ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.2

Utilizing ISC Mathematics Class 12 Solutions Chapter 4 Determinants Ex 4.2 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.2

Question 1.
Without expanding, find the values of
(i) \(\left|\begin{array}{rrr}
102 & 18 & 36 \\
1 & 3 & 4 \\
17 & 3 & 6
\end{array}\right|\) (NCERT)
(ii) \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\) (NCERT)
(iii) \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
7 & 9 & 11 \\
13 & 15 & 17
\end{array}\right|\)
Solution:
(i) Let ∆ = \(\left|\begin{array}{rrr}
102 & 18 & 36 \\
1 & 3 & 4 \\
17 & 3 & 6
\end{array}\right|\)
Taking 6 common from R₁ ; we have
= 6 \(\left|\begin{array}{rrr}
17 & 3 & 6 \\
1 & 3 & 4 \\
17 & 3 & 6
\end{array}\right|\)
= 6 × 0 = 0
[∵ R₁ and R₃ are identical]

(ii) Let ∆ = \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
operate C₃ → C₃ – C₁
= \(\left|\begin{array}{lll}
2 & 7 & 63 \\
3 & 8 & 72 \\
5 & 9 & 81
\end{array}\right|\) ;
operate C₃ → \(\frac{1}{9}\) C₃
= \(\left|\begin{array}{lll}
2 & 7 & 7 \\
3 & 8 & 8 \\
5 & 9 & 9
\end{array}\right|\) = 0
[∵ C₂ and C₃ are identical]

(iii) Let ∆ = \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
7 & 9 & 11 \\
13 & 15 & 17
\end{array}\right|\) ;
operate R₃ → R₃ + R₁
= \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
7 & 9 & 11 \\
14 & 18 & 22
\end{array}\right|\) ;
Taking 2 common from R₃
= \(\left|\begin{array}{rrr}
1 & 3 & 5 \\
7 & 9 & 11 \\
7 & 9 & 11
\end{array}\right|\)
= 0
[∵ R₂ and R₃ are identical]

Question 1 (old).
Show that |A| = |B|, where A = \(\left[\begin{array}{rrr}
1 & -3 & 2 \\
2 & 3 & -1 \\
0 & 5 & -4
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
-3 & 3 & 5 \\
2 & -1 & -4
\end{array}\right]\).
Solution:
Given A = \(\left[\begin{array}{rrr}
1 & -3 & 2 \\
2 & 3 & -1 \\
0 & 5 & -4
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
-3 & 3 & 5 \\
2 & -1 & -4
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rrr}
1 & -3 & 2 \\
2 & 3 & -1 \\
0 & 5 & -4
\end{array}\right|\) ;
expanding along R₁
= \(\left|\begin{array}{ll}
3 & -1 \\
5 & -4
\end{array}\right|+3\left|\begin{array}{ll}
2 & -1 \\
0 & -4
\end{array}\right|+2\left|\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right|\)
= 1 (- 12 + 5) + 3 (- 8 – 0) + 2 (10 – 0)
= – 7 – 24 + 20
= – 11
|B| = \(\left|\begin{array}{rrr}
1 & 2 & 0 \\
-3 & 3 & 5 \\
2 & -1 & -4
\end{array}\right|\) ;
Expanding along R₁
= \(1\left|\begin{array}{rr}
3 & 5 \\
-1 & -4
\end{array}\right|-2\left|\begin{array}{rr}
-3 & 5 \\
2 & -4
\end{array}\right|+0\left|\begin{array}{rr}
-3 & 3 \\
2 & -1
\end{array}\right|\)
= 1 (- 12 + 5) – 2 (12 – 10) + 0 (3 – 6)
= – 7 – 4
= – 11
Thus, |A| = |B|.

Question 2.
Without expanding, find the values of
(i) \(\left|\begin{array}{lll}
x & a & x+a \\
y & b & y+b \\
z & c & z+c
\end{array}\right|\) (NCERT)
(ii) \(\left|\begin{array}{ccc}
a & b & c \\
a+2 x & b+2 y & c+2 z \\
x & y & z
\end{array}\right|\) (NCERT)
Solution:
(i) Let Δ = \(\left|\begin{array}{ccc}
x & a & x+a \\
y & b & y+b \\
z & c & z+c
\end{array}\right|\)
operate C₁ → C₁ + C₂
= \(\left|\begin{array}{lll}
x+a & a & x+a \\
y+b & b & y+b \\
z+c & c & z+c
\end{array}\right|\)
= 0
[∵ C₁ and C₃ are identical]

(ii) Let Δ = \(\left|\begin{array}{ccc}
a & b & c \\
a+2 x & b+2 y & c+2 z \\
x & y & z
\end{array}\right|\) ;
operate R₁ → R₁ + 2R₃
= \(\left|\begin{array}{ccc}
a+2 x & b+2 y & c+2 z \\
a+2 x & b+2 y & c+2 z \\
x & y & z
\end{array}\right|\)
= 0
[∵R₁ and R₂ are identical]

Question 3.
Without expanding, find the values of
(i) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
(ii) \(\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & b(c+a) \\
1 & a b & c(a+b)
\end{array}\right|\) (NCERT)
Solution:
(i) Let Δ = \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-c & r-p
\end{array}\right|\)
= 0
[∵ C₁ be a zero column]

(ii) Let Δ = \(\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & b(c+a) \\
1 & a b & c(a+b)
\end{array}\right|\) ;
operate C₃ → C₃ + C₂
= \(\left|\begin{array}{lll}
1 & b c & a b+b c+a c \\
1 & c a & b c+a b+c a \\
1 & a b & c a+b c+a b
\end{array}\right|\) ;
Taking (ab + bc + ca) common from C₃
= (ab + bc + ca) \(\left|\begin{array}{ccc}
1 & b c & 1 \\
1 & c a & 1 \\
1 & a b & 1
\end{array}\right|\)
[∵ C₁ and C₃ are identical]

Question 4.
(i) \(\left|\begin{array}{lll}
\frac{1}{a} & 1 & b c \\
\frac{1}{b} & 1 & c a \\
\frac{1}{c} & 1 & a b
\end{array}\right|\)
(ii) Without expanding, find the value of: \(\left|\begin{array}{rrr}
a+b & 2 a+b & 3 a+b \\
2 a+b & 3 a+b & 4 a+b \\
4 a+b & 5 a+b & 6 a+b
\end{array}\right|\)
Solution:
(i) Let Δ = \(\left|\begin{array}{lll}
\frac{1}{a} & 1 & b c \\
\frac{1}{b} & 1 & c a \\
\frac{1}{c} & 1 & a b
\end{array}\right|\) ;
Multiplying R₁ by a ; R₂ by b ; R₃ by c
and divide the whole det by abc.
= \(\frac{1}{a b c}\left|\begin{array}{lll}
1 & a & a b c \\
1 & b & a b c \\
1 & c & a b c
\end{array}\right|\) ;
Taking abc common from C₃
= \(\left|\begin{array}{lll}
1 & a & 1 \\
1 & b & 1 \\
1 & c & 1
\end{array}\right|\)
= 0
[∵ C₁ and C₂ are identical]

