0 𝑧
]
(D) 1
𝑥𝑦𝑧[1 0 00 1 00 0 1
]
Solution:
Given matrix is 𝐴 = [𝑥 0 00 𝑦 00 0 𝑧
]
Now, |𝐴| = 𝑥(𝑦𝑧 − 0) − 0(0 − 0) + 0(0 − 0) = 𝑥𝑦𝑧 ≠ 0 ⇒ 𝐴−1 exists. Therefore,
𝐴11 = 𝑦𝑧 𝐴12 = 0 𝐴13 = 0𝐴21 = 0 𝐴22 = 𝑥𝑧 𝐴23 = 0𝐴31 = 0 𝐴32 = 0 𝐴33 = 𝑥𝑦
We know, 𝐴−1 =1
|𝐴|𝑎𝑑𝑗 𝐴
=1
𝑥𝑦𝑧[
𝑦𝑧 0 00 𝑥𝑧 00 0 𝑥𝑦
]
=
[ 1
𝑥0 0
01
𝑦0
0 01
𝑧]
= [𝑥−1 0 00 𝑦−1 0
0 0 𝑧−1
]
Hence, the option (𝐴) is correct.
19. Choose the correct answer.
Let 𝐴 = [1 sin 𝜃 1
−sin 𝜃 1 sin 𝜃−1 − sin 𝜃 1
], where 0 ≤ 𝜃 ≤ 2𝜋 then:
(A) det(𝐴) = 0
(B) det(𝐴) ∈ (2,∞)
(C) det(𝐴) ∈ (2, 4)
(D) det(𝐴) ∈ [2, 4]
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Solution:
Given matrix is 𝐴 = [1 sin 𝜃 1
−sin 𝜃 1 sin 𝜃−1 −sin 𝜃 1
]
= 1(1 + sin2 𝜃) + sin𝜃(− sin 𝜃 + sin𝜃) + 1(sin2 𝜃 + 1) [Expanding along 𝐶1]
= 2(1 + sin2 𝜃)
Now, given that: 0 ≤ 𝜃 ≤ 2𝜋
⇒ 0 ≤ sin𝜃 ≤ 1
⇒ 0 ≤ sin2 𝜃 ≤ 1
⇒ 1 ≤ 1 + sin2 𝜃 ≤ 2
⇒ 2 ≤ 2(1 + sin2 𝜃) ≤ 4
⇒ det(𝐴) ∈ [2, 4]
Hence, the option (𝐷) is correct.