ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Chapter Test

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Chapter Test

Question 1.
If A and B are square matrices of order 3 such that |A| = – 2 and |B| = 3, find the value of |- 5 AB|.
Solution:
Given A and B are square matrices of order 3 such that
|A| = – 2 and |B| = 3 …………(1)
We know that if A be a square matrix of order n
Then |kA| = kⁿ A …………..(2)
Now |- 5 AB| = (- 5)³ |A| |B| [using (1) and (2)
= – 125 × (- 2) × 3
= 750

Question 2.
Using properties of determinants, prove the following:
(i) \(\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
2 & 3+2 p & 4+3 p+2 q \\
3 & 6+3 p & 10+6 p+3 q
\end{array}\right|\) = 1
(ii) \(\left|\begin{array}{lll}
p a & p^2+a^2 & 1 \\
p b & p^2+b^2 & 1 \\
p c & p^2+c^2 & 1
\end{array}\right|\) = p (a – b) (b – c) (c – a)
Solution:
(i) Let Δ = \(\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
2 & 3+2 p & 4+3 p+2 q \\
3 & 6+3 p & 10+6 p+3 q
\end{array}\right|\)
operate R₂ → R₂ – 2R₁ ;
R₃ → R₃ – 3R₁
= \(\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
0 & 1 & 2+p \\
0 & 3 & 7+3 p
\end{array}\right|\)
Expanding along C₁ ; we get
= 1 \(\left|\begin{array}{cc}
1 & 2+p \\
3 & 7+3 p
\end{array}\right|\)
= (7 + 3p) – 3 (2 + p)
= 7 + 3p – 6 – 3p = 1

(ii) Let Δ = \(\left|\begin{array}{lll}
p a & p^2+a^2 & 1 \\
p b & p^2+b^2 & 1 \\
p c & p^2+c^2 & 1
\end{array}\right|\)
Taking p common from C₁ ; we have
= p \(\left|\begin{array}{ccc}
a & p^2+a^2 & 1 \\
b & p^2+b^2 & 1 \\
c & p^2+c^2 & 1
\end{array}\right|\) ;
operate R₂ → R₂ – 2R₁ ;
R₃ → R₃ – 3R₁
= p \(\left|\begin{array}{ccc}
a & p^2+a^2 & 1 \\
b-a & b^2-a^2 & 0 \\
c-a & c^2-a^2 & 0
\end{array}\right|\)
Taking (b – a) common from r₂ and (c – a) common from R₃
= p (b – a) (c – a) \(\left|\begin{array}{ccc}
a & p^2+a^2 & 1 \\
1 & b+a & 0 \\
1 & c+a & 0
\end{array}\right|\) ;
Expanding along C₃ ;
= p (b – a) (c – a) (c + a – b – a)
= p (a – b) (b – c) (c – a).

Question 3.
If a, b and c are real numbers and ∆ = \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\) = 0, then show that either a + b + c = 0 or a = b = c.
Solution:
Since, we have \(\left|\begin{array}{lll}
b+c & c+a & a+b \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\) = 0
operate C₁ → C₁ + C₂ + C₃
⇒ \(\left|\begin{array}{lll}
2(a+b+c) & c+a & a+b \\
2(a+b+c) & a+b & b+c \\
2(a+b+c) & b+c & c+a
\end{array}\right|\) = 0
Taking 2 (a + b + c) common from C₁
2 (a + b + c) \(\left|\begin{array}{lll}
1 & c+a & a+b \\
1 & a+b & b+c \\
1 & b+c & c+a
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
⇒ 2 (a + b + c) \(\left|\begin{array}{lll}
1 & c+a & a+b \\
0 & b-c & c-a \\
0 & b-a & c-b
\end{array}\right|\) = 0
Expanding along C₁ ; we have
⇒ 2 (a + b + c) \(\left|\begin{array}{ll}
b-c & c-a \\
b-a & c-b
\end{array}\right|\) = 0
⇒ 2 (a + b + c) [- (b – c)² – (b – a) (c – a)] = 0
⇒ 2 (a + b + c) [- b² – c² + 2bc – bc + ab + ac – a²] = 0
⇒ – 2 (a + b + c) [a² + b² + c² – ab – bc – ca] = 0
⇒ (a + b + c) [2a² + 2b² + 2c² – 2ab – 2bc – 2ca] = 0
⇒ (a + b + c) [(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ac)] = 0
⇒ (a + b + c) [(a – b)² + (b – c)² + (c – a)² ]= 0
∴ either a + b + c = 0 or
(a – b)² + (b – c)² + (c a)² = 0
i.e. a + b + c = 0 or
a = b = c.

