RD Chapter 6 Determinants Ex 6.1 Solutions
Question - 1 : - Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:
Answer - 1 : -
(i) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn.The minor of the matrix can be obtained for a particular element byremoving the row and column where the element is present. Then finding theabsolute value of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
From the given matrix we have,
M11 = –1
M21 = 20
C11 = (–1)1+1 × M11
= 1 × –1
= –1
C21 = (–1)2+1 × M21
= 20 × –1
= –20
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21
= 5× (–1) + 0 × (–20)
= –5
(ii) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of matrix can be obtained for particular element by removingthe row and column where the element is present. Then finding the absolutevalue of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given
From the above matrix we have
M11 = 3
M21 = 4
C11 = (–1)1+1 × M11
= 1 × 3
= 3
C21 = (–1)2+1 × 4
= –1 × 4
= –4
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21
= –1× 3 + 2 × (–4)
= –11
(iii) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of the matrix can be obtained for a particular element byremoving the row and column where the element is present. Then finding theabsolute value of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
M31 = –3 × 2 – (–1) × 2
M31 = –4
C11 = (–1)1+1 × M11
= 1 × –12
= –12
C21 = (–1)2+1 × M21
= –1 × –16
= 16
C31 = (–1)3+1 × M31
= 1 × –4
= –4
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21+ a31× C31
= 1× (–12) + 4 × 16 + 3× (–4)
= –12 + 64 –12
= 40
(iv) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of the matrix can be obtained for a particular element byremoving the row and column where the element is present. Then finding theabsolute value of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
M31 = a × c a – b × bc
M31 = a2c – b2c
C11 = (–1)1+1 × M11
= 1 × (ab2 – ac2)
= ab2 – ac2
C21 = (–1)2+1 × M21
= –1 × (a2b – c2b)
= c2b – a2b
C31 = (–1)3+1 × M31
= 1 × (a2c – b2c)
= a2c – b2c
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21+ a31× C31
= 1× (ab2 – ac2) + 1 × (c2b– a2b) + 1× (a2c – b2c)
= ab2 – ac2 + c2b – a2b+ a2c – b2c
(v) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of matrix can be obtained for particular element by removingthe row and column where the element is present. Then finding the absolutevalue of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
M31 = 2×0 – 5×6
M31 = –30
C11 = (–1)1+1 × M11
= 1 × 5
= 5
C21 = (–1)2+1 × M21
= –1 × –40
= 40
C31 = (–1)3+1 × M31
= 1 × –30
= –30
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21+ a31× C31
= 0× 5 + 1 × 40 + 3× (–30)
= 0 + 40 – 90
= 50
(vi) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of matrix can be obtained for particular element by removingthe row and column where the element is present. Then finding the absolutevalue of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
M31 = h × f – b × g
M31 = hf – bg
C11 = (–1)1+1 × M11
= 1 × (bc– f2)
= bc– f2
C21 = (–1)2+1 × M21
= –1 × (hc – fg)
= fg – hc
C31 = (–1)3+1 × M31
= 1 × (hf – bg)
= hf – bg
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21+ a31× C31
= a× (bc– f2) + h× (fg – hc) + g× (hf – bg)
= abc– af2 + hgf – h2c +ghf – bg2
(vii) Let Mij and Cij representsthe minor and co–factor of an element, where i and j represent the row andcolumn. The minor of matrix can be obtained for particular element by removingthe row and column where the element is present. Then finding the absolutevalue of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
Given,
M31 = –1(1 × 0 – 5 × (–2)) – 0(0 × 0 – (–1) ×(–2)) + 1(0 × 5 – (–1) × 1)
M31 = –9
M41 = –1(1×1 – (–1) × (–2)) – 0(0 × 1 – 1 ×(–2)) + 1(0 × (–1) – 1 × 1)
M41 = 0
C11 = (–1)1+1 × M11
= 1 × (–9)
= –9
C21 = (–1)2+1 × M21
= –1 × 9
= –9
C31 = (–1)3+1 × M31
= 1 × –9
= –9
C41 = (–1)4+1 × M41
= –1 × 0
= 0
Now expanding along the first column we get
|A| = a11 × C11 + a21×C21+ a31× C31 + a41× C41
= 2 × (–9) + (–3) × –9 + 1 × (–9) + 2 × 0
= – 18 + 27 –9
= 0
Question - 2 : - Evaluate thefollowing determinants:
Answer - 2 : -
(i) Given
⇒ |A|= x (5x + 1) – (–7) x
|A| = 5x2 + 8x
(ii) Given
⇒ |A|= cos θ × cos θ – (–sin θ) x sin θ
|A| = cos2θ + sin2θ
We know that cos2θ + sin2θ = 1
|A| = 1
(iii) Given
⇒ |A|= cos15° × cos75° + sin15° x sin75°
We know that cos (A – B) = cos A cos B + Sin A sin B
By substituting this we get, |A| = cos (75 – 15)°
|A| = cos60°
|A| = 0.5
(iv) Given
⇒ |A|= (a + ib) (a – ib) – (c + id) (–c + id)
= (a + ib) (a – ib) + (c + id) (c – id)
= a2 – i2 b2 + c2 –i2 d2
We know that i2 = -1
= a2 – (–1) b2 + c2 –(–1) d2
= a2 + b2 + c2 +d2
Question - 3 : - Evaluate:
Answer - 3 : -
Since |AB|= |A||B|
= 2(17 × 12 – 5 × 20) – 3(13 × 12 – 5 × 15) + 7(13 × 20 – 15 ×17)
= 2 (204 – 100) – 3 (156 – 75) + 7 (260 – 255)
= 2×104 – 3×81 + 7×5
= 208 – 243 +35
= 0
Now |A|2 = |A|×|A|
|A|2= 0
Question - 4 : - Show that
Answer - 4 : -
Given
Let the given determinant as A
Using sin (A+B) = sin A × cos B + cos A × sin B
⇒ |A|= sin 10° × cos 80° + cos 10° x sin 80°
|A| = sin (10 + 80)°
|A| = sin90°
|A| = 1
Hence Proved
Question - 5 : - 
Answer - 5 : -
Given,
= 2(1 × 1 – 4 × (–2)) – 3(7 × 1 – (–2) × (–3)) – 5(7 × 4 – 1 ×(–3))
= 2(1 + 8) – 3(7 – 6) – 5(28 + 3)
= 2 × 9 – 3 × 1 – 5 × 31
= 18 – 3 – 155
= –140
Now by expanding along the second column
= 2(1 × 1 – 4 × (–2)) – 7(3 × 1 – 4 × (–5)) – 3(3 × (–2) – 1 ×(–5))
= 2 (1 + 8) – 7 (3 + 20) – 3 (–6 + 5)
= 2 × 9 – 7 × 23 – 3 × (–1)
= 18 – 161 +3
= –140
Question - 6 : - 
Answer - 6 : -
Question - 7 : -
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Question - 8 : - 
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Question - 10 : - 
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