OP Malhotra Class 12 Maths Solutions Chapter 5 Determinants Ex 5(d)

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S Chand Class 12 ICSE Maths Solutions Chapter 5 Determinants Ex 5(d)

Question 1.
(a) Without evaluating problems (i) to (x) state why each statement is true.
(i) \(\left|\begin{array}{rrr}
2 & 3 & 1 \\
0 & 0 & 0 \\
-1 & 2 & 0
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{lll}
3 & 1 & 3 \\
0 & 1 & 0 \\
1 & 2 & 1
\end{array}\right|=0\)
(iii) \(\left|\begin{array}{rrr}
-2 & 1 & 0 \\
3 & 4 & 1 \\
-4 & 2 & 0
\end{array}\right|=0\)
(iv) \(\left|\begin{array}{llll}
7 & 3 & 2 & 0 \\
2 & 1 & 2 & 0 \\
4 & 1 & 1 & 0 \\
0 & 2 & 1 & 0
\end{array}\right|=0\)
(v) \(\left|\begin{array}{llll}
2 & 3 & 1 & 1 \\
2 & 0 & 1 & 2 \\
2 & 3 & 1 & 1 \\
0 & 1 & 2 & 0
\end{array}\right|=0\)
(vi) \(\left|\begin{array}{rrrr}
6 & 1 & 3 & 2 \\
-2 & 0 & 1 & 4 \\
3 & 6 & 1 & 2 \\
-4 & 0 & 2 & 8
\end{array}\right|=0\)
(vii) \(\left|\begin{array}{lll}
3 & 2 & 1 \\
4 & 0 & 7 \\
5 & 3 & 4
\end{array}\right|=\left|\begin{array}{lll}
3 & 4 & 5 \\
2 & 0 & 3 \\
1 & 7 & 4
\end{array}\right|\)
(viii) \(\left|\begin{array}{rrr}
2 & 3 & 21 \\
11 & 4 & 7 \\
6 & 15 & 8
\end{array}\right|=\left|\begin{array}{rrr}
21 & 2 & 3 \\
7 & 11 & 4 \\
8 & 6 & 15
\end{array}\right|\)
(ix) \(\left|\begin{array}{rrr}
1 & 2 & 7 \\
6 & 0 & 13 \\
8 & 3 & 5
\end{array}\right|=-\left|\begin{array}{rrr}
6 & 0 & 13 \\
1 & 2 & 7 \\
8 & 3 & 5
\end{array}\right|\)
(x) \(\left|\begin{array}{rrr}
2+3 & -1 & 2 \\
3+4 & 0 & 1 \\
4+5 & 3 & 0
\end{array}\right|\) = \(\left|\begin{array}{rrr}
2 & -1 & 2 \\
3 & 0 & 1 \\
4 & 3 & 0
\end{array}\right|+\left|\begin{array}{rrr}
3 & -1 & 2 \\
4 & 0 & 1 \\
5 & 3 & 0
\end{array}\right|\)

(b) Without actually expanding the determinants but starting and using the theorems on determinants, show that
\(\left|\begin{array}{rrr}
1 & 2 & 3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right|=\left|\begin{array}{rrr}
1 & 2 & 3 \\
1 & 1 & 1 \\
1 & 0 & -1
\end{array}\right|\).
Solution:

Question 2.
Without expanding the determinants show that
(i) \(\left|\begin{array}{lll}
42 & 1 & 6 \\
28 & 7 & 4 \\
14 & 3 & 2
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{rrr}
5 & 15 & -25 \\
7 & 21 & 30 \\
8 & 24 & 42
\end{array}\right|=0\)
(iii) \(\left|\begin{array}{ccc}
x+y & x & x \\
5 x+4 y & 4 x & 2 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|=0\)
(iv) \(\left|\begin{array}{lll}
1 / a & a^2 & b c \\
1 / b & b^2 & c a \\
1 / c & c^2 & a b
\end{array}\right|=0\)
(v) \(\left|\begin{array}{lll}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega
\end{array}\right|=0\), where ω is one of the cube roots of unity.
Solution:
(i) Let ∆ = \(\left|\begin{array}{lll}
42 & 1 & 6 \\
28 & 7 & 4 \\
14 & 3 & 2
\end{array}\right|\)
Taking 7 common from C₁
= \(7\left|\begin{array}{lll}
6 & 1 & 6 \\
4 & 7 & 4 \\
2 & 3 & 2
\end{array}\right|=7 \times 0=0\)
[∵ C₁ and C₃ are identical]

(ii) Let ∆ = \(\left|\begin{array}{rrr}
5 & 15 & -25 \\
7 & 21 & 30 \\
8 & 24 & 42
\end{array}\right|\)
Taking 3 common from C₂

Taking y common from C₁ and x common from C₂ in 2nd det.
= \(\left|\begin{array}{ccc}
x & x & x \\
5 x & 4 x & 2 x \\
10 x & 8 x & 3 x
\end{array}\right|+x y\left|\begin{array}{ccc}
1 & 1 & 1 \\
4 & 4 & 2 \\
8 & 8 & 3
\end{array}\right|\)
operate R₂ → R₂ – 5R₁ ;
R₃ → R₃ – 10R₁ in 1st det
= \(\left|\begin{array}{ccc}
x & x & x \\
0 & -x & -3 x \\
0 & -2 x & -7 x
\end{array}\right|+x y \times 0\)
[∵ C₁ and C₂ are identical in det II]
Expanding along C₁
= \(x\left|\begin{array}{cc}
-x & -3 x \\
-2 x & -7 x
\end{array}\right|=x\left(+7 x^2-6 x^2\right)=x^3\)

Question 3.
Using the properties of determinants, show that:
(i) \(\left|\begin{array}{ccc}
0 & p-q & p-r \\
q-p & 0 & q-r \\
r-p & r-q & 0
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{ccc}
1 & x+y & x^2+y^2 \\
1 & y+z & y^2+z^2 \\
1 & z+x & z^2+x^2
\end{array}\right|\) = (x – y)(y – z)(z – x)
(iii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
b+c & c+a & a+b
\end{array}\right|=0\)
(iv) \(\left|\begin{array}{ccc}
2 & 3 & 7 \\
13 & 17 & 5 \\
15 & 20 & 12
\end{array}\right|=0\)
(v) \(\left|\begin{array}{ccc}
x+\lambda & x & x \\
x & x+\lambda & x \\
x & x & x+\lambda
\end{array}\right|=\lambda^2(3 \lambda+\lambda)\)
(vi) \(\left|\begin{array}{ccc}
\sin ^2 x & \cos ^2 x & 1 \\
\cos ^2 x & \sin ^2 x & 1 \\
-10 & 12 & 2
\end{array}\right|=0\)
Solution:

Question 4.
Without expanding the determinants at any stage, 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, ß, γ are in A.P.
Solution:
Let ∆ = \(\left|\begin{array}{lll}
x-3 & x-4 & x-\alpha \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\)
operate R₁ → R₁ + R₂ – 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, ß, γ are in A.P
∴ 2ß = α + y
= \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
x-2 & x-3 & x-\beta \\
x-1 & x-2 & x-\gamma
\end{array}\right|\) = 0
[∵ R₁ be a zero row]

Using properties of Determients, evaluate

Question 5.
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0
[∵ R₁ be a zero row]

Question 6.
\(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\);
operate R₂ → R₂ + R₁
R₃ → R₃ + R₁
= \(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
0 & 0 & 2 \\
0 & 2 & 0
\end{array}\right|\) ; Expanding along C₁
= \(-1\left|\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right|\) = – 1(0 – 4) = 4