Question 4.
If p + q + r = 0, prove that \(\left|\begin{array}{lll}
p a & q b & r c \\
q c & r a & p b \\
r b & p c & q a
\end{array}\right|\) = pqr \(\left|\begin{array}{lll}
a & b & c \\
c & a & b \\
b & c & a
\end{array}\right|\). (NCERT Exampler)
Solution:
L.H.S. = \(\left|\begin{array}{lll}
p a & q b & r c \\
q c & r a & p b \\
r b & p c & q a
\end{array}\right|\) ;
expanding along R₁
= pa [qra² – p²bc] – qb [q²ac – prb²] + rc [pqc² – r²ab]
= pqr (a³ + b³ + c³) – abc (p³ + q³ + r³) ………..(1)
Since p + q + r = 0
⇒ p³ + q³ + r³ = 3pqr
∴ from (1) ; we have
L.H.S. = pqr (a³ + b³ + c³) – 3pqrabc
= pqr (a³ + b³ + c³ – 3abc) ……………(2)
R.H.S. = pqr \(\left|\begin{array}{lll}
a & b & c \\
c & a & b \\
b & c & a
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + C₃
= pqr \(\left|\begin{array}{lll}
a+b+c & b & c \\
a+b+c & a & b \\
a+b+c & c & a
\end{array}\right|\) ;
Taking (a + b + c) common from C₁
= pqr (a + b + c) \(\left|\begin{array}{lll}
1 & b & c \\
1 & a & b \\
1 & c & a
\end{array}\right|\) ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= pqr (a + b + c) \(\left|\begin{array}{rrr}
1 & b & c \\
0 & a-b & b-c \\
0 & c-b & a-c
\end{array}\right|\) ;
expanding along C₁
= pqr (a + b + c) [(a – b) (a – c) + (b – c)²]
= pqr (a + b + c) [a² + b² + c² – ab – ac – bc]
= pqr [a³ + b³ + c³ – 3abc] …………(3)
From eqn. (2) and eqn. (3) ; we have
L.H.S. = R.H.S.
Thus, \(\left|\begin{array}{lll}
p a & q b & r c \\
q c & r a & p b \\
r b & p c & q a
\end{array}\right|\) = pqr \(\left|\begin{array}{lll}
a & b & c \\
c & a & b \\
b & c & a
\end{array}\right|\).

Question 5.
(i) Show that \(\left|\begin{array}{ccc}
\cos ^2 x & \sin x \cos x & -\sin x \\
\sin x \cos x & \sin ^2 x & \cos x \\
\sin x & -\cos x & 0
\end{array}\right|\) = 1, for all real values of x.
(ii) Show that \(\left|\begin{array}{ccc}
1 & a & a^2 \\
\cos (n-1) x & \cos n x & \cos (n+1) x \\
\sin (n-1) x & \sin n x & \sin (n+1) x
\end{array}\right|\) is independent of n.
Solution:
Let Δ = \(\left|\begin{array}{ccc}
\cos ^2 x & \sin x \cos x & -\sin x \\
\sin x \cos x & \sin ^2 x & \cos x \\
\sin x & -\cos x & 0
\end{array}\right|\) ;
operate C₁ → C₁ – sin x C₃ ;
C₂ → C₂ + cos x C₃
= \(\left|\begin{array}{ccc}
1 & 0 & -\sin x \\
0 & 1 & \cos x \\
\sin x & -\cos x & 0
\end{array}\right|\) ;
expanding along R₁
= 1 (0 + cos² x) + 0 – sin x (0 – sin x)
= cos² x + sin² x
= 1 ∀ x ∈ R

(ii) Let Δ = \(\left|\begin{array}{ccc}
1 & a & a^2 \\
\cos (n-1) x & \cos n x & \cos (n+1) x \\
\sin (n-1) x & \sin n x & \sin (n+1) x
\end{array}\right|\) ;
expanding along R₁
= 1 [cos nx sin (n + 1) x – sin nx cos (n + 1) x] – a [sin (n + 1) x cos (n – 1) x – sin (n – 1) x cos (n + 1) x] + a² [sin nx cos (n – 1) x – cos nx sin (n – 1) x]
= sin (n + 1 – n) x – a sin (n + 1 – n + 1) x + a² sin (n – n + 1) x
[∵ sin (A ± B) = sin A cos B ± cos A sin B)]
= sin x – a sin 2x + a² sin x
= (1 + a²) sin x – a sin 2x
Thus, Δ is independent of n.