(ii) Let Δ = \(\left|\begin{array}{rrr}
a+b & 2 a+b & 3 a+b \\
2 a+b & 3 a+b & 4 a+b \\
4 a+b & 5 a+b & 6 a+b
\end{array}\right|\)
operate C₁ → C₁ + C₃ ; we have
= \(\left|\begin{array}{lll}
2(2 a+b) & 2 a+b & 3 a+b \\
2(3 a+b) & 3 a+b & 4 a+b \\
2(5 a+b) & 5 a+b & 6 a+b
\end{array}\right|\)
Taking 2 common from C₁
= 2 \(\left|\begin{array}{ccc}
2 a+b & 2 a+b & 3 a+b \\
3 a+b & 3 a+b & 4 a+b \\
5 a+b & 5 a+b & 6 a+b
\end{array}\right|\)
= 2 × 0 = 0
[∵ C₁ and C₂ are identical]

Question 5.
If A is a square matrix of order 2 and |A| = – 5, find the value of | 3A |.
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given |A| = – 5,
where A be a square matrix of order 2
∴ |3A| = 3² |A|
= 9 × (- 5)
= – 45

Question 6.
If A is a square matrix of order 3 and | A | = 4, then write the value of |2A| .
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given A be a matrix of order 3 s.t. |A| = 4
∴ |2A| = 2³ |A|
= 8 × 4 = 32.

Question 6 (old).
If A = \(=\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 1 \\
3 & 6
\end{array}\right]\), then find the value of | AB|.
Solution:
Given A = \(=\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\)
and B = \(\left[\begin{array}{rr}
-1 & 1 \\
3 & 6
\end{array}\right]\)
∴ AB = \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\left[\begin{array}{rr}
-1 & 1 \\
3 & 6
\end{array}\right]\)
= \(\left[\begin{array}{rr}
5 & 22 \\
13 & 32
\end{array}\right]\)
∴ |AB| = \(\left[\begin{array}{rr}
5 & 22 \\
13 & 32
\end{array}\right]\)
= 160 – 286
= – 126.

Question 7.
If A is a square matrix of order 3 and |A| = 5, find the value of |3A| .
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given A be a matrix of order 3 s.t. |A| = 5
∴ |3A| = 3³ |A|
= 27 × 5
= 135.

Question 7 (old).
If A = \(\left[\begin{array}{rrr}
5 & 5 \alpha & \alpha \\
0 & \alpha & 5 \alpha \\
0 & 0 & 5
\end{array}\right]\) and |A²| = 25, find the value(s) of α.
Solution:
Now A² = \(\left[\begin{array}{ccc}
5 & 5 \alpha & \alpha \\
0 & \alpha & 5 \alpha \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 \alpha & \alpha \\
0 & \alpha & 5 \alpha \\
0 & 0 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5 & 25 \alpha+5 \alpha^2 & 5 \alpha+25 \alpha^2+5 \alpha \\
0 & \alpha^2 & 5 \alpha^2+25 \alpha \\
0 & 0 & 25
\end{array}\right]\)

∴ |A²| = \(\left|\begin{array}{ccc}
5 & 25 \alpha+5 \alpha^2 & 25 \alpha^2+10 \alpha \\
0 & \alpha^2 & 5 \alpha^2+25 \alpha \\
0 & 0 & 25
\end{array}\right|\) ;
Expanding along C₁
= 5 \(\left|\begin{array}{cc}
\alpha^2 & 5 \alpha^2+25 \alpha \\
0 & 25
\end{array}\right|\)
= 5 × 25 α²
= 125 α²
Also given |A²| = 25
⇒ 125α² = 25
⇒ α² = \(\frac{1}{5}\)
⇒ α = ± \(\frac{1}{\sqrt{5}}\)

Question 8.
If A is a square matrix of order 3 and |A| = – 2, find the value of |- 5A| .
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given A be a square matrix of order 3 s.t. |A| = – 2
∴ |- 5A| = (- 5)³ |A|
= – 125 × (- 2)
= 250.

Question 9.
If A is a matrix of order 3 × 3 and its determinant is 4. then find | 3A |.
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given A be a square matrix of order 3 × 3 s.t. |A| = 4
∴ | 3A | = 3³ |A|
= 27 × 4
= 108

Question 10.
If A is a 3 × 3 matrix and | 3A | = k |A|, then write the value of k.
Solution:
We know that det (kA) = kⁿ |A| = kⁿ det A
where A be a square matrix of order n.
Given A be a square matrix of order 3 × 3 s.t. | 3A | = k |A|
⇒ 3³ |A| = k |A|
⇒ 27 |A| = k |A|
⇒ k = 27

Question 11.
(i) If A is a square matrix such that |A| = 5, write the value of |AA^T|.
(ii) If A is a square matrix satisfying AA’ = I, write the value of |A|.
Solution:
(i) |AA^T| = |A| |A^T|
[∵ |AB| = |A| |B|]
= |A| |A|
[∵ |A^T| = |A|]
= 5 × 5 = 25

(ii) Given AA’ = I
⇒ |AA’| = |I| = I
⇒ |A| |A’| = I
⇒ |A|² = I
⇒ |A| = ± 1.

Question 12.
If A is a square matrix of order 3 such that A² = 2A, then find the value of |A|.
Solution:
Given A² = 2A
⇒ |A²| = |2 A|
⇒ |A|² = 2³ |A|
[If A be a square matrix of order n.
Then |kA| = kⁿ |A|]
⇒ |A| = 8, 0.