Question 7.
\(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right|\);
operate R₂ → R₂ – 4R₁
R₃ → R₃ – 9R₁
= \(\left|\begin{array}{ccc}
1 & 4 & 9 \\
0 & -7 & -20 \\
0 & -20 & -56
\end{array}\right|\) ; Expanding along C₁
= 7 x 56 – 20 x 20
= 392 – 400 = – 8

Question 8.
Solve the following equations :
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|\) = 0
(ii) \(\left|\begin{array}{ccc}
15-2 x & 11 & 10 \\
11-3 x & 17 & 16 \\
7-x & 14 & 13
\end{array}\right|\) = 0
(iii) \(\left|\begin{array}{ccc}
x-1 & 1 & 1 \\
1 & x-1 & 1 \\
1 & 1 & x-1
\end{array}\right|\) = 0
(iv) \(\left|\begin{array}{ccc}
3-\lambda & -1 & 1 \\
-1 & 5-\lambda & -1 \\
1 & -1 & 3-\lambda
\end{array}\right|\) = 0
Solution:
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁
R₃ → R₃ – R₁
Expanding along C₁
⇒ – 6(5x² – 20) + 15(2x – 4) = 0
⇒ – 30x² + 30x + 60 = 0
⇒ x² – x – 2 = 0
⇒ (x + 1) (x – 2) = 0
⇒ x = – 1, 2

(ii) \(\left|\begin{array}{ccc}
15-2 x & 11 & 10 \\
11-3 x & 17 & 16 \\
7-x & 14 & 13
\end{array}\right|\) = 0;
operate R₂ → R₂ – R₁
R₃ → R₃ + R₁

Question 9.
\(\left|\begin{array}{lll}
a & a^2 & b+c \\
b & b^2 & a+c \\
c & c^2 & a+b
\end{array}\right|\) = (b – c)(c – a)(a – b)(a + b + c)
Solution:

Question 10.
\(\left|\begin{array}{lll}
1 & a & b+c \\
1 & b & a+c \\
1 & c & a+b
\end{array}\right|=0\)
Solution:

Question 11.
\(\left|\begin{array}{lll}
\sin \alpha & \cos \alpha & \cos (\alpha+\beta) \\
\sin \beta & \cos \beta & \cos (\beta+\delta) \\
\sin \gamma & \cos \gamma & \cos (\gamma+\delta)
\end{array}\right|=0\)
Solution:

Question 12.
\(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\) = 4xyz
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\)
operate C₁ → C₁ – C₂ – C₃
= \(\left|\begin{array}{ccc}
0 & z & y \\
-2 x & z+x & x \\
-2 x & x & x+y
\end{array}\right|\);
operate R₂ → R₂ – R₃
= \(\left|\begin{array}{ccc}
0 & z & y \\
0 & z & -y \\
-2 x & x & x+y
\end{array}\right|\); Expanding along C₃
= \(-2 x\left|\begin{array}{cc}
z & y \\
z & -y
\end{array}\right|\) = – 2 x(- yz – yz) = 4xyz

Question 13.
\(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2 (a + b)(b + c)(c + a)
Solution:

Question 14.
\(\left|\begin{array}{lll}
1 & x & x^3 \\
1 & y & y^3 \\
1 & z & z^3
\end{array}\right|\) = (x-y)(y-z)(z-x)(x + y + z)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
1 & x & x^3 \\
1 & y & y^3 \\
1 & z & z^3
\end{array}\right|\);
operate R₂ → R₂ + R₁; R₃ → R₃ + R₁
= \(\left|\begin{array}{rrr}
1 . & x & x^3 \\
0 & y-x & y^3-x^3 \\
0 & z-x & z^3-x^3
\end{array}\right|\);
Taking (y – A) common from R₂
and (z – x) common from R₃
= \((y-x)(z-x)\left|\begin{array}{rrr}
1 & x & x^3 \\
0 & 1 & y^2+x y+x^2 \\
0 & 1 & z^2+x^2+x z
\end{array}\right|\);
Expanding along C₁
= (y-x)(z-x) (z² + x² + xz-y² -xy-x²)
= (y-x)(z- x)[(z -y)(z+y) + x(z- y)]
= (x – y)(y ~ z) (z – x) (x + y + z)

Question 15.
\(\left|\begin{array}{rrr}
1 & 1 & 1 \\
\alpha & \beta & \gamma \\
\beta \gamma & \gamma \alpha & \alpha \beta
\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\)
Solution:
Let ∆ = \(\left|\begin{array}{rrr}
1 & 1 & 1 \\
\alpha & \beta & \gamma \\
\beta \gamma & \gamma \alpha & \alpha \beta
\end{array}\right|\);
operate C₂ → C₂ – C₁; C₃ → C₃ – C₁
= \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
\alpha & \beta-\alpha & \gamma-\alpha \\
\beta \gamma & \gamma(\alpha-\beta) & \beta(\alpha-\gamma)
\end{array}\right|\);
Taking (ß – α) common from C₂
and (γ – α) common from C₃
= \((\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{rrr}
1 & 0 & 0 \\
\alpha & 1 & 1 \\
\beta \gamma & -\gamma & -\beta
\end{array}\right|\);
Expanding along R₁
= (ß – α)(γ – α)(-ß + γ)
= (α – ß)(ß – γ)(γ – α)

Question 16.
\(\left|\begin{array}{lll}
1 & b c & b c(b+c) \\
1 & c a & c a(c+a) \\
1 & a b & a b(a+b)
\end{array}\right|\) = 0
Solution:

Question 17.
\(\left|\begin{array}{lll}
1 & b+c & b^2+c^2 \\
1 & c+a & c^2+a^2 \\
1 & a+b & a^2+b^2
\end{array}\right|\) = (b – c)(c – a)(a – b)
Solution:

Question 18.
(i) \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\) = = 2 (a + b + c)(ab + bc + ca – a² – b² – c²)
(ii) \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|=2\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\)
Solution:

Question 19.
\(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
y z & z x & x y
\end{array}\right|\) = (x-y)(y-z)(z-x)(xy + yz + zx)
Solution:

Question 20.
\(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3
\end{array}\right|=x y z(x-y)(y-z)(z-x)\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3
\end{array}\right|=x y z(x-y)(y-z)(z-x)\)
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; C₃ → C₃ – C₁
= xyz\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & y-x & z-x \\
x^2 & y^2-x^2 & z^2-x^2
\end{array}\right|\)
Taking (y – x) common from C₂ and (z – x) common from C₃
= xyz\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
x^2 & y+x & z+x
\end{array}\right|\)
Expanding along R₁
= xyz(y – x) (z – x) (z + x – x – y)
= xyz (x – y) (y – z) (z – x)

Question 21.
\(\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|=(a+b+c)^3\)
Solution:

Question 22.
(i) \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|=x^2(x+a+b+c)\)
(ii) \(\left|\begin{array}{lll}
x & p & q \\
p & x & q \\
p & q & x
\end{array}\right|=(x+p+q)(x-p)(x-q)\)
(iii) \(\left|\begin{array}{lll}
x & a & a \\
a & x & a \\
a & a & x
\end{array}\right|=(x+2 a)(x-a)^2\)
(iv) \(\left|\begin{array}{ccc}
1+a_1 & a_2 & a_3 \\
a_1 & 1+a_2 & a_3 \\
a_1 & a_2 & 1+a_3
\end{array}\right|=1+a_1+a_2+a_3\)
Solution:
(i) Let ∆ = \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|=x^2(x+a+b+c)\); operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
x+a+b+c & b & c \\
x+a+b+c & x+b & c \\
x+a+b+c & b & x+c
\end{array}\right|\)
Taking (x + a + b + c) common from C₁
= (x + a + b + c)\(\left|\begin{array}{ccc}
1 & b & c \\
1 & x+b & c \\
1 & b & x+c
\end{array}\right|\);
operate R₂ → R₂ – R₁; R₃ → R₃ – R₁
= (x + a + b + c)\(\left|\begin{array}{lll}
1 & b & c \\
0 & x & 0 \\
0 & 0 & x
\end{array}\right|\);
Expanding along C₁
= (x + a + b + c)x²