Question 6.
Solve the following for x :
(i) \(\left|\begin{array}{lll}
1 & x & x^3 \\
1 & a & a^3 \\
1 & b & b^3
\end{array}\right|\) = 0, a ≠ b
(ii) \(\left|\begin{array}{ccc}
a^2+x & a b & a c \\
a b & b^2+x & b c \\
a c & b c & c^2+x
\end{array}\right|\) = 0.
Solution:
(i) Given, \(\left|\begin{array}{lll}
1 & x & x^3 \\
1 & a & a^3 \\
1 & b & b^3
\end{array}\right|\) = 0
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
\(\left|\begin{array}{rrr}
1 & x & x^3 \\
0 & a-x & a^3-x^3 \\
0 & b-x & b^3-x^3
\end{array}\right|\) = 0
Taking (a – x) common from R₂ and (b – x) common from R₃
(a – x) (b – x) \(\left|\begin{array}{ccc}
1 & x & x^3 \\
0 & 1 & a^2+a x+x^2 \\
0 & 1 & b^2+b x+x^2
\end{array}\right|\) = 0 ;
Expanding along C₁ ; we have
(a – x) (b – x) \(\left|\begin{array}{ll}
1 & a^2+a x+x^2 \\
1 & b^2+b x+x^2
\end{array}\right|\) = 0
⇒ (a – x) (b – x) [b² + bx + x² – a² – ax – a²] = 0
⇒ (a – x) (b – x) [(b – a) (b + a) + x (b – a)]
⇒ (a – x) (b – x) (b – a) (x + b + a) = 0
a – x = 0 or
b – x = 0 or
x + b + a = 0 or
b – a = 0
⇒ x = a or
x = b or
x = – (a + b) [but b ≠ a]
∴ x = a, b, – (a + b)

(ii) Given \(\left|\begin{array}{ccc}
a^2+x & a b & a c \\
a b & b^2+x & b c \\
a c & b c & c^2+x
\end{array}\right|\) = 0
Multiply C₁ by a, C₂ by b and C₃ by c ; we get
⇒ \(\frac{1}{a b c}\left|\begin{array}{ccc}
a\left(a^2+x\right) & a b^2 & a c^2 \\
a^2 b & b\left(b^2+x\right) & b c^2 \\
a^2 c & b^2 c & c\left(c^2+x\right)
\end{array}\right|\) = 0 ;
operate C₁ → C₁ + C₂ + C₃
⇒ \(\frac{1}{a b c}\left|\begin{array}{lcc}
a\left(a^2+x+b^2+c^2\right) & a b^2 & a c^2 \\
b\left(a^2+b^2+x+c^2\right) b\left(b^2+x\right) & b c^2 \\
c\left(a^2+b^2+c^2+x\right) & b^2 c & c\left(c^2+x\right)
\end{array}\right|\) = 0
Taking (a² + x + b² + c²) common from C₁; we have
⇒ \(\frac{\left(a^2+b^2+c^2+x\right)}{a b c}\left|\begin{array}{ccc}
a & a b^2 & a c^2 \\
b & b\left(b^2+x\right) & b c^2 \\
c & b^2 c & c\left(c^2+x\right)
\end{array}\right|\) = 0
Taking a, b and c common from R₁, R₂ and R₃ respectively
⇒ (a² + b² + c² + x) \(\left|\begin{array}{ccc}
1 & b^2 & c^2 \\
1 & b^2+x & c^2 \\
1 & b^2 & c^2+x
\end{array}\right|\) = 0 ;
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
⇒ (a² + b² + c² + x) \(\left|\begin{array}{ccc}
1 & b^2 & c^2 \\
0 & x & 0 \\
0 & 0 & x
\end{array}\right|\) = 0 ;
Expanding along C₁ ; we have
⇒ (x + a² + b² + c²) x² = 0
⇒ x = 0, 0, – (a² + b² + c²)