Question 13.
If A = \(\left[\begin{array}{rr}
1 & 2 \\
3 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & -4 \\
3 & -2
\end{array}\right]\), find |AB|.
Solution:
A = \(\left[\begin{array}{rr}
1 & 2 \\
3 & -1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{rr}
1 & 2 \\
3 & -1
\end{array}\right|\)
= – 1 – 6
= – 7

and B = \(\left[\begin{array}{ll}
1 & -4 \\
3 & -2
\end{array}\right]\)
∴ |B| = \(\left|\begin{array}{ll}
1 & -4 \\
3 & -2
\end{array}\right|\)
= – 2 + 12
= 10
∴ |AB| = |A| |B|
= – 7 × 10
= – 70

Question 14.
If A = \(\left[\begin{array}{cc}
x & 2 \\
2 & x
\end{array}\right]\) and |A³| = 125, find the value?(s) of x.
Solution:
Given A = \(\left[\begin{array}{cc}
x & 2 \\
2 & x
\end{array}\right]\)
⇒ |A| = \(\left|\begin{array}{ll}
x & 2 \\
2 & x
\end{array}\right|\)
= x² – 4
and |A³| = 125
⇒ |A|³ = 125
⇒ (x² – 4)³ = 125
⇒ x² – 4 = 5
⇒ x = ± 3 [∵ x ∈ R].

Question 15.
Using properties of determinants, solve the following equation for x:
\(\left|\begin{array}{ccc}
x+a & x & x \\
x & x+a & x \\
x & x & x+a
\end{array}\right|\) = 0.
Solution:
Let D = \(\left|\begin{array}{ccc}
x+a & x & x \\
x & x+a & x \\
x & x & x+a
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃ ; we get
= \(\left|\begin{array}{ccc}
3 x+a & x & x \\
3 x+a & x+a & x \\
3 x+a & x & x+a
\end{array}\right|\)
Taking (3x + a) common from C₁
= (3x + a) \(\left|\begin{array}{ccc}
1 & x & x \\
1 & x+a & x \\
1 & x & x+a
\end{array}\right|\)
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= (3x + a) \(\left|\begin{array}{lll}
1 & x & x \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\) ;
Expanding along C₁
= (3x + a) a²

Question 16.
Show that one root of the equation \(\left|\begin{array}{ccc}
x+a & b & c \\
b & x+c & a \\
c & a & x+b
\end{array}\right|\) = 0 is – (a + b + c).
Solution:
Given eqn. be, \(\left|\begin{array}{ccc}
x+a & b & c \\
b & x+c & a \\
c & a & x+b
\end{array}\right|\) = 0
Operate C₁ → C₁ + C₂ + C₃
⇒ \(\left|\begin{array}{ccc}
x+a+b+c & b & c \\
x+b+c+a & x+c & a \\
x+c+a+b & a & x+b
\end{array}\right|\) = 0
Taking (x + a + b + c) common from C₁
⇒ (x + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
1 & x+c & a \\
1 & a & x+b
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
⇒ (x + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
0 & x+c-b & a-c \\
0 & a-b & x+b-c
\end{array}\right|\) = 0
Expanding along C₁ ; we have
(x + a + b + c) [(x + c – b) (x + b – c) – (a – b) (a – c)] = 0
⇒ (x + a + b + c) [x² – (b – c)² – a² + ac + ab – bc] = 0
⇒ (x + a + h + c)[x² – a² – b² – c² + ac + ab + bc] = 0
∴ x + a + b + c = 0
or x² = a² + b² + c² – ac – bc – ab = 0
⇒ x = – (a + b + c) be one of the root of given eqn.

Question 17.
Without expanding, prove that \(\left|\begin{array}{lll}
a & b & c \\
x & y & z \\
p & q & r
\end{array}\right|=\left|\begin{array}{lll}
x & y & z \\
p & q & r \\
a & b & c
\end{array}\right|=\left|\begin{array}{ccc}
y & b & q \\
x & a & p \\
z & c & r
\end{array}\right|\).
Solution:
Let D = \(\left|\begin{array}{lll}
a & b & c \\
x & y & z \\
p & q & r
\end{array}\right|\) ;
operate R₁ → R₂
= – \(\left|\begin{array}{lll}
x & y & z \\
a & b & c \\
p & q & r
\end{array}\right|\) ;
operate R₂ ↔ R₃
= \(-(-1)\left|\begin{array}{lll}
x & y & z \\
p & q & r \\
a & b & c
\end{array}\right|=\left|\begin{array}{lll}
x & y & z \\
p & q & r \\
a & b & c
\end{array}\right|\)
[Since |A| = |A^T]
= \(\left|\begin{array}{lll}
x & p & a \\
y & q & b \\
z & r & c
\end{array}\right|\)
operate R₂ ↔ R₁
= – \(\left|\begin{array}{lll}
y & q & b \\
x & p & a \\
z & r & c
\end{array}\right|\)
operate C₃ ↔ C₂
= \(\left|\begin{array}{lll}
y & b & q \\
x & a & p \\
z & c & r
\end{array}\right|\)

Question 18.
Without expanding, prove that \(\left|\begin{array}{ccc}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\).
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|\) ;
Multiplying R₁ by a ;
R₂ by b
and R₃ by c ;
we have
= \(\frac{1}{a b c}\left|\begin{array}{lll}
a^2 & a^3 & a b c \\
b^2 & b^3 & a b c \\
c^2 & c^3 & a b c
\end{array}\right|\) ;
Taking abc common from C₃
= \(\frac{a b c}{a b c}\left|\begin{array}{ccc}
a^2 & a^3 & 1 \\
b^2 & b^3 & 1 \\
c^2 & c^3 & 1
\end{array}\right|\) ;
pass C₃ over first two columns
= \(\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\)
Hence proved.

Question 19.
Use properties of determinants to solve for x:
(i) \(\left|\begin{array}{ccc}
x+a & b & c \\
c & x+b & a \\
a & b & x+c
\end{array}\right|\) = 0
(ii) \(\left|\begin{array}{ccc}
3-x & -1 & 1 \\
-1 & 5-x & -1 \\
1 & -1 & 3-x
\end{array}\right|\) = 0
Solution:
(i) Given, \(\left|\begin{array}{ccc}
x+a & b & c \\
c & x+b & a \\
a & b & x+c
\end{array}\right|\) = 0
operate C₁ → C₁ + C₂ + C₃
\(\left|\begin{array}{ccc}
x+a+b+c & b & c \\
x+a+b+c & x+b & a \\
x+a+b+c & b & x+c
\end{array}\right|\) = 0
Taking (x + a + b + c) common from C₁
⇒ (x + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
1 & x+b & a \\
1 & b & x+c
\end{array}\right|\) = 0
operate R₂ → R – R₁ ;
R₃ → R₃ – R₁
⇒ (x + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
0 & x & a-c \\
0 & 0 & x
\end{array}\right|\) = 0
Expanding along C₁
⇒ (x + a + b + c) \(\left|\begin{array}{cc}
x & (a-c) \\
0 & x
\end{array}\right|\) = 0
⇒ x² (x + a + b + c) = 0
⇒ x = 0, 0, – (a + b + c)

(ii) Given eqn. be
\(\left|\begin{array}{ccc}
3-x & -1 & 1 \\
-1 & 5-x & -1 \\
1 & -1 & 3-x
\end{array}\right|\) = 0
operate C₁ → C₁ + C₂ + C₃
\(\left|\begin{array}{ccc}
3-x & -1 & 1 \\
3-x & 5-x & -1 \\
3-x & -1 & 3-x
\end{array}\right|\) = 0
Taking (3 – x) common from C₁
⇒ (3 – x) \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
1 & 5-x & -1 \\
1 & -1 & 3-x
\end{array}\right|\)
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
⇒ (3 – x) \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
0 & 6-x & -2 \\
0 & 0 & 2-x
\end{array}\right|\)
Expanding along C₁
⇒ (3 – x) . 1. \(\left|\begin{array}{cc}
6-x & -2 \\
0 & 2-x
\end{array}\right|\) = 0
⇒ (3 – x) (6 – x) (2 – x) = 0
⇒ x = 3, 6, 2.