(ii) Let ∆ = \(\left|\begin{array}{lll}
x & p & q \\
p & x & q \\
p & q & x
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{lll}
x+p+q & p & q \\
x+p+q & x & q \\
x+p+q & q & x
\end{array}\right|\);
Taking (x + p + q) common from C₁
= \((x+p+q)\left|\begin{array}{lll}
1 & p & q \\
1 & x & q \\
1 & q & x
\end{array}\right|\)
operate R₂ → R₂ – R₁ ; R₃ → R₃ – R₁
= \((x+p+q)\left|\begin{array}{ccc}
1 & p & q \\
0 & x-p & 0 \\
0 & q-p & x-q
\end{array}\right|\);
Expanding along C₁
= (x + p + q) (x – p) (x – q)

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

(iv) Let ∆ = \(\left|\begin{array}{ccc}
1+a_1 & a_2 & a_3 \\
a_1 & 1+a_2 & a_3 \\
a_1 & a_2 & 1+a_3
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
1+a_1+a_2+a_3 & a_2 & a_3 \\
1+a_1+a_2+a_3 & 1+a_2 & a_3 \\
1+a_1+a_2+a_3 & a_2 & 1+a_3
\end{array}\right|\);
Taking 1 + a₁ + a₂ + a₃ common from C₁
= \(\left(1+a_1+a_2+a_3\right)\left|\begin{array}{ccc}
1 & a_2 & a_3 \\
1 & 1+a_2 & a_3 \\
1 & a_2 & 1+a_3
\end{array}\right|\);
operate R₂ → R₂ – R₁ ; R₃ → R₃ – R₁
= \(\left(1+a_1+a_2+a_3\right)\left|\begin{array}{ccc}
1 & a_2 & a_3 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\);
Expanding along C₁
= (1 + a₁ + a₂ + a₃)(1 – 0)
= 1 + a₁ + a₂ + a₃

Question 23.
(i) \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|\) = a³ + b³ + c³ – 3abc
(ii) \(\left|\begin{array}{ccc}
a & b-c & c-b \\
a-c & b & c-a \\
a-b & b-a & c
\end{array}\right|\) = (a + b – c)(b + c – a)(c + a – b)
Solution:
(i) Let ∆ = \(\left|\begin{array}{ccc}
a & b & c \\
a-b & b-c & c-a \\
b+c & c+a & a+b
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
a+b+c & b & c \\
0 & b-c & c-a \\
2(a+b+c) & c+a & a+b
\end{array}\right|\);
Taking (a + b + c) common from C₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
2 & c+a & a+b
\end{array}\right|\);
operate R₂ → R₃ – 2R₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
0 & c+a-2 b & a+b-2 c
\end{array}\right|\);
operate R₃ → R₃ + R₂
= \((a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & b-c & c-a \\
0 & a-b & b-c
\end{array}\right|\);
Expanding along C₁
= (a + b + c)[(b – c)² – (a – b) (c – a))
= (a + b + c) (b² + c² + a² – ab – bc – ca)
= a³ + b³ + c³ – 3abc

(ii) Let ∆ = \(\left|\begin{array}{ccc}
a & b-c & c-b \\
a-c & b & c-a \\
a-b & b-a & c
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
a+b-c & b-c & c-b \\
a+b-c & b & c-a \\
0 & b-a & c
\end{array}\right|\);
Taking (a + b + c) common from C₁
= \((a+b-c)\left|\begin{array}{ccc}
1 & b-c & c-b \\
1 & b & c-a \\
0 & b-a & c
\end{array}\right|\);
operate R₂ → R₂ – R₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & b-c & c-b \\
0 & c & b-a \\
0 & b-a & c
\end{array}\right|\);
operate R₂ → R₂ – R₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & b-c & c-b \\
0 & c & b-a \\
0 & b-a & c
\end{array}\right|\);
Expanding along C₁
= (a + b + c)\(\left|\begin{array}{cc}
c & b-a \\
b-a & c
\end{array}\right|\)
= (a + b – c)[c² – (b – a)²]
= (a + b – c)(c – b + a)(c + b – a)

Question 24.
\(\left|\begin{array}{ccc}
a^2 & b c & a c+c^2 \\
a^2+a b & b^2 & a c \\
a b & b^2+b c & c^2
\end{array}\right|\) = 4a²b²c²
Solution:
\(\left|\begin{array}{ccc}
a^2 & b c & a c+c^2 \\
a^2+a b & b^2 & a c \\
a b & b^2+b c & c^2
\end{array}\right|\);
Taking a common from C₁, b common from C₂, and c common from C₃,
= \((a b c)\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
b & b+c & c
\end{array}\right|\);
operate R₂ → R₂ – R₁ – R₃
= \((a b c)\left|\begin{array}{ccc}
a & c & a+c \\
0 & -2 c & -2 c \\
b & b+c & c
\end{array}\right|\)
operate C₂ → C₂ – C₃
= \((a b c)\left|\begin{array}{ccc}
a & -a & a+c \\
0 & 0 & -2 c \\
b & b & c
\end{array}\right|\)
Expanding along R₂
= \((a b c)(+2 c)\left|\begin{array}{cc}
a & -a \\
b & b
\end{array}\right|\)
= (abc)(2c) (ab + ab)
= 4a²b²c²

Question 25.
Using the properties of determinants, prove that \(\left|\begin{array}{ccc}
1 & x & x^2 \\
x^2 & 1 & x \\
x & x^2 & 1
\end{array}\right|=\left(1-x^3\right)^2\)
Solution:

Question 26.
Solve the equation
\(\left|\begin{array}{ccc}
x+a & x & x \\
x & x+a & x \\
x & x & x+a
\end{array}\right|=0, a \neq 0\)
Solution:

Question 27.
Without expanding the determinants, show that
\(\left|\begin{array}{lll}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|=\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

Question 28.
Using properties of determinants, find the value of the following determinant :
\(\left|\begin{array}{ccc}
x^3 & x^2 & x \\
y^3 & y^2 & y \\
z^3 & z^2 & z
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
x^3 & x^2 & x \\
y^3 & y^2 & y \\
z^3 & z^2 & z
\end{array}\right|\)
Taking x common from R₁, y common from R₂ and z common from R₃
= \(x y z\left|\begin{array}{lll}
x^2 & x & 1 \\
y^2 & y & 1 \\
z^2 & z & 1
\end{array}\right|\)
operate R₂ → R₂ – R₁; R₃ → R₃ – R₁
= \(x y z\left|\begin{array}{ccc}
x^2 & x & 1 \\
y^2-x^2 & y-x & 0 \\
z^2-x^2 & z-x & 0
\end{array}\right|\)
Taking (y – x) common from R₂
and (z – x) common from R₃
= xyz(y – x) (z – x)\(\left|\begin{array}{ccc}
x^2 & x & 1 \\
y+x & 1 & 0 \\
z+x & 1 & 0
\end{array}\right|\);
Expanding along C₃
= xyz(y – x) (z – x) (y + x – z – x)
= – xyz (x – y) (y – z) (z – x)

Examples

Question 1.
Show that the value of the following determinant is negative, if a, b and c are positive and unequal \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\).
Solution:
Let ∆ = \(\frac { 1 }{ 2 }\); operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{lll}
a+b+c & b & c \\
b+c+a & c & a \\
c+a+b & a & b
\end{array}\right|\);
Taking (a + b + c) common from C₁
= \((a+b+c)\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|\); operate R₂ → R₂ – R₂ ; R₃ → R₃ – R₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & c-b & a-c \\
0 & a-b & b-c
\end{array}\right|\);
Expanding along C₁
= (a + b + c) [-(b – c)² – (a – b) (a – c)]
= (a + b + c) [ab + bc + ca – a² – b² – c²]
= \(\frac { -1 }{ 2 }\) (a + b + c) [2a² + 2b² + 2c² – 2ab – 2bc – 2ca]
= \(\frac {- 1 }{ 2 }\)(a + b + c) [(a – b)² + (b – c)² + (c – a)²]
Given a, b, c > 0 & a ≠ b ≠ c
∴ a – b ≠ b – c ≠ c – a # 0
From (1) ; ∆ < 0 if a, b, c are positive and a, b & c are all unequal.