Question 7.
(i) Factorise \(\left|\begin{array}{ccc}
x^2 & (y-z)^2 & y z \\
y^2 & (z-x)^2 & z x \\
z^2 & (x-y)^2 & x y
\end{array}\right|\).
(ii) \(\left|\begin{array}{ccc}
0 & x y z & x-z \\
y-x & 0 & y-z \\
z-x & z-y & 0
\end{array}\right|\) (Exampler)
Solution:

Taking (x – y), (x – z) common from R₂ and R₃ respectively
= (x² + y² + z²) (x – y) (x – z) \(\left|\begin{array}{ccc}
1 & y^2+z^2 & y z \\
0 & x+y & z \\
0 & x+z & y
\end{array}\right|\)
= (x² + y² + z²) (x – y) (x – z) [xy + y² – zx – z²]
= (x² + y² + z²) (x – y) (x – z) [x (y – z) + (y – z) (y + z)]
= (x² + y² + z²) (x – y) (x – z) (y – z) (x + y + z)
= – (x² + y² + z²) (x – y) (y – z) (z – x) (x + y + z)

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

Question 7 (old).
Prove that \(\left|\begin{array}{ccc}
a & b & c \\
a\left(a^2+1\right) & b\left(b^2+1\right) & c\left(c^2+1\right) \\
a+1 & b+1 & c+1
\end{array}\right|\) = (a – b) (b – c) (c – a) (a + b + c)
Solution:
Let = \(\left|\begin{array}{ccc}
a & b & c \\
a\left(a^2+1\right) & b\left(b^2+1\right) & c\left(c^2+1\right) \\
a+1 & b+1 & c+1
\end{array}\right|\) ;
operate R₂ → R₂ – R₁
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
a & b & c \\
a^3 & b^3 & c^3 \\
1 & 1 & 1
\end{array}\right|\) ;
operate C₂ → C₂ – C₁ ;
C₃ → C₃ – C₁
= \(\left|\begin{array}{ccc}
a & b-a & c-a \\
a^3 & b^3-a^3 & c^3-a^3 \\
1 & 0 & 0
\end{array}\right|\)
Taking (b – a) common from C₂ and (c – a) common from C₃ respectively
= (b – a) (c – a) \(\left|\begin{array}{ccc}
a & 1 & 1 \\
a^3 & b^2+a b+a^2 & c^2+a c+a^2 \\
1 & 0 & 0
\end{array}\right|\) ;
expanding along R₃
= (b – a) (c – a) [c² + ac + a² – b² – ab – a²]
= (b – a) (c – a) [(c² – b²) + a (c – b)]
= (b – a) (c – a) [(c – b) (c + b) + a (c – b)]
= (b – a) (c – a) (c – b) (c + b + a)
= (a – b) (b – c) (c – a) (a + b + c)

Question 8.
If Δ = \(\left|\begin{array}{ccc}
(x-2)^2 & (x-1)^2 & x^2 \\
(x-1)^2 & x^2 & (x+1)^2 \\
x^2 & (x+1)^2 & (x+2)^2
\end{array}\right|\), then show that Δ is a negative even integer.
Solution:
Given Δ = \(\left|\begin{array}{ccc}
(x-2)^2 & (x-1)^2 & x^2 \\
(x-1)^2 & x^2 & (x+1)^2 \\
x^2 & (x+1)^2 & (x+2)^2
\end{array}\right|\)
operate C₁ → C₁ – C₃ ;
C₂ → C₂ – C₃
= \(\left|\begin{array}{ccc}
-4 x+4 & -2 x+1 & x^2 \\
-4 x & -2 x-1 & (x+1)^2 \\
-4 x-4 & -2 x-3 & (x+2)^2
\end{array}\right|\)
Taking – 4 common from C₁
= – 4 \(\left|\begin{array}{ccc}
x-1 & -2 x+1 & x^2 \\
x & -2 x-1 & (x+1)^2 \\
x+1 & -2 x-3 & (x+2)^2
\end{array}\right|\) ;
operate R₂ → R₂ – R₁
R₃ → R₃ – R₁
= – 4 \(\left|\begin{array}{ccc}
x-1 & -2 x+1 & x^2 \\
1 & -2 & 2 x+1 \\
2 & -4 & 4 x+4
\end{array}\right|\) ;
operate R₃ → R₃ – 2R₂
= – 4 \(\left|\begin{array}{ccc}
x-1 & -2 x+1 & x^2 \\
1 & -2 & 2 x+1 \\
0 & 0 & 2
\end{array}\right|\) ;
expanding along R₃
∴ Δ = – 4 × 2 \(\left|\begin{array}{cc}
x-1 & -2 x+1 \\
1 & -2
\end{array}\right|\)
= – 8 [- 2x + 2 + 2x – 1]
= – 8
Hence Δ be clearly a negative even integer.