Question 20.
If s = a + b + c, then prove that \(\left|\begin{array}{ccc}
s+c & a & b \\
c & s+a & b \\
c & a & s+b
\end{array}\right|\) = 2s³.
Solution:
Given s = a + b + c
Let ∆ = \(\left|\begin{array}{ccc}
s+c & a & b \\
c & s+a & b \\
c & a & s+b
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
s+c+a+b & a & b \\
s+c+a+b & s+a & b \\
s+c+a+b & a & s+b
\end{array}\right|\) ;
Taking (s + a + b + c) common from C₁
= (s + a + b + c) \(\left|\begin{array}{ccc}
1 & a & b \\
1 & s+a & b \\
1 & a & s+b
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= (s + a + b + c) \(\left|\begin{array}{lll}
1 & a & b \\
0 & s & 0 \\
0 & 0 & s
\end{array}\right|\) ;
expanding along C₁
= (s + a + b + c) s²
= (2s) s²
[∵ s = a + b + c]
= 2s³.

Question 21.
If a, b and c are in A.P., then find the value of \(\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|\). (NCERT)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
2 y+4 & 5 y+7 & 8 y+a \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|\)
Operate R₁ → R₁ + R₃ – 2R₂
= \(\left|\begin{array}{ccc}
0 & 0 & a+c-2 b \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|\)
since a, b, c are in A.P.
∴ a + c = 2b
= \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
3 y+5 & 6 y+8 & 9 y+b \\
4 y+6 & 7 y+9 & 10 y+c
\end{array}\right|\)
= 0
[∵ R₁ is a zero matrix]

Question 22.
Using the properties of determinant, prove that \(\left|\begin{array}{ccc}
x-3 & x-4 & x-\alpha \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\) = 0, where α, β and γ are in A.P.
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
x-3 & x-4 & x-\alpha \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\)
Operate R₁ → R₁ + R₃ – 2R₂
= \(\left|\begin{array}{ccc}
0 & 0 & -(\alpha+\gamma-2 \beta) \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\)
Since α, β and γ are in A.P.
∴ α + γ = 2β
= \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\)
= 0
[∵ R₁ be zero row]

Question 23.
Using properties of determinants, prove that:
\(\left|\begin{array}{lll}
a+b & b+c & c+a \\
b+c & c+a & a+b \\
c+a & a+b & b+c
\end{array}\right|=\mathbf{2}\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\).
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a+b & b+c & c+a \\
b+c & c+a & a+b \\
c+a & a+b & b+c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
a & b+c & c+a \\
b & c+a & a+b \\
c & a+b & b+c
\end{array}\right|+\left|\begin{array}{lll}
b & b+c & c+a \\
c & c+a & a+b \\
a & a+b & b+c
\end{array}\right|\)

Question 24.
(i) \(\left|\begin{array}{ccc}
1+a & b & c \\
a & 1+b & c \\
a & b & 1+c
\end{array}\right|\) = 1 + a + b + c
(ii) \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\) = a² (a + x + y + z)
(iii) \(\left|\begin{array}{ccc}
y+k & y & y \\
y & y+k & y \\
y & y & y+k
\end{array}\right|\) = k² (3y + k)
(iv) \(\left|\begin{array}{ccc}
x+4 & 2 x & 2 x \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\) = (5x + 4) (4 – x)²
Solution:
(i) Let ∆ = \(\left|\begin{array}{ccc}
1+a & b & c \\
a & 1+b & c \\
a & b & 1+c
\end{array}\right|\) ;
Operate C₁ → C₁ + C₂ + C₃
= \(\) ;
Taking a + a + b + c common from C₁
= (1 + a + b + c) \(\left|\begin{array}{ccc}
1 & b & c \\
1 & 1+b & c \\
1 & b & 1+c
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= (1 + a + b + c) \(\left|\begin{array}{lll}
1 & b & c \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) ;
expanding along C₁
= (1 + a + b + c) (1 – 0)
= 1 +a + b + c

(ii) Let ∆ = \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
a+x+y+z & y & z \\
a+x+y+z & a+y & z \\
a+x+y+z & y & a+z
\end{array}\right|\) ;
Taking a + x + y + z common C₁ ; we get
= (a + x + y + z) \(\left|\begin{array}{ccc}
1 & y & z \\
1 & a+y & z \\
1 & y & a+z
\end{array}\right|\) ;
Operate R₁ → R₂ – R₁,
R₃ → R₃ – R₁
= (a + x + y + z) \(\left|\begin{array}{lll}
1 & y & z \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\) ;
expanding along C₁ = a² (a + x + y + z)

(iii) Let ∆ = \(\left|\begin{array}{ccc}
y+k & y & y \\
y & y+k & y \\
y & y & y+k
\end{array}\right|\) ;
Operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
3 y+k & y & y \\
3 y+k & y+k & y \\
3 y+k & y & y+k
\end{array}\right|\) ;
Taking (3y + k) common from C₁
= (3y + k) \(\left|\begin{array}{ccc}
1 & y & y \\
1 & y+k & y \\
1 & y & y+k
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= (3y + k) \(\left|\begin{array}{lll}
1 & y & y \\
0 & k & 0 \\
0 & 0 & k
\end{array}\right|\) ;
expanding along C₁
= (3y + k) k²

(iv) L.H.S. = \(\left|\begin{array}{ccc}
x+4 & 2 x & 2 x \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\)
Operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
5 x+4 & 2 x & 2 x \\
5 x+4 & x+4 & 2 x \\
5 x+4 & 2 x & x+4
\end{array}\right|\) ;
Taking (5x + 4) common from C₁
= (5x 4) \(\left|\begin{array}{ccc}
1 & 2 x & 2 x \\
1 & x+4 & 2 x \\
1 & 2 x & x+4
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= (5x 4) \(\left|\begin{array}{ccc}
1 & 2 x & 2 x \\
0 & 4-x & 0 \\
0 & 0 & 4-x
\end{array}\right|\) ;
expanding along C₁
= (5x + 4) (4 – x)²
= R.H.S.