Question 2.
Use properties of determinants to solve for x :
\(\left|\begin{array}{ccc}
x+a & b & c \\
c & x+b & a \\
a & b & x+c
\end{array}\right|\)
Solution:

Question 3.
Show that
\(\left|\begin{array}{ccc}
\sin ^2 \mathrm{~A} & \sin \mathrm{A} & \cos ^2 \mathrm{~A} \\
\sin ^2 \mathrm{~B} & \sin \mathrm{B} & \cos ^2 \mathrm{~B} \\
\sin ^2 \mathrm{C} & \sin \mathrm{C} & \cos ^2 \mathrm{C}
\end{array}\right|\) = – (sin A – sin B)(sinB – sin C)(sin C – tan A)
Solution:

Question 4.
Prove that
\(\left|\begin{array}{ccc}
y+z & z & y \\
z & z+x & x \\
y & x & x+y
\end{array}\right|\) = 4xyz
Solution:

Using properties of determinants, prove that
Question 5.
\(\left|\begin{array}{lll}
y+z & x+y & x \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\) = x³ + y³ + z² – 3xyz
Solution:
Let ∆ = \(\left|\begin{array}{lll}
y+z & x+y & x \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\)
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
2(x+y+z) & 2(x+y+z) & x+y+z \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\);
Taking (x + y + z) common from R₁, we have
= \((x+y+z)\left|\begin{array}{ccc}
2 & 2 & 1 \\
z+x & y+z & y \\
x+y & z+x & z
\end{array}\right|\); operate C₁ → C₁ – 2C₃ ; C₂ → C₂ – 2C₃
= \((x+y+z)\left|\begin{array}{ccc}
0 & 0 & 1 \\
z+x-2 y & z-y & y \\
x+y-2 z & x-z & z
\end{array}\right|\);
= (x + y + z)[(z + x – 2y)(x – z) – (z – y)(x + y – 2z]
= (x + y + z)[x² – z² – 2xy + 2yz – xz – yz + 2z² + xy + y² – 2yz]
= (x + y + z)[x² + y² + z² – xy – yz – zx]
= x³ + y³ + z³ – 3xyz

Question 6.
\(\left|\begin{array}{ccc}
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3 \\
x y z & x y z & x y z
\end{array}\right|\) = xyz(x – y){y – z)(z – x)(xy + yz + zx)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3 \\
x y z & x y z & x y z
\end{array}\right|\)
Taking xyz common from R₃
\(x y z\left|\begin{array}{ccc}
x^2 & y^2 & z^2 \\
x^3 & y^3 & z^3 \\
1 & 1 & 1
\end{array}\right|\); operate C₁ → C₁ – C₃ ; C₂ → C₂ – C₃
= \(x y z\left|\begin{array}{ccc}
x^2-z^2 & y^2-z^2 & z^2 \\
x^3-z^3 & y^3-z^3 & z^3 \\
0 & 0 & 1
\end{array}\right|\);
Taking (x – z) common from C₁ & (y – z) common from C₂
= \((x y z)(x-z)(y-z)\left|\begin{array}{ccc}
x+z & y+z & z^2 \\
x^2+x z+z^2 & y^2+y z+z^2 & z^3 \\
0 & 0 & 1
\end{array}\right|\); operate C₂ → C₂ – C₁
= \((x y z)(x-z)(y-z)\left|\begin{array}{ccc}
x-y & y+z & z^2 \\
(x-y)(x+y+z) & y^2+y z+z^2 & z^3 \\
0 & 0 & 1
\end{array}\right|\);
Taking (x – y) common from C₁
= (xyz)(x – y)(y – z)(x – z)\(\left|\begin{array}{ccc}
1 & y+z & z^2 \\
x+y+z & y^2+y z+z^2 & z^3 \\
0 & 0 & 1
\end{array}\right|\); Expanding along R₃
= (xyz)(x – y)(y – z)(x – z)\(\left|\begin{array}{cc}
1 & y+z \\
x+y+z & y^2+y z+z^2
\end{array}\right|\)
= (xyz)(x – y)(y – z)(x – z)(y² + yz + z² – xy – xz – y² – z² – 2yz)
= (xyz)(x – y)(y – x)(x – z)(- xy – yz – zx)
= (xyz)(x – y)(y – z)(z – x)(xy + yz + zx)

Question 7.
Using properties of determinants, find the value of the following determinants :
D = \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
b a & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\)
Solution:
D = \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
b a & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\)
Taking a common from R₁, b common from R₂ and c common from R₃
= abc\(\left|\begin{array}{ccc}
-a & b & c \\
a & -b & c \\
a & b & -c
\end{array}\right|\)
Taking a common from C₁, b common from C₂ and c common from C₃
= (abc)(abc)\(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\)
operate R₂ → R₂ + R₁ & R₃ → R₃ + R₁
= \(a^2 b^2 c^2\left|\begin{array}{ccc}
-1 & 1 & 1 \\
0 & 0 & 2 \\
0 & 2 & 0
\end{array}\right|\); Expanding along C₁
= \(a^2 b^2 c^2(-1)\left|\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right|=4 a^2 b^2 c^2\)

Question 8.
If x ≠ y ≠ z and \(\left|\begin{array}{ccc}
x & x^2 & 1+x^3 \\
y & y^2 & 1+y^3 \\
z & z^2 & 1+z^3
\end{array}\right|\) = 0, show that xyz = 1.
Solution:
Given \(\left|\begin{array}{ccc}
x & x^2 & 1+x^3 \\
y & y^2 & 1+y^3 \\
z & z^2 & 1+z^3
\end{array}\right|\) = 0
⇒ \(\left|\begin{array}{ccc}
x & x^2 & 1 \\
y & y^2 & 1 \\
z & z^2 & 1
\end{array}\right|+\left|\begin{array}{ccc}
x & x^2 & x^3 \\
y & y^2 & y^3 \\
z & z^2 & z^3
\end{array}\right|\) = 0
Taking x common from R₁, y common from R₂ and z common from R₃ in 2nd determinant Also, pass C₃ over first two columns in Ist det.