Question 9.
If x, y, z are all different and \(\left|\begin{array}{ccc}
x^3 & (x+p)^3 & (x-p)^3 \\
y^3 & (y+p)^3 & (y-p)^3 \\
z^3 & (z+p)^3 & (z-p)^3
\end{array}\right|\) = 0, prove that p² (x + y + z) = 3xyz.
Solution:

Taking (y – x), (z – x) common from R₂ and R₃ resp. and expanding along C₃ ; we have
= xyz (y – x) (z – x) \(\left|\begin{array}{ll}
y+x & 1 \\
z+x & 1
\end{array}\right|\)
= xyz (y – x) (y – z) (z – x) ………….(3)
putting (2) and (3) in eqn. (1) ; we get
∴ 0 = 6p⁵ (x – y) (y – z) (z – x) (x + y + z) – 18³ xyz (x – y) (y – z) (z – x)
[∴ Δ = 0 given]
i.e. 0 = (x – y) (y – z) (z – x) [p² (x + y + z) – 3xyz]
since x, y, z are all different i.e. x ≠ y ≠ z
∴ (x – y) (y – z) (z – x) ≠ 0
i.e. p² (x + y + z) = 3xyz is the required result.

Question 10.
If a² + b² + c² = – 2 and f(x) = \(\left|\begin{array}{ccc}
1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
\left(1+a^2 x\right) & 1+b^2 x & \left(1+c^2\right) x \\
\left(1+a^2 x\right) & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|\), then prove that f(x) is a polynomial of degree 2.
Solution:
Given f(x) = \(\left|\begin{array}{ccc}
1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
\left(1+a^2 x\right) & 1+b^2 x & \left(1+c^2\right) x \\
\left(1+a^2 x\right) & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|\) ;
operate C₁ → C₁ + C₂ + c₃
= \(\left|\begin{array}{ccc}
1+2 x+\left(a^2+b^2+c^2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1+2 x+\left(a^2+b^2+c^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\
1+2 x+\left(a^2+b^2+c^2\right) x & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|\)
since a² + b² + c² = – 2 (given)
= \(\left|\begin{array}{ccc}
1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1 & 1+b^2 x & \left(1+c^2\right) x \\
1 & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right|\)
operate R₂ → R₂ – R₁ ;
R₃ → R₃ – R₁
= \(\left|\begin{array}{ccc}
1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\
0 & 1-x & 0 \\
0 & 0 & 1-x
\end{array}\right|\) ;
Expanding along C₁
= (1 – x)² which is clearly a polynomial of degree 2.