Question 25.
\(\left|\begin{array}{ccc}
a+b+n c & n a-a & n b-b \\
n c-c & b+c+n a & n b-b \\
n c-c & n a-a & c+a+n b
\end{array}\right|\) = n (a + b + c)³.
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
a+b+n c & n a-a & n b-b \\
n c-c & b+c+n a & n b-b \\
n c-c & n a-a & c+a+n b
\end{array}\right|\) ;
Operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
n a+n b+n c & n a-a & n b-b \\
n a+n b+n c & b+c+n a & n b-b \\
n a+n b+n c & n a-a & c+a+n b
\end{array}\right|\) ;
Taking n (a + b + c) common from C₁
= n (a + b + c) \(\left|\begin{array}{ccc}
1 & n a-a & n b-b \\
1 & b+c+n a & n b-b \\
1 & n a-a & c+a+n b
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
= n (a + b + c) \(\left|\begin{array}{ccc}
1 & n a-a & n b-b \\
0 & a+b+c & 0 \\
0 & 0 & c+a+b
\end{array}\right|\) ;
expanding along C₁ ; we have
= n (a + b + c) (a + b + c)²
= n (a + b + c)³

Question 26.
(i) \(\left|\begin{array}{ccc}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|=\left|\begin{array}{ccc}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{array}\right|\) = (x – y) (y – z) (z – x)
(ii) \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3
\end{array}\right|\) = xyz (x – y) (y – z) (z – x)
Solution:
(i) Let Δ = \(\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|\)
Multiply R₁ by x, R₂ by and R₃ by z
Δ = \(\frac{1}{x y z}\left|\begin{array}{ccc}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{array}\right|\)
Taking xyz common from c₃
= \(\frac{x y z}{x y z}\left|\begin{array}{ccc}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|\)
pass c₃ our first two columns
= \(\left|\begin{array}{lll}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ → R₃ – R₁
= \(\left|\begin{array}{rrr}
1 & x & x^2 \\
0 & y-x & y^2-x^2 \\
0 & z-x & z^2-x^2
\end{array}\right|\)
Taking (y – x) common from R₂ and (z – x) common from R₃
= (y – x) (x – z) \(\left|\begin{array}{ccc}
1 & x & x^2 \\
0 & 1 & y+x \\
0 & 0 & z+x
\end{array}\right|\);
Expanding along C₁
= (y – x) (x – z) (z + x – y – x)
= (x – y) (y – z) (z – x)

(ii) L.H.S. = \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3
\end{array}\right|\) ;
Taking x common from C₁,
y common from C₂
and z common from C₃
= xyz \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
x & y & z \\
x^2 & y^2 & z^2
\end{array}\right|\) ;
operate C₂ → C₂ – C₁
and C₃ → C₃ – C₁
= xyz \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
x & y-x & z-x \\
x^2 & y^2-x^2 & z^2-x^2
\end{array}\right|\) ;
Taking (y – x) common from C₂ and (z – x) common from C₃
= xyz (y – x) (z – x) \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
x^2 & y+x & z+x
\end{array}\right|\);
expanding along R₁
= xyz (y – x) (z – x) (z + x – y – x)
= xyz (x – y) (y – z) (z – x)

Question 27.
(i) \(\left|\begin{array}{lll}
x & y & 1 \\
\alpha & x & 1 \\
\alpha & \beta & 1
\end{array}\right|\) = (x – α) (x – β)
(ii) \(\) = xy (NCERT)
(iii) \(\) = (x + y + z) (x – z)²
Solution:
Let Δ = \(\left|\begin{array}{lll}
x & y & 1 \\
\alpha & x & 1 \\
\alpha & \beta & 1
\end{array}\right|\)
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
x & y & 1 \\
\alpha-x & x-y & 0 \\
\alpha-x & \beta-y & 0
\end{array}\right|\) ;
expanding along C₁
= (α – x) (β – y) – (x – y) (α – x)
= (α – x) [β – y – x + y]
= (α – x) (β – x)
= (x – α) (x – β)

(ii) Let Δ = \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{lll}
1 & x & y \\
0 & y & 0 \\
0 & 0 & x
\end{array}\right|\) ;
expanding along C₁
= 1 . \(\left|\begin{array}{ll}
y & 0 \\
0 & x
\end{array}\right|\)
= xy

(iii) Let Δ = \(\left|\begin{array}{lll}
y+z & x & y \\
z+x & z & x \\
x+y & y & z
\end{array}\right|\) ;
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
2(x+y+z) & x+y+z & x+y+z \\
z+x & z & x \\
x+y & y & z
\end{array}\right|\)
Taking (x + y + z) common from R₁
= (x + y + z) \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
z+x & z & x \\
x+y & y & z
\end{array}\right|\)
operate C₁ → C₁ – 2C₃ ;
C₂ → C₂ – C₃
= (x + y + z) \(\left|\begin{array}{ccc}
0 & 0 & 1 \\
z-x & z-x & x \\
x+y-2 z & y-z & z
\end{array}\right|\)
Expanding along R₁ ; we have
= (x + y + z) \(\left|\begin{array}{cc}
z-x & z-x \\
x+y-2 z & y-z
\end{array}\right|\)
= (x + y + z) [(z – x) (y – z) – (z – x) (x + y + 2z)]
= (x + y + z) (z – x) [y – z – x – y + 2z]
= (x – y + z) (z – x)²

Question 28.
\(\left|\begin{array}{rrr}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
c a & b c & -c^2
\end{array}\right|\) = 4a²b²c².
Solution:
Set ∆ = \(\left|\begin{array}{rrr}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
c a & b c & -c^2
\end{array}\right|\)
Taking a common from R₁,
b common from R₂
and c common from R₃
= abc \(\left|\begin{array}{rrr}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right|\) ;
Taking a common from C₁,
b common from C₂
and c common from C₃
= a²b²c² \(\left|\begin{array}{rrr}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\) ;
operate R₂ → R₂ + R₁ ;
R₃ → R₃ + R₁
= a²b²c² \(\left|\begin{array}{rrr}
-1 & 1 & 1 \\
0 & 0 & 2 \\
0 & 2 & 0
\end{array}\right|\) ;
expanding along C₁
D = a²b²c² (- 1) \(\left|\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right|\)
= a²b²c² (- 1) (0 – 4)
= 4a²b²c²