Expanding along C₁
(1 + xyz)(y – x)(z – x) (z + x – y – x) = 0
⇒ (1 + xyz)(x – y)(y – z) (z – x) = 0 … (1)
since x ≠ y ≠ z
∴ x – y ≠ 0; y – z ≠ 0 & z – x ≠ 0
∴ (x – y)(y – z) (z – x) ≠ 0
Thus, from (1); 1 + xyz = 0 ⇒ xyz = – 1

Question 9.
Using properties of determinants, show that \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
b a & b^2+1 & b c \\
c a & c b & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
b a & b^2+1 & b c \\
c a & c b & c^2+1
\end{array}\right|\);
Multiply C₁ by a, C₂ by b and C₃ by c
= \(\frac{1}{a b c}\left|\begin{array}{rrr}
a\left(a^2+1\right) & a b^2 & a c^2 \\
b a^2 & b\left(b^2+1\right) & b c^2 \\
c a^2 & c b^2 & c\left(c^2+1\right)
\end{array}\right|\)
Taking a common from R₁, b common from R₂ and c common from R₃
= \(\frac{a b c}{a b c}\left|\begin{array}{ccc}
a^2+1 & b^2 & c^2 \\
a^2 & b^2+1 & c^2 \\
a^2 & b^2 & c^2+1
\end{array}\right|\)
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{ccc}
1+a^2+b^2+c^2 & b^2 & c^2 \\
1+a^2+b^2+c^2 & b^2+1 & c^2 \\
1+a^2+b^2+c^2 & b^2 & c^2+1
\end{array}\right|\)
Taking (1 + a² + b² + c²) common from C₁
= \(\left(1+a^2+b^2+c^2\right)\left|\begin{array}{ccc}
1 & b^2 & c^2 \\
1 & b^2+1 & c^2 \\
1 & b^2 & c^2+1
\end{array}\right|\)
operate R₂ → R₂ – R₁; R₃ → R₃ – R₁
= (1 + a² + b² + c²)\(\left|\begin{array}{ccc}
1 & b^2 & c^2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
Expanding along C₁
= (1 + a² + b² + c²)\(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)
= 1 + a² + b² + c²

Question 10.
Show that \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
\alpha^2 & \beta^2 & \gamma^2 \\
\alpha^3 & \beta^3 & \gamma^3
\end{array}\right|\) = (α – ß)(ß – γ)(γ – α)(αß + ßγ + γα).
Solution:

Question 11.
Using properties of determinants, show that \(\left|\begin{array}{ccc}
x-y-z & 2 x & 2 x \\
2 y & y-z-x & 2 y \\
2 z & 2 z & z-x-y
\end{array}\right|\) = (x + y + z)³
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
x-y-z & 2 x & 2 x \\
2 y & y-z-x & 2 y \\
2 z & 2 z & z-x-y
\end{array}\right|\)
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
x+y+z & x+y+z & z+x+y \\
2 y & y-z-x & 2 y \\
2 z & 2 z & z-x-y
\end{array}\right|\);
Taking (x + y + z) common from R₁ = (x + y + z)
\(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 y & y-z-x & 2 y \\
2 z & 2 z & z-x-y
\end{array}\right|\)
operate C₁ → C₁ – C₃; C₂ → C₂ – C₃ = (x + y + z)
\(\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & -(x+y+z) & 2 y \\
x+y+z & z+x+y & z-x-y
\end{array}\right|\)
Expanding along R₁
= (x + y + z) \(\left|\begin{array}{cc}
0 & -(x+y+z) \\
x+y+z & z+x+y
\end{array}\right|\)
= (x + y + z)³

Question 12.
Prove that
\(\left|\begin{array}{ccc}
a & b & a x+b y \\
b & c & b x+c y \\
a x+b y & b x+c y & 0
\end{array}\right|\) = (b² – ac)(ax² + 2bxy + cy²).
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
a & b & a x+b y \\
b & c & b x+c y \\
a x+b y & b x+c y & 0
\end{array}\right|\);
operate R₃ → R₃ – xR₁ – yR₂
= \(\left|\begin{array}{ccc}
a & b & a x+b y \\
b & c & b x+c y \\
0 & 0 & -\left(a x^2+2 b x y+c y^2\right)
\end{array}\right|\)
Expanding along R₃
= – (ax² + 2bxy + cy²) \(\left|\begin{array}{ll}
a & b \\
b & c
\end{array}\right|\)
= (ax² + 2bxy + cy²) (b² – ac)

Question 13.
By using properties of determinants, prove that
\(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin 2 x
\end{array}\right|\) = 2 + 4sin2x
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin 2 x
\end{array}\right|\);
operate C₁ → C₁ + C₂
∴ ∆ = \(\left|\begin{array}{ccc}
2 & \cos ^2 x & 4 \sin 2 x \\
2 & 1+\cos ^2 x & 4 \sin 2 x \\
1 & \cos ^2 x & 1+4 \sin 2 x
\end{array}\right|\)
operate R₂ → R₂ – 2R₃; R₁ → R₁ – 2R₃
= \(\left|\begin{array}{ccc}
0 & -\cos ^2 x & -2-4 \sin 2 x \\
0 & 1-\cos ^2 x & -2-4 \sin 2 x \\
1 & \cos ^2 x & 1+4 \sin 2 x
\end{array}\right|\);
Expanding along C₁
= +1\(\left|\begin{array}{ll}
-\cos ^2 x & -2-4 \sin 2 x \\
\sin ^2 x & -2-4 \sin 2 x
\end{array}\right|\)
= (2 + 4 sin2x)(cos²x + sin²x)
= 2 + 4 sin2x

Question 14.
By using properties of determinants, prove that the determinant
\(\left|\begin{array}{ccc}
a & \sin x & \cos x \\
-\sin x & -a & 1 \\
\cos x & 1 & a
\end{array}\right|\) is independent of x.
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
a & \sin x & \cos x \\
-\sin x & -a & 1 \\
\cos x & 1 & a
\end{array}\right|\);
operate R₂ → R₂ + aR₃
= \(\left|\begin{array}{ccc}
a & \sin x & \cos x \\
-\sin x+a \cos x & 0 & 1+a^2 \\
\cos x & 1 & 0
\end{array}\right|\);
operate C₃ → C₃ – aC₂
= \(\left|\begin{array}{ccc}
a & \sin x & \cos x-a \sin x \\
-\sin x+a \cos x & 0 & 1+a^2 \\
\cos x & 1 & 0
\end{array}\right|\);
Expanding along R₃
= cos x (1 + a²) sin x – 1 {a(1 + a²) + (sin x – acos x) (cos x – a sin x)}
= (1 + a²) sin x cos x
– {a( 1 + a²) – a + (1 + a²) sin x cos x} = – a – a³ + a = – a³
which is clearly independent of x

Question 15.
Using properties of determinants, show that pα² + 2qα + r = 0, given that p, q, r are not in G.P and
\(\left|\begin{array}{ccc}
1 & \frac{q}{p} & \alpha+\frac{q}{p} \\
1 & \frac{r}{q} & \alpha+\frac{r}{q} \\
p \alpha+q & q \alpha+r & 0
\end{array}\right|\) = 0
Solution:

Question 16.
Using properties of determinants, prove that
\(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|\) = a³
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|\)
operate R₂ → R₂ – 2R₁; R₃ → R₃ – 3R₁
= \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 3 a & 7 a+3 b
\end{array}\right|\);
Expanding along C₁ = a \(\left|\begin{array}{cc}
a & 2 a+b \\
3 a & 7 a+3 b
\end{array}\right|\)
= a[a(7a + 3b) – 3a(2a + b)]
= a(a²) = a³

Question 17.
Using properties of determinants, prove that
\(\left|\begin{array}{ccc}
x & y & z \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\) = (x – y)(y – z)(z – x)(x + y + z)
Solution:

Question 18.
Using properties of determinants, prove that
\(\left|\begin{array}{ccc}
1 \neq a^2-b^2 & 2 a b & -2 b \\
2 g b & 1=a^2 \neq b^2 & 2 a \\
2 h & -2 a & 1-a^2-b^2
\end{array}\right|\) = (1 + a² + b²)³
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
1 \neq a^2-b^2 & 2 a b & -2 b \\
2 a b & 1-a^2+b^2 & 2 a \\
2 b & -2 a & 1-a^2-b^2
\end{array}\right|\); operate C₁ → C₁ – bC₃; C₂ → C₂ + aC₃
= \(\left|\begin{array}{ccc}
1+a^2 \neq b^2 & 0 & -2 b \\
0 & 1+a^2+b^2 & 2 a \\
2\left(2-1+a^2 \neq b^2\right) & -a\left(2-1+a^2+b^2\right) & 1-a^2-b^2
\end{array}\right|\)
Taking 1 + a² + b² common from C₁ & C₂
= (1 + a² + b²)² \(\left|\begin{array}{ccc}
1 & 0 & -2 b \\
0 & 1 & 2 a \\
b & -a & 1-a^2-b^2
\end{array}\right|\)
operate R₃ → R₃ – bR₁
= (1 + a² + b²)² \(\left|\begin{array}{ccc}
1 & 0 & -2 b \\
0 & 1 & 2 a \\
0 & -a & 1-a^2+b^2
\end{array}\right|\)
Expanding along C₁
= (1 + a² + b²)² (1 – a² + b² + 2a²) = (1 + a² + b²)³