Question 11.
If a, b, c are all different and \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
(x-a)^2 & (x-b)^2 & (x-c)^2 \\
(x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b)
\end{array}\right|\) = 0, then find the value(s) of x.
Solution:
Let Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
(x-a)^2 & (x-b)^2 & (x-c)^2 \\
(x-b)(x-c) & (x-c)(x-a) & (x-a)(x-b)
\end{array}\right|\) ………………(1)
putting x – a = α,
x – b = β and
x – c = γ in eqn. (1) ; we get
= \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
\alpha^2 & \beta^2 & \gamma^2 \\
\beta \gamma & \gamma \alpha & \alpha \beta
\end{array}\right|\) ;
Operate C₂ → C₂ – C₁ ;
C₃ → C₃ – C₁
= \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
\alpha^2 & \beta^2-\alpha^2 & \gamma^2-\alpha^2 \\
\beta \gamma & \gamma(\alpha-\beta) & \beta(\alpha-\gamma)
\end{array}\right|\)
Taking (β – α) common from C₂ and (γ – α) common from C₃
= (β – α) (γ – α) \(\left|\begin{array}{ccc}
1 & 0 & 0 \\
\alpha^2 & \beta+\alpha & \gamma+\alpha \\
\beta \gamma & -\gamma & -\beta
\end{array}\right|\) ;
Expanding along R₁ ; we have
= (β – α) (γ – α) \(\left|\begin{array}{cc}
\beta+\alpha & \gamma+\alpha \\
-\gamma & -\beta
\end{array}\right|\)
= (β – α) (γ – α) [- β (β + α) + γ (γ + α)]
= (β – α) (γ – α) [γ² – β² + α (γ – β)]
= (β – α) (γ – α) (γ – β) (α + β + γ)
= (α – β) (β – γ) (γ – α) (α + β + γ)
Also Δ has value given to be 0.
∴ (α – β) (β – γ) (γ – α) (α + β + γ) = 0
⇒ [(x – a) – (x – b) [(x – b) – (x – c)] [(x – c) – (x – a)] [3x – a – b – c] = 0
⇒ (b – a) (c – b) (a – c) (3x – a – b – c) = 0
⇒ a ≠ b ≠ c
⇒ a – b ≠ 0, b – c ≠ 0, c – a ≠ 0
⇒ (b – a) (c – b) (c – a) ≠ 0
∴ from (2) ; 3x – a – b – c = 0
⇒ x = \(\frac{a+b+c}{3}\)

Question 12.
If A = \(\left[\begin{array}{cc}
1 & \tan \frac{x}{2} \\
-\tan \frac{x}{2} & 1
\end{array}\right]\), then show that A’A^-1 = \(\left[\begin{array}{rr}
\cos x & -\sin x \\
\sin x & \cos x
\end{array}\right]\).
Solution:
Given A = \(\left[\begin{array}{cc}
1 & \tan \frac{x}{2} \\
-\tan \frac{x}{2} & 1
\end{array}\right]\)
∴ |A| = 1 + tan² \(\frac{x}{2}\)
= sec² \(\frac{x}{2}\) ≠ 0 ∀ x ≠ odd multiple of π
Here A₁₁ = 1 ;
A₁₂ = tan \(\frac{x}{2}\)
A₂₁ = – tan \(\frac{x}{2}\)
A₂₂ = 1
∴ adj A = \(\left[\begin{array}{cc}
1 & \tan \frac{x}{2} \\
-\tan \frac{x}{2} & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & -\tan \frac{x}{2} \\
\tan \frac{x}{2} & 1
\end{array}\right]\)
Since |A| ≠ 0
∴ A^-1 exists
and A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A
∴ A^-1 = \(\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc}
1 & -\tan \frac{x}{2} \\
\tan \frac{x}{2} & 1
\end{array}\right]\)
Thus A’A^-1 = \(\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc}
1 & -\tan \frac{x}{2} \\
\tan \frac{x}{2} & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\tan \frac{x}{2} \\
\tan \frac{x}{2} & 1
\end{array}\right]\)

= \(=\frac{1}{1+\tan ^2 \frac{x}{2}}\left[\begin{array}{cc}
1-\tan ^2 \frac{x}{2} & -2 \tan \frac{x}{2} \\
2 \tan \frac{x}{2} & 1-\tan ^2 \frac{x}{2}
\end{array}\right]\)

= \(\left[\begin{array}{cc}
\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} & \frac{-2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \\
\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} & \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}
\end{array}\right]\)

= \(\left[\begin{array}{rr}
\cos x & -\sin x \\
\sin x & \cos x
\end{array}\right]\)

Question 13.
For the matrix A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\), show that A² – 4A + 5I = O. Hence, obtain A^-1.
Solution:
Given A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Here |A| = 3 + 2 = 5 ≠ 0
∴ A^-1 exists
L.H.S. = A² – 4A + 5I
= \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]-4\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]+5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-1 & -4 \\
8 & 7
\end{array}\right]-\left[\begin{array}{cc}
4 & -4 \\
8 & 12
\end{array}\right]+\left[\begin{array}{cc}
5 & 0 \\
0 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
= O = R.H.S.
Thus A² – 4A + 5I = O
⇒ 5I = 4A – A² …………….(1)
pre-multiplying both sides of eqn. (1) by A^-1 ; we have
5A^-1I = 4A^-1 A – (A^-1 A) A
⇒ 5A^-1 = 4I – IA
[∵ A^-1 A = I]
⇒ 5A^-1 = \(4\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
⇒ 5A^-1 = \(\left[\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right]-\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3 & 1 \\
-2 & 1
\end{array}\right]\)
⇒ A^-1 = \(\frac{1}{5}\left[\begin{array}{rr}
3 & 1 \\
-2 & 1
\end{array}\right]\)