Question 29.
\(\left|\begin{array}{ccc}
1 & \alpha & \alpha^2+\beta \gamma \\
1 & \beta & \beta^2+\gamma \alpha \\
1 & \gamma & \gamma^2+\alpha \beta
\end{array}\right|\) = 2 (α – β) (β – γ) (γ – α)
Solution:
Let Δ = \(\left|\begin{array}{ccc}
1 & \alpha & \alpha^2+\beta \gamma \\
1 & \beta & \beta^2+\gamma \alpha \\
1 & \gamma & \gamma^2+\alpha \beta
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁

= 2 (β – α) (β – γ) (α – γ)
2 (α – β) (β – γ) (γ – α)

Question 30.
\(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|=\left|\begin{array}{ccc}
a^2+1 & b^2 & c^2 \\
a^2 & b^2+1 & c^2 \\
a^2 & b^2 & c^2+1
\end{array}\right|\) = 1 + a² + b² + c². (NCERT)
Solution:
Let D = \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
c a & c b & c^2+1
\end{array}\right|\)
Multiply R₁ by a;
R₂ by b
and R₃ by c

Question 31.
\(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
b c & c a & a b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca)
Solution:
L.H.S. = \(\left|\begin{array}{lll}
a & a^2 & b c \\
b & b^2 & c a \\
c & c^2 & a b
\end{array}\right|\) ;
Multiply R₁ by a;
R₂ by b
and R₃ by c
and taking \(\frac{1}{a b c}\) outside the determinant

= (b – a) (c – a) (c – b) [(b + a) (b + a + c) – b² – ab – a²]
= (b – a) (c – a) (c – b) [b² + a² + 2ab + bc + ac – b² – ab – a²]
= (a – b) (b – c) (c – a) (ab + bc + ca)

Question 32.
\(\left|\begin{array}{ccc}
a^2 & a^2-(b-c)^2 & b c \\
b^2 & b^2-(c-a)^2 & c a \\
c^2 & c^2-(a-b)^2 & a b
\end{array}\right|\) = (a – b) (b – c) (c – a) (a + b + c) (a² + b² + c²)
Solution:
Let D = \(\left|\begin{array}{ccc}
a^2 & a^2-(b-c)^2 & b c \\
b^2 & b^2-(c-a)^2 & c a \\
c^2 & c^2-(a-b)^2 & a b
\end{array}\right|\) ;
operate C₂ → C₂ – C₁ – 2C₃
= \(\mid \begin{array}{lll}
a^2 & -\left(b^2+c^2\right) & b c \\
b^2 & -\left(c^2+a^2\right) & c a \\
c^2 & -\left(a^2+b^2\right) & a b
\end{array}\) ;
Taking (- 1) common from C₂
= – \(\left|\begin{array}{lll}
a^2 & b^2+c^2 & b c \\
b^2 & c^2+a^2 & c a \\
c^2 & a^2+b^2 & a b
\end{array}\right|\) ;
operate C₂ → C₂ + C₁
and then taking a² + b² + c² common from C₂ ; we have
= – (a² + b² + c²) \(\left|\begin{array}{lll}
a^2 & 1 & b c \\
b^2 & 1 & c a \\
c^2 & 1 & a b
\end{array}\right|\) ;
operate R₂ → R₂ – R₁
R₃ → R₃ – R₁
= – (a² + b² + c²) \(\left|\begin{array}{ccc}
a^2 & 1 & b c \\
b^2-a^2 & 0 & c(a-b) \\
c^2-a^2 & 0 & b(a-c)
\end{array}\right|\) ;
Taking b – a common from R₂ and (c – a) common from R₃ ;
we get
= – (a² + b² + c²) \(\left|\begin{array}{ccc}
a^2 & 1 & b c \\
b+a & 0 & -c \\
c+a & 0 & -b
\end{array}\right|\) ;
Expanding along C₂ ; we get
= (a² + b² + c²) \(\left|\begin{array}{ll}
b+a & -c \\
c+a & -b
\end{array}\right|\)
= (a² + b² + c²) (b – a) (c – a) [- b² – ab + c² + ac]
= (a² + b² + c²) (b – a) (c – a) [(c – b) (c + b) + a (c – b)]
= (a² + b² + c²) (a – b) (b – c) (c – a) (a + b + c).

Question 33.
(i) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a – b) (b – c) (c – a) (a + b + c) (a² + b² + c²)
(ii) \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\) = a² (a + x + y + z)
Solution:
Let D = \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) ;
operate C₁ → C₁ – C₁ – 2C₃ ;
we get

Expanding along C₁ ; we get
= (a² + b² + c²) (b – a) (c – a) \(\left|\begin{array}{ll}
b+a & -c \\
c+a & -b
\end{array}\right|\)
= (a² + b² + c²) (b – a) (c – a) [- b² – ab + c² + ac]
= (a² + b² + c²) (b – a) (c – a) [(c -b) (c + b) + a (c – b)]
= (a² + b² + c²) (b – a) (c – a) (c – b) (a + b + c)
= (a² + b² + c²) (a -b) (b – c) (c – a) (a + b + c)

(ii) Let ∆ = \(\left|\begin{array}{ccc}
a+x & y & z \\
x & a+y & z \\
x & y & a+z
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
a+x+y+z & y & z \\
a+x+y+z & a+y & z \\
a+x+y+z & y & a+z
\end{array}\right|\) ;
Taking a + x + y + z common C₁ ; we get
= (a + x + y + z) \(\left|\begin{array}{ccc}
1 & y & z \\
1 & a+y & z \\
1 & y & a+z
\end{array}\right|\) ;
Operate R₁ → R₂ – R₁,
R₃ → R₃ – R₁
= (a + x + y + z) \(\left|\begin{array}{lll}
1 & y & z \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right|\) ;
expanding along C₁ = a² (a + x + y + z)

Question 34.
(i) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2 (a + b) (b + c) (c + a)
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) ;
operate R₁ → R₁ + R₂
= \(\left|\begin{array}{ccc}
a+b & a+b & -(a+b) \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) ;
Taking (a + b) common from R₁
= (a + b) \(\left|\begin{array}{ccc}
1 & 1 & -1 \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) ;
operate C₁ → C₁ + C₃ ;
C₂ → C₂ + C₃
= (a + b) \(\left|\begin{array}{ccc}
0 & 0 & -1 \\
-(a+c) & b+c & -a \\
c+a & b+c & a+b+c
\end{array}\right|\) ;
Taking (a + c) common from C₂
and (b + c) common from C₂
= (a + b) (a + c) (b + c) \(\left|\begin{array}{ccc}
0 & 0 & -1 \\
-1 & 1 & -a \\
1 & 1 & a+b+c
\end{array}\right|\) ;
expanding along R₁
= – (a + b) (b + c) (a + c) (- 2)
= 2 (a + b) (b + c) (c + a).