Question 19.
Using properties of determinants, Prove that
\(\left|\begin{array}{lll}
a & b & b + c \\
c & a & c + a \\
b & c & a + b
\end{array}\right|=(a+b+c)(a-c)^2\)
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a & b & b + c \\
c & a & c + a \\
b & c & a + b
\end{array}\right|\)
operate R₁ → R₁ + R₂ + R₃
= \(\left|\begin{array}{ccc}
a+b+c & a+b+c & 2(a+b+c) \\
c & a & c+a \\
b & c & a+b
\end{array}\right|\);
Taking (a + b + c) common from R₁
= \((a+b+c)\left|\begin{array}{ccc}
1 & 1 & 2 \\
c & a & c+a \\
b & c & a+b
\end{array}\right|\)
operate C₂ → C₂ – C₁; C₃ → C₃ – 2C₁
= (a + b + c)\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
c & a-c & a-c \\
b & c-b & a-b
\end{array}\right|\);
Expanding along R₁
= (a + b + c) [(a – c) (a – b) – (a – c) (c – b)]
= (a + b + c) (a – c) [a – b – c + b]
= (a + b + c) (a – c)²

Question 20.
If any two rows (or columns) of a determinant are identical then the value of the determinant is ………….
Solution:
zero

Question 21.
The value of the determinant \(\left|\begin{array}{cc}
p & p+1 \\
p-1 & p
\end{array}\right|\) is ………………
Solution:
Let ∆ = \(\left|\begin{array}{cc}
p & p+1 \\
p-1 & p
\end{array}\right|\)
= p² – (p – 1) (p + 1)
= p² – (p² – 1)
= p² – p² + 1 = 1

Question 22.
The value of the determinant \(\left|\begin{array}{rrr}
0 & 0 & 2 \\
0 & 2 & 4 \\
-5 & 1 & 6
\end{array}\right|\) is …………..
Solution:
\(\left|\begin{array}{rrr}
0 & 0 & 2 \\
0 & 2 & 4 \\
-5 & 1 & 6
\end{array}\right|\); Expanding along C₁
= – 5\(\left|\begin{array}{ll}
0 & 2 \\
2 & 4
\end{array}\right|\) = – 5(0 – 4) = 20

Question 23.
The value of the determinant
\(\left|\begin{array}{rrr}
2 & 3 & 4 \\
-4 & 5 & 7 \\
8 x & 12 x & 16 x
\end{array}\right|\) is …………..
Solution:
Let ∆ = \(\left|\begin{array}{rrr}
2 & 3 & 4 \\
-4 & 5 & 7 \\
8 x & 12 x & 16 x
\end{array}\right|\)
Taking 4x common from R₃
= 4x × 0 = 0 [∵ R₁ and R₃ are identical]

Question 24.
The minor of the element of second row and third column (a₂₃) is the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\) is ……………..
Solution:
M₂₃ = \(\left|\begin{array}{rr}
2 & -3 \\
1 & 5
\end{array}\right|\) = 10 + 3 = 13

Question 25.
If A = \(\left|\begin{array}{rrr}
5 & 6 & -3 \\
-4 & 3 & 2 \\
-4 & -7 & 3
\end{array}\right|\), then the cofactor of the elements a₂₁ of its 2nd row = …………….
Solution:
A₂₁ = \(\left|\begin{array}{rr}
6 & -3 \\
-7 & 3
\end{array}\right|\) = (18 – 21) = 3

Question 26.
If the points P (7, 5), Q (a, 2a) and R (12, 30) are collinear, then the value of a is …………….
Solution:
Here \(\left|\begin{array}{ccc}
7 & 5 & 1 \\
a & 2 a & 1 \\
12 & 30 & 1
\end{array}\right|\) = 0
[∵ given points are collinear]
⇒ 7 (2a – 30) – 5 (a – 12) + 1 (30a – 24a) = 0
⇒ 14a – 210 – 5a + 60 + 6a = 0
⇒ 15a – 150 = 0
⇒ a = 10

Question 27.
If \(\left|\begin{array}{rrr}
1 & 2 & -1 \\
-3 & 4 & k \\
-4 & 2 & 6
\end{array}\right|\) = 0, then k = ……………..
Solution:
\(\left|\begin{array}{rrr}
1 & 2 & -1 \\
-3 & 4 & k \\
-4 & 2 & 6
\end{array}\right|\) = 0; Expanding along R₁
1 (24 – 2k) – 2 (- 18 + 4k) – 1 (- 6 + 16) = 0
⇒ 24 – 2k + 36 – 8k – 10 = 0
⇒ 50 – 10k = 0
⇒ k = 5

Question 28.
If \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|=\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\) = 0, then k = ……………..
Solution:
\(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|=\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\)
⇒ 3 – x² = 3 – 8 ⇒ 3 – x² = – 5
⇒ x² = 8 ⇒ x = ±2\(\sqrt{2}\)

Question 29.
\(\left|\begin{array}{ccc}
0 & x y^2 & x z^2 \\
x^2 y & 0 & y z^2 \\
x^2 z & z y^2 & 0
\end{array}\right|\) = ……………..
Solution:

Question 30.
If x ∈ N and \(\left|\begin{array}{cc}
x+3 & -2 \\
-3 x & 2 x
\end{array}\right|\) = 8, then find the value of x is
(a) – 2
(b) 2
(c) 0
(d) None of these
Solution:
\(\left|\begin{array}{cc}
x+3 & -2 \\
-3 x & 2 x
\end{array}\right|\) = 8
⇒ 2x (x + 3) – 6x = 8
⇒ x² = 4
⇒ x = ± 2 but x ∈N
∴ x = 2

Question 31.
If A_ij is the co-factor or of the element a_ij the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\)
then write the value of a₃₂.a₃₂.
(a) 110
(b) – 110
(c) 130
(d) None of these
Solution:
A₃₂ = (- 1)³⁺²\(\left|\begin{array}{ll}
2 & 5 \\
6 & 4
\end{array}\right|\)
= – (8 – 30) = 22 and a₃₂ = 5
∴ a₃₂.A₃₂ = 5 x 22 = 110

Question 32.
The area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 square units. The value of k will be
(a) 9
(b) 3
(c) – 9
(d) 6
Solution:
Area of triangle = \(\frac { 1 }{ 2 }\)\(\left|\begin{array}{rrr}
-3 & 0 & 1 \\
3 & 0 & 1 \\
0 & k & 1
\end{array}\right|\);
expanding along R₁
= \(\mid \frac{1}{2}[-3(0-k)+0+1(3 k) \mid\) = |3k| sq. units
also given area of triangle = 9 sq. units
∴ | 3Ak | = 9 ⇒ k = ± 3

Question 33.
Consider the following statements.
I. If any two rows or column of a determinant are identical, then the value of the determinant is zero.
II. If the corresponding rows and column of a determinant are interchanged, then the value of determinants does not change.
III. If any two rows (or columns) of a determinant are interchanged, the value of the determinant changes in sign. Which of these are correct?
(a) I and II
(b) II and III
(c) I, II and III
(d) I and III
Solution:
(c) I, II and III