Question 14.
Using matrix method, solve the following system of linear equations:
(i) 3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4

(ii) 2x + y + z = 1
x – 2y – z = \(\frac{3}{2}\)
3y – 5z = 9.
Solution:
(i) Given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) ;
B = \(\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]\)
Here |A|= \(\left|\begin{array}{rrr}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right|\)
= 3 (2 – 3) + 2 (4 + 4) + 3 (- 6 – 4)
= – 3 + 16 – 30
= – 17 ≠ 0
∴ A^-1 exists and given system has unique solution.
The cofactors of R₁ are

Thus x = 1 ; y = 2 and z = 3
Hence, the required solution of given system be x = 1, y = 2 and z = 3.

(ii) Given system of equations is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
2 & 1 & 1 \\
1 & -2 & -1 \\
0 & 3 & -5
\end{array}\right]\) ;
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(\left[\begin{array}{r}
1 \\
3 / 2 \\
9
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
2 & 1 & 1 \\
1 & -2 & -1 \\
0 & 3 & -5
\end{array}\right|\) ;
Expanding along R₁
= 2 (10 + 3) – 1 (- 5 – 0) + 1 (3 – 0)
= 26 + 5 + 3
= 34 ≠ 0
∴ A^-1 exists and given system be unique solution.

Question 15.
If A = \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
2 & 1 & -3 \\
-1 & 2 & 1
\end{array}\right]\), find A^-1. Using A, solve the following system of equations:
3x – 2y + z = 2, 2x + y – 3z = – 5, – x + 2y + z = 6.
Solution:
Given A = \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
2 & 1 & -3 \\
-1 & 2 & 1
\end{array}\right]\)
Here |A| = \(\left|\begin{array}{rrr}
3 & -2 & 1 \\
2 & 1 & -3 \\
-1 & 2 & 1
\end{array}\right|\)
= 3 (1 + 6) + 2 (2 – 3) + 1 (4 + 1)
= 21 – 2 + 5
= 24 ≠ 0
∴ A^-1 exists and
A^-1 = \(\frac{1}{|\mathrm{~A}|}\) adj A ………..(1)
The cofactors of R₁ are ;

The given system of eqn’s is equivalent to AX = B
where A = \(\left[\begin{array}{rrr}
3 & -2 & 1 \\
2 & 1 & -3 \\
-1 & 2 & 1
\end{array}\right]\)
X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B = \(=\left[\begin{array}{r}
2 \\
-5 \\
6
\end{array}\right]\)
Here |A| = 24 ≠ 0
∴ given system has unique solution given by
X = A^-1B
⇒ X = \(\frac{1}{24}\left[\begin{array}{rrr}
7 & 4 & 5 \\
1 & 4 & 11 \\
5 & -4 & 7
\end{array}\right]\left[\begin{array}{r}
2 \\
-5 \\
6
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{24}\left[\begin{array}{r}
14-20+30 \\
2-20+66 \\
10+20+42
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{24}\left[\begin{array}{l}
24 \\
48 \\
72
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Thus x = 1 ; y = 2 and z = 3.

Question 16.
Evaluate \(\left[\begin{array}{rrr}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\left[\begin{array}{rrr}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\). Hence, solve the system of equations
3x – 2y + 3z = 8, 2x + y = 1, 4x – 3y + 2z = 4
Solution:
Let A = \(\left[\begin{array}{rrr}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
and B = \(\left[\begin{array}{rrr}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\)
Here AB = \(\left[\begin{array}{rrr}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\left[\begin{array}{rrr}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-3+16-30 & -15+12+3 & -3-18+21 \\
-2-8+10 & -10-6-1 & -2+9-7 \\
-4+24-20 & -20+18+2 & -4-27+14
\end{array}\right]\)

Expanding along R₁
= 3 (2 – 3) + 2 (4 + 4) + 3 (- 6 – 4)
= – 3 + 16 – 30
= – 17 ≠ 0
∴ A^-1 exists.
Now AX = C
⇒ X = A^-1C
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{rrr}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]\)
= \(-\frac{1}{17}\left[\begin{array}{r}
-8-5-4 \\
-64-6+36 \\
-80+1+28
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{l}
-17 \\
-34 \\
-51
\end{array}\right]\)
= \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
∴ x = 1 ; y = 2 and z = 3.