Question 35.
\(\left|\begin{array}{ccc}
y+z & x+y & x \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\) = x³ + y³ + z³ – 3 xyz
Solution:
Let Δ = \(\left|\begin{array}{ccc}
y+z & x+y & x \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\) ;
Operate C₁ → C₁ + C₃
= \(\left|\begin{array}{ccc}
x+y+z & x+y & x \\
x+y+z & y+z & y \\
x+y+z & z+x & z
\end{array}\right|\) ;
Taking (x + y + z) common from C₁
= (x + y + z) \(\left|\begin{array}{ccc}
1 & x+y & x \\
1 & y+z & y \\
1 & z+x & z
\end{array}\right|\) ;
Operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= (x + y + z) \(\left|\begin{array}{ccc}
1 & x+y & x \\
0 & z-x & y-x \\
0 & z-y & z-x
\end{array}\right|\)
= (x + y + z) \(\left|\begin{array}{cc}
z-x & y-x \\
z-y & z-x
\end{array}\right|\)
= (x + y + z) [(z – x)² – (y – x) (z – y)]
= (x + y + z) (z² + x² + y² – xz – xy – zx)
= x³ + y³ + z³ – 3xyz

Question 36.
\(\left|\begin{array}{ccc}
a & b-c & c-b \\
a-c & b & c-a \\
a-b & b-a & c
\end{array}\right|\) = (a + b – c) (b + c – a) (c + a – b)
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
a & b-c & c-b \\
a-c & b & c-a \\
a-b & b-a & c
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
a & b-c & c-b \\
b & b & c-a \\
c & b-a & c
\end{array}\right|\) ;
operate C₃ → C₃ + C₂
= \(\left|\begin{array}{ccc}
a & b-c & 0 \\
b & b & c-a+b \\
c & b-a & c+b-a
\end{array}\right|\) ;
Taking (b + c – a) common from C₃
= (c + b – a) \(\left|\begin{array}{ccc}
a & b-c & 0 \\
b & b & 1 \\
c & b-a & 1
\end{array}\right|\) ;
operate R₃ → R₃ – R₂
= (b + c – a) \(\left|\begin{array}{ccc}
a & b-c & 0 \\
b & b & 1 \\
c-b & -a & 0
\end{array}\right|\) ;
expanding along C₃
= – (b + c – a) [- a² + (b – c)²]
= – (b + c – a) [(b – c) – a]
= – (b + c – a) (b – c – a) (b – c + a)
= (b + c – a) (c + a – b) (a + b – c)

Question 37.
\(\left|\begin{array}{ccc}
1 & b c+a d & b^2 c^2+a^2 d^2 \\
1 & c a+b d & c^2 a^2+b^2 d^2 \\
1 & a b+c d & a^2 b^2+c^2 d^2
\end{array}\right|\) = (a – b) (a – c) (a – d) (b – c) (b – d) (c – d).
Solution:
L.H.S. = \(\left|\begin{array}{ccc}
1 & b c+a d & b^2 c^2+a^2 d^2 \\
1 & c a+b d & c^2 a^2+b^2 d^2 \\
1 & a b+c d & a^2 b^2+c^2 d^2
\end{array}\right|\)
Operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁

Question 38.
If a, b, c are all non-zero and \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = 0, then prove that \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) + 1 = 0.
Solution:
Given \(\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|\) = 0
Taking a common from R₁,
b from R₂
and c from R₃
abc \(\left|\begin{array}{ccc}
\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\
\frac{1}{b} & \frac{1}{b}=1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|\) = 0;
operate R₁ → R₁ + R₂ + R₃

Question 39.
Without expanding any of the following determinants given below, prove that
\(\mid \begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+1)^2 & (b+1)^2 & (c+1)^2 \\
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array}\) = 4 \(\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right|\).
Solution:
Let Δ = \(\mid \begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+1)^2 & (b+1)^2 & (c+1)^2 \\
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array}\)
Operate R₂ → R₂ – R₃
= \(\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
4 a & 4 b & 4 c \\
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array}\right|\) ;
Taking 4 common from R₂
= 4 \(\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
(a-1)^2 & (b-1)^2 & (c-1)^2
\end{array}\right|\) ;
Operate R₃ → R₃ – R₁ + 2 R₂
= 4 \(\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right|\)
Hence proved.

Question 40.
(i) Using properties of determinants, solve the following equations for x:
(i) \(\left|\begin{array}{ccc}
3 x-8 & 3 & 3 \\
3 & 3 x-8 & 3 \\
3 & 3 & 3 x-8
\end{array}\right|\) = 0
(ii) \(\left|\begin{array}{ccc}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0
(iii) \(\left|\begin{array}{ccc}
x & -6 & -1 \\
2 & -3 x & x-3 \\
-3 & 2 x & x+2
\end{array}\right|\) = 0
(iv) \(\left|\begin{array}{ccc}
15-2 x & 11 & 10 \\
11-3 x & 17 & 16 \\
7-x & 14 & 13
\end{array}\right|\) = 0
Solution:
(i) Given \(\left|\begin{array}{ccc}
3 x-8 & 3 & 3 \\
3 & 3 x-8 & 3 \\
3 & 3 & 3 x-8
\end{array}\right|\) = 0;
operate C₁ → C₁ + C₂ + C₃
\(\left|\begin{array}{ccc}
3 x-2 & 3 & 3 \\
3 x-2 & 3 x-8 & 3 \\
3 x-2 & 3 & 3 x-8
\end{array}\right|\) = 0;
Taking 3x – 2 common from C₁; we have
⇒ (3x – 2) \(\left|\begin{array}{ccc}
1 & 3 & 3 \\
1 & 3 x-8 & 3 \\
1 & 3 & 3 x-8
\end{array}\right|\) = 0 ;
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
⇒ (3x – 2) \(\left|\begin{array}{ccc}
1 & 3 & 3 \\
0 & 3 x-11 & 0 \\
0 & 0 & 3 x-11
\end{array}\right|\) = 0
Expanding along C₁, we have
⇒ (3x – 2) \(\left|\begin{array}{cc}
3 x-11 & 0 \\
0 & 3 x-11
\end{array}\right|\) = 0
⇒ (3x – 2) (3x – 11)² = 0
⇒ x = \(\frac{2}{3}, \frac{11}{3}, \frac{11}{3}\)

(ii) Given \(\left|\begin{array}{ccc}
x+1 & 3 & 5 \\
2 & x+2 & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0
operate C₁ → C₁ + C₂ + C₃
\(\left|\begin{array}{ccc}
x+9 & 3 & 5 \\
x+9 & x+2 & 5 \\
x+9 & 3 & x+4
\end{array}\right|\) = 0 ;
Taking x + 9 common from C₁
⇒ (x + 9) \(\left|\begin{array}{ccc}
1 & 3 & 5 \\
1 & x+2 & 5 \\
1 & 3 & x+4
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁,
R₃ → R₃ – R₁
⇒ (x + 9) \(\left|\begin{array}{ccc}
1 & 3 & 5 \\
0 & x-1 & 0 \\
0 & 0 & x-1
\end{array}\right|\) = 0
Expanding along C₁, we have
⇒ (x + 9) \(\left|\begin{array}{cc}
x-1 & 0 \\
0 & x-1
\end{array}\right|\) = 0
⇒ (x + 9) (x – 1)² = 0
⇒ x = – 9, 1, 1.

(iii) Given \(\left|\begin{array}{ccc}
x & -6 & -1 \\
2 & -3 x & x-3 \\
-3 & 2 x & x+2
\end{array}\right|\) = 0
Expanding along R₁, we get
\(x\left|\begin{array}{cc}
-3 x & x-3 \\
2 x & x+2
\end{array}\right|+6\left|\begin{array}{cc}
2 & x-3 \\
-3 & x+2
\end{array}\right|\) – 1 \(\left|\begin{array}{cc}
2 & -3 x \\
-3 & 2 x
\end{array}\right|\) = 0
⇒ x [- 3x (x + 2) – 2x (x – 3)] + 6 (2x + 4 + 3x – 9) – 1 (4x – 9) = 0
⇒ x² (- 3x – 6 – 2x + 6) + 6 (5x – 5) + 5x = 0
⇒ – 5x³ + 35x – 30 = 0
⇒ x³ – 7x + 6 = 0
⇒ (x – 2) (x² + 2x – 3) = 0
⇒ (x – 2) (x + 3) (x – 1) = 0
⇒ x = 2, 1, – 3
[clearly 2 be the root of given eqn.]
Moreover when x = 2;
∆ = \(\left|\begin{array}{rrr}
2 & -6 & -1 \\
2 & -6 & -1 \\
-3 & 4 & 4
\end{array}\right|\) = 0
[∵ R and R are identical rows]

(iv) Given \(\left|\begin{array}{ccc}
15-2 x & 11-3 x & 7-x \\
11 & 17 & 14 \\
10 & 16 & 13
\end{array}\right|\) = 0
operate R₂ → R₂ – R₃,
⇒ \(\left|\begin{array}{ccc}
15-2 x & 11-3 x & 7-x \\
1 & 1 & 1 \\
10 & 16 & 13
\end{array}\right|\) = 0
operate C₂ → C₂ – C₁ ;
C₃ → C₃ – C₁
⇒ \(\left|\begin{array}{ccc}
15-2 x & -4-x & -8+x \\
1 & 0 & 0 \\
10 & 6 & 3
\end{array}\right|\) = 0
expanding along C₂
⇒ – 1 \(\left|\begin{array}{cc}
-4-x & -8+x \\
6 & 3
\end{array}\right|\) = 0
⇒ – 12 – 3x – 6 (- 8 + x) = 0
⇒ – 12 – 3x + 48 – 6x = 0
⇒ 9x = 36
⇒ x = 4.

Question 41.
If a, b, c are all different and \(\left|\begin{array}{lll}
a & a^3 & a^4-1 \\
b & b^3 & b^4-1 \\
c & c^3 & c^4-1
\end{array}\right|\) = 0, prove that abc (ab + bc + ca) = a + b + c.
Solution:
Given \(\left|\begin{array}{lll}
a & a^3 & a^4-1 \\
b & b^3 & b^4-1 \\
c & c^3 & c^4-1
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{ccc}
a & a^3 & a^4 \\
b & b^3 & b^4 \\
c & c^3 & c^4
\end{array}\right|-\left|\begin{array}{ccc}
a & a^3 & 1 \\
b & b^3 & 1 \\
c & c^3 & 1
\end{array}\right|\) = 0
Taking a, b and c common from R₁, R₂ and R₃ respectively in 1st determinant.

Taking b – a common from C₂ and (c – a) common from C₃
= (b – a) (c – a) \(\left|\begin{array}{ccc}
a & a^3 & 1 \\
1 & b^2+a b+a^2 & 0 \\
1 & c^2+a c+a^2 & 0
\end{array}\right|\) ;
Expanding along C₃
= (b – a) (c – a) \(\left|\begin{array}{ll}
1 & b^2+a b+a^2 \\
1 & c^2+a c+a^2
\end{array}\right|\)
= (b – a) (c – a) [c² + ac + a² – b² – ab – a²]
= (b – a) (c – a) [(c – b) (c + b) + a (c – b)]
= (a – b) (b – c) (c – a) (a + b + c) ……………..(3)
Using eqn. (2) and eqn. (3) in eqn. (1) ; we get
abc (a – b) (b – c) (c – a) (a + b + c) – (a – b) (b – c) (c – a) (a + b + c) = 0
⇒ (a – b) (b – c) (c – a) [abc (ab + bc + ca) – (a + b + c)] = 0 …………….(4)
Since a, b, c are all different
∴ a ≠ b ≠ c
⇒ a – b ≠ 0 ; b – c ≠ 0 and c – a ≠ 0
∴ (a – b) (b – c) (c – a) ≠ 0
Thus eqn. (4) becomes ;
abc (ab + bc + ca) – (a + b + c) = 0
i.e., abc (ab + bc + ca) = a + b + c

Question 42.
In a triangle ABC, if \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1+\cos A & 1+\cos B & 1+\cos C \\
\cos A+\cos ^2 B & \cos B+\cos ^2 B & \cos C+\cos ^2 C
\end{array}\right|\) = 0, then prove that ABC is an isosceless triangle.
Solution:

expanding along R₁
⇒ (cos B – cos A) (cos C – cos A) [1 + cos C + cos A – 1 – cos B – cos A] = 0
⇒ (cos B – cos A) (cos C – cos A) (cos C – cos B) = 0
cos B – cos A = 0 or
cos C – cos A = 0 or
cos C – cos B = 0
⇒ cos B = cos A or cos C = cos A or cos C = cos B
⇒ B = A or C = A or C = B [∵ A, B and C are angles of ΔABC]
Thus ΔABC be an isosceless triangle.