Question 34.
If there are two values of a which makes determinant
∆ = \(\left|\begin{array}{rrr}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|\) = 86,
then the sum of these numbers is
(a) 4
(b) 5
(c) – 4
(d) 9
Solution:
∆ = \(\left|\begin{array}{rrr}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|\) = 86
Expanding along R₁
1 (2a² + 4) + 2 (4a – 0) + 5 (8 – 0) = 86
⇒ 2a² + 8a – 42 = 0
⇒ a² + 4a – 21 = 0
⇒ (a + 7) (a – 3) = 0
⇒ a = – 7, 3
∴ sum of values of a = – 7 + 3 = – 4

Question 35.
If cos 2θ = 0, then
\(\left|\begin{array}{ccc}
0 & \cos \theta & \sin \theta \\
\cos \theta & \sin \theta & 0 \\
\sin \theta & 0 & \cos \theta
\end{array}\right|\) is equal to
(a) – \(\frac { 1 }{ 2 }\)
(b) 2
(c) – 2
(d) \(\frac { 1 }{ 2 }\)
Solution:

Question 36.
If ∆ = \(\left|\begin{array}{ccc}
0 & \sin \alpha & -\cos \alpha \\
-\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right|\) then ∆ is equal to
(a) – 1
(b) 1
(c) 0
(d) – 2
Solution:
Given ∆ = \(\left|\begin{array}{ccc}
0 & \sin \alpha & -\cos \alpha \\
-\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right|\);
Expanding along R₁
= – sin α {0 – cus α sin ß} – cos α {sin α sin ß} = 0

Question 37.
The value of the determinant
\(\left|\begin{array}{ll}
x^2+x y+y^2 & x+y \\
x^2-x y+y^2 & x-y
\end{array}\right|\) is
(a) 2y³
(b) – 2y³
(c) 2y³
(d) – 2y²
Solution:
Let ∆ = \(\left|\begin{array}{ll}
x^2+x y+y^2 & x+y \\
x^2-x y+y^2 & x-y
\end{array}\right|\)
= (x – y) (x² + xy + y²) – (x + y) (x² – xy + y²)
= (x³ – y³) – (x³ + y³) = – 2y³

Question 38.
If ∆ = \(\left|\begin{array}{rr}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right|\), then ∆ is equal to
(a) a² + b² + c² + d²
(b) a² – b² – c² – d²
(c) a² – b² + c² – d²
(d) a² – b² – c² + d²
Solution:
∆ = \(\left|\begin{array}{rr}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right|\)
= (a + ib) (a – ib) – (c + id) (- c + id)
= a² – (ib)² + (c² – (id)²)
= a² + b² + c² + d²

Question 39.
If ∆ = \(\left|\begin{array}{cc}
1 & \log _b a \\
\log _a b & 1
\end{array}\right|\), then ∆ is equal to
(a) 1
(b) – 1
(c) 0
(d) 2
Solution:
∆ = \(\left|\begin{array}{cc}
1 & \log _b a \\
\log _a b & 1
\end{array}\right|\)
= 1 – log_b a log_ab
= 1 – \(\frac{\log a}{\log b} \times \frac{\log b}{\log a}\) = 1 – 1 = 0

Question 40.
Let f(x) = \(\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|\). If x = – 9 is as root of f(x) = 0, then the other roots are
(a) 2 and 7
(b) 3 and 6
(c) 7 and 3
(d) 6 and 2
Solution:
Given f(x) = \(\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|\)
Expanding along R₁
= x(x² – 12) – 3 (2x – 14) + 7 (12 – 7x)
= x³ – 67x + 126
= (x + 9) (x² – 9x + 14)
= (x + 9) (x – 2) (x – 7)
Thus f(x) = 0 ⇒ (x + 9) (x – 2) (x – 7) = 0
⇒ x = – 9, 2, 7
∴ 2 and 7 are the other roots of f(x) = 0

Question 41.
If \(\left|\begin{array}{rrr}
3 i & -9 i & 1 \\
2 & 9 i & -1 \\
10 & 9 & i
\end{array}\right|\) = x + ly, then
(a) x = 1, y = 1
(b) x = 0, y = 1
(c) x = 0, y = 1
(d) x = 1, y = 0
Solution:
Let ∆ = \(\left|\begin{array}{rrr}
3 i & -9 i & 1 \\
2 & 9 i & -1 \\
10 & 9 & i
\end{array}\right|\); Expanding along R₁
= 3i(- 9 + 9) + 9i(2i + 0) + 1(18 – 90i)
= – 18 + 90i + 18 – 90i = 0 + i0
also, ∆ = x + y
∴ x = y = 0

Question 42.
If A = \(\left|\begin{array}{lll}
4 & k & k \\
0 & k & k \\
0 & 0 & k
\end{array}\right|\) and det (A) = 256, then | k | equals
(a) 4
(b) 5
(c) 6
Solution:
Given A = \(\left|\begin{array}{lll}
4 & k & k \\
0 & k & k \\
0 & 0 & k
\end{array}\right|\); Expanding along C₁
= 4\(\left|\begin{array}{ll}
k & k \\
0 & k
\end{array}\right|\) = 4k²
Given det (A) = 256
∴ 4k² = 256
⇒ k = ± 8
⇒ k² = 64
⇒ | k | = | ± 8 | = 8

Question 43.
If f(x) = \(\left|\begin{array}{ccc}
x & x^2 & x^3 \\
1 & 2 x & 3 x^2 \\
0 & 2 & 6 x
\end{array}\right|\), then f'(x) is equal to
(a) x³ + 6x²
(b) 6x²
(c) 3x
(d) 6x³
Solution:
Given f(x) = \(\left|\begin{array}{ccc}
x & x^2 & x^3 \\
1 & 2 x & 3 x^2 \\
0 & 2 & 6 x
\end{array}\right|\)
∴ f'(x) = \(\left|\begin{array}{ccc}
1 & 2 x & 3 x^2 \\
1 & 2 x & 3 x^2 \\
0 & 2 & 6 x
\end{array}\right|+\left|\begin{array}{ccc}
x & x^2 & x^3 \\
0 & 2 & 6 x \\
0 & 2 & 6 x
\end{array}\right|+\left|\begin{array}{ccc}
x & x^2 & x^3 \\
1 & 2 x & 3 x^2 \\
0 & 0 & 6
\end{array}\right|=0+0+\left|\begin{array}{ccc}
x & x^2 & x^3 \\
1 & 2 x & 3 x^2 \\
0 & 0 & 6
\end{array}\right|\)
(Here R₁ and R₂ are identical in first det, R₂ and R₃ are identical in 2nd det]
Expanding along R₃
= 6\(\left|\begin{array}{ll}
x & x^2 \\
1 & 2 x
\end{array}\right|\) = 6(2x² – x²) = 6x²

Question 44.
If a, b, c are distinct positive real numbers, then the value of the determinant \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\) is
(a) < 0 (b) > 0
(c) 0
(d) ≥ 0
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\); operate C₁ → C₁ + C₂ + C₃
\(\left|\begin{array}{lll}
a+b+c & b & c \\
a+b+c & c & a \\
a+b+c & a & b
\end{array}\right|\), Taking (a + b + c) common from C₁
= \(=(a+b+c)\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right| ; \text { operate } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1\);
= \((a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & c-b & a-c \\
0 & a-b & b-c
\end{array}\right| ; \text { Expanding along } \mathrm{C}_1\);
= (a + b + c) [- (b – c)² – (a – b) (a – c)]
= – (a + b + c) [b² + c² + a² – ab – bc – ca]
= – \(\frac { 1 }{ 2 }\) (a + b + c) [(a – b)² + (b – c)² + (c – a)²]
Since a, b, c are distinct positive real numbers.
∴ a > 0, b > 0, c > 0 and a ≠ b ≠ c
∴ a – b ≠ 0; b – c ≠ 0; c – a ≠ 0
∴ (c – a)² > 0 ; (b – c)² > 0
and (a – b)² > 0 and a + b + c > 0
∴ from (1); ∆ < 0