Question 17.
Examine whether or not the system of equations 2x – 5y = 3, 4x – 10y = 6 is inconsistent. If consistent, solve it.
Solution:
Given system of eqns is 2x – 5y = 3, 4x – 10y = 6
Here D = \(\left|\begin{array}{rr}
2 & -5 \\
4 & -10
\end{array}\right|\)
= – 20 + 20 = 0
D₁ = \(\left|\begin{array}{cc}
3 & -5 \\
6 & -10
\end{array}\right|\)
= – 30 + 30 = 0
D₂ = \(\left|\begin{array}{ll}
2 & 3 \\
4 & 6
\end{array}\right|\)
= 12 – 12 = 0
Thus, D = 0 = D₁ = D₂
∴ Given system of eqn’s has infinitely many solutions and given system is consistent.

For solution:
putting y = k in first eqn. of given system , we get
2x – 5k = 3
⇒ x = \(\frac{3+5 k}{2}\)
Clearly x = \(\frac{3+5 k}{2}\), y = k satisfies the second eqn. of given system.
Hence the required solution of given system be x = \(\frac{3+5 k}{2}\); y = k , where k be any number.

Question 18.
Examine whether or not the system of equations
x – y + z = 3
2x + y – z = 2
x + 2y – 2z = – 1
Here, D = \(\left|\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -1 \\
1 & 2 & -2
\end{array}\right|\)
= 1 (- 2 + 2) + 1 (- 4 + 1) + 1 (4 – 1)
= 0 – 3 + 3 = 0
D₁ = \(\left|\begin{array}{rrr}
3 & -1 & 1 \\
2 & 1 & -1 \\
-1 & 2 & -2
\end{array}\right|\)
= 3 (- 2 + 2) + 1 (- 4 – 1) + 1 (4 + 1)
= 0 – 5 + 5 = 0
D₂ = \(\left|\begin{array}{rrr}
1 & 3 & 1 \\
2 & 2 & -1 \\
1 & -1 & -2
\end{array}\right|\)
= 1 (- 4 – 1) – 3 (- 4 + 1) + 1 (- 2 – 2)
= – 5 + 9 – 4 = 0
and D₃ = \(\left|\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 2 \\
1 & 2 & -1
\end{array}\right|\)
= 1 (- 1 – 4) – 3 (- 4 + 1) + 3 (4 – 1)
= – 5 – 4 + 9 = 0
Thus D = D₃ = D₁ = D₂ = 0
Then by cramer’s rule, given system is consistent and have infinitely many solutions.

For solution:
putting z = k, where k be any real number in first two eqn’s of given system, we get
x – y = 3 – k
2x + y = 2 + k
Here |A| = \(\left|\begin{array}{rr}
1 & -1 \\
2 & 1
\end{array}\right|\)
= 1 + 2
= 3 ≠ 0
∴ given system has unique soln.
|A₁| = \(\left|\begin{array}{rr}
3-k & -1 \\
2+k & 1
\end{array}\right|\)
= 3 – k + 2 + k = 5
|A₂| = \(\left|\begin{array}{cc}
1 & 3-k \\
2 & 2+k
\end{array}\right|\)
= 2 + k – 6 + 2k
= 3k – 4
Then by Cramer’s rule, we have
x = \(\frac{\left|A_1\right|}{|A|}=\frac{5}{3}\)
and y = \(\frac{\left|\mathrm{A}_2\right|}{|\mathrm{A}|}=\frac{3 k-4}{3}\)
= k – \(\frac{4}{3}\)
Also x = \(\frac{5}{3}\), y = k – \(\frac{4}{3}\) and z = k,
where k be any real number satisfies the 3rd eqn. of given system.
Hence required soln. of given system be x = \(\frac{5}{3}\), y = k – \(\frac{4}{3}\) and z = k,
where k be any real number.