Question 45.
If A = \(\left|\begin{array}{ccc}
8 & 27 & 125 \\
2 & 3 & 5 \\
1 & 1 & 1
\end{array}\right|\), then A² is equal to
(a) 0
(b) 36
(c) 2400
(d) 3600
Solution:
A = \(\left|\begin{array}{ccc}
8 & 27 & 125 \\
2 & 3 & 5 \\
1 & 1 & 1
\end{array}\right|\); operate C₂ → C₂ – C₁; C₃ → C₃ – C₁
= \(\left|\begin{array}{rrr}
8 & 19 & 117 \\
2 & 1 & 3 \\
1 & 0 & 0
\end{array}\right|\); Expanding along R₃
= 57 – 117 = – 60
∴ A² = (- 60)² = 3600

Question 46.
If Z = \(\left|\begin{array}{ccc}
1 & 1+2 i & -5 i \\
1-2 i & -3 & 5+3 i \\
5 i & 5-3 i & 7
\end{array}\right|\), then
(a) Z is purely real
(b) Z is purely imaginary
(c) Z + \(\overline{z}\) = 0
(d) (Z – \(\overline{z}\)) is purely imaginary.
Solution:
Given Z = \(\left|\begin{array}{ccc}
1 & 1+2 i & -5 i \\
1-2 i & -3 & 5+3 i \\
5 i & 5-3 i & 7
\end{array}\right|\); Expanding along R₁
= 1 {- 21 – (5 + 3i) (5 – 3i)} – (1 + 2i) {7 (1 – 2i) – 5i (5 + 3i)} – 5i {(1 – 2i) (5 – 3i) + 15i}
= {- 21 – (25 + 9)} – (1 + 20 {- 39i + 22} – 5i {5 – 13i – 6 + 15i}
= – 55 – (- 39i + 22 + 78 – 44i) – 5i (2i – 1)
= – 55 – 5i – 100 + 10 + 5c = – 145
which is purely real

Question 47.
f(θ) = \(\left|\begin{array}{ccc}
\cos ^2 \theta & \cos \theta \cdot \sin \theta & -\sin \theta \\
\cos \theta \cdot \sin \theta & \sin ^2 \theta & \cos \theta \\
\sin \theta & -\cos \theta & 0
\end{array}\right|\), then for all θ
(a) f(θ) = 1
(b) f(θ) = 2
(c) f(θ) = 0
(d) none of these
Solution:
Given f(θ) = \(\left|\begin{array}{ccc}
\cos ^2 \theta & \cos \theta \cdot \sin \theta & -\sin \theta \\
\cos \theta \cdot \sin \theta & \sin ^2 \theta & \cos \theta \\
\sin \theta & -\cos \theta & 0
\end{array}\right|\); Expanding along R₁
= cos² θ (θ + cos² θ) – cos θ sin θ (0 – cos θ sin θ) + (- sin θ) (- cos² θ sin θ – sin³ θ)
= cos⁴4 θ + cos² θ sin² θ + sin² θ (cos² θ + sin² θ)
= cos² θ (cos² θ + sin² θ) + sin² θ x 1
= cos² θ + sin² θ = 1

Question 48.
The roots of the equation \(\left|\begin{array}{ccc}
1+x & 3 & 5 \\
2 & 2+x & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0 are
(a) 2, 1, – 9
(b) 1, 1, – 9
(c) – 1, 1, – 9
(d) – 2, – 1, – 8
Solution:
Given \(\left|\begin{array}{ccc}
1+x & 3 & 5 \\
2 & 2+x & 5 \\
2 & 3 & x+4
\end{array}\right|\) = 0; operate C₁ → C₁ + C₂ + C₃
⇒ \(\left|\begin{array}{ccc}
9+x & 3 & 5 \\
9+x & 2+x & 5 \\
9+x & 3 & x+4
\end{array}\right|\) = 0; Taking (9 + x) common from C₁
⇒ (9 + x) \(\left|\begin{array}{ccc}
1 & 3 & 5 \\
1 & 2+x & 5 \\
1 & 3 & x+4
\end{array}\right|\) = 0;
operate R₂ → R₂ – R₁; R₃ → R₃ – R₁
⇒ (9 + x) \(\left|\begin{array}{ccc}
1 & 3 & 5 \\
0 & x-1 & 0 \\
0 & 0 & x-1
\end{array}\right|\);
Expanding along C₁
⇒ (9 + x) (x – 1)² = 0
⇒ x = – 9, 1, 1

Question 49.
Solution of the equation
\(\left|\begin{array}{ccc}
1 & 1 & x \\
p+1 & p+1 & p+x \\
3 & x+1 & x+2
\end{array}\right|\) = 0 are
(a) x = 1, 2
(b) x = 2, 3
(c) x = 1, p, 2
(d) x = 1, 2, – p
Solution:
\(\left|\begin{array}{ccc}
1 & 1 & x \\
p+1 & p+1 & p+x \\
3 & x+1 & x+2
\end{array}\right|\)
operate C₁ → C₁ – C₂
⇒ \(\left|\begin{array}{ccc}
0 & 1 & x \\
0 & p+1 & p+x \\
2-x & x+1 & x+2
\end{array}\right|\)
Expanding along C₁
⇒ \((2-x)\left|\begin{array}{cc}
1 & x \\
p+1 & p+x
\end{array}\right|\) = 0
⇒ (2 – x) [p + x – px – x] = 0
⇒ (2 – x) (p (1 – x)) = 0
⇒ x = 2, 1

Question 50.
What positive value of x makes the determinants \(\left|\begin{array}{cc}
2 x & 3 \\
5 & x
\end{array}\right| \text { and }\left|\begin{array}{cc}
16 & 3 \\
5 & 2
\end{array}\right|\) equal?
Solution:
Given, \(\left|\begin{array}{cc}
2 x & 3 \\
5 & x
\end{array}\right|=\left|\begin{array}{cc}
16 & 3 \\
5 & 2
\end{array}\right|\)
⇒ 2x² – 15 = 32 – 15
⇒ x² = 16 ⇒ x = ≠ 4
∴ Required positive value of x be 4.

Question 51.
Evaluate : \(\left|\begin{array}{cc}
\cos 15^{\circ} & \sin 15^{\circ} \\
\sin 75^{\circ} & \cos 75^{\circ}
\end{array}\right|\).
Solution:
Let ∆ = \(\left|\begin{array}{cc}
\cos 15^{\circ} & \sin 15^{\circ} \\
\sin 75^{\circ} & \cos 75^{\circ}
\end{array}\right|\)
= cos 15° cos 75° – sin 15° sin 75° = cos (75° + 15°)
= cos 90° = 0
[∵ Cos (A + B) = cos A cos B – sin A sin B]

Question 52.
What is the value of the determinant \(\left|\begin{array}{lll}
0 & 2 & 0 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right|\)?
Solution:
Let ∆ = \(\left|\begin{array}{lll}
0 & 2 & 0 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right|\); Expanding along C₂
= – 2 \(\left|\begin{array}{ll}
2 & 4 \\
4 & 6
\end{array}\right|\) = – 2 (12 – 16) = 8

Question 53.
Write the value of the determinant
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\);
operate C₁ → C₁ + C₂ + C₃
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0^* & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\)
[∵ C₁ be a zero column]

Question 54.
Write the value of the determinant
\(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
-3 & -3 & -3
\end{array}\right|\)
Solution:

Question 55.
Find the maximum value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|\).
Solution: