NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE PDF Download

Using the properties of determinants in Exercises 1 to 6, evaluate:
Q.1.
Ans.
Let
C₁ → C₁ - C₂

= (x + 1) (x² – 2x + 2) – 0
= x³ – 2x² + 2x + x² – 2x + 2 = x³ – x² + 2

Q.2.
Ans.
Let
C₁ → C₁ + C₂ + C₃

(Taking a + x + y + z common)
R₁ → R₁ - R₂, R₂ → R₂ - R₃

Expanding along C₁ = (a + x+y +z) = a²(a + x + y + z)

Q.3.
Ans.
Let
Taking x², y² and z² common from C₁, C₂ and C₃ respectively

Expanding along R₁

= x²y²z² [- x(0 - yz) + x( yz - 0)]
= x²y²z²(xyz + xyz) = x²y²z²(2xyz) = 2x³y³z³

Q.4.
Ans.
Let
C₁ → C₁ + C₂ + C₃

Taking (x + y + z) common from C₁

R₁ → R₁ - R₂, R₂ → R₂ - R₃

Expanding along C₁

= (x+y+ z) [(-x - 2y)(-y - 2z) -(2y +z)(-x +y)]
= (x + y + z) (xy + 2zx + 2y² + 4yz + 2xy – 2y² + zx – zy)
= (x + y + z) (3xy + 3zx + 3yz) = 3(x + y + z) (xy + yz + zx)

Q.5.
Ans.
Let
C₁ → C₁ + C₂ + C₃

Taking (3x + 4) common from C₁

R₁ → R₁ - R₂, R₂ → R₂ - R₃

Expanding along C₁


Q.6.
Ans.
Let
R₁ → R₁ + R₂ + R₃

Taking (a + b + c) common from R₁

C₁ → C₁ - C₂, C₂ → C₂ - C₃

Taking (b + c + a) from C₁ and C₂

Expanding along R₁


Using the properties of determinants in Exercises 7 to 9, prove that:
Q.7.
Ans.
L.H.S. =
R₁ → xR₁, R₂ → yR₂, R₃ → zR₃ and dividing the determinant by xyz.

Taking xyz common from C₁ and C₂

C₃ → C₃ + C₁

Taking (xy + yz + zx) common from C₃

[∵ C₂ and C₃ are identical]
L.H.S. = R.H.S. Hence proved.

Q.8.
Ans.
L.H.S. =
C₁ → C₁ - (C₂ + C₃)

Taking –2 common from C₁

R₂ → R₂ - R₃

Expanding along C₁
= - 2 [ x - zy- zy ] = – 2(– 2xyz) = 4xyz R.H.S.
L.H.S. = R.H.S.
Hence, proved.

Q.9.
Ans.
L.H.S.=
R₁ → R₁ - R₂, R₂ → R₂ - R₃

Taking (a – 1) common from C₁ and C₂

Expanding along C₃

= (a – 1)² (a + 1 – 2) = (a – 1)² (a – 1) = (a – 1)³ R.H.S.
L.H.S.= R.H.S.
Hence, proved.

Q.10. If A + B + C = 0, then prove that
Ans.
L.H.S. =
Expanding along C₁

= 1(1 – cos² A) – cos C(cos C – cos A cos B) + cos B (cos A cos C – cos B)
= sin² A – cos² C + cos A cos B cos C + cos A cos B cos C – cos² B
= sin² A – cos² B – cos² C + 2 cos A cos B cos C
= – cos (A + B) × cos (A - B) - cos² C + 2 cos A cos B cos C
[∵ sin² A - cos² B= - cos (A +B)× cos (A - B)]     
= - cos ( - C) × cos (A - B) + cos C (2 cos A cos B - cos C) [∵ A + B + C = 0]
= – cos C(cos A cos B + sin A sin B) + cos C(2 cos A cos B – cos C)
= – cos C(cos A cos B + sin A sin B – 2 cos A cos B + cos C)
= – cos C(– cos A cos B + sin A sin B + cos C)
= cos C(cos A cos B – sin A sin B – cos C)
= cos C[cos(A + B) – cos C ]     
= cos C[cos (– C) – cos C]  [∵ A + B = - C]
= cos C[cos C - cos C] = cos C × 0 = 0 R.H.S.     
L.H.S. = R.H.S.
Hence, proved.

Q.11. If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x₁, y₁), (x₂, y₂), (x₃, y₃), then
Ans.
Area of triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃)

Let⇒
But area of equilateral triangle whose side is ‘a‘ =


Hence, proved.

Q.12. Find the value of θ satisfying
Ans.
Let

C₁ → C₁ - C₂
⇒
Taking 7 common from C₁
⇒
⇒
Expanding along C₁
⇒
⇒ –2 + 7 sin 3θ + 2 (cos 2θ – 3 sin 3θ) = 0
⇒ – 2 + 7 sin 3θ + 2 cos 2θ – 6 sin 3θ = 0
⇒ – 2 + 2 cos 2θ + sin 3θ = 0
⇒ – 2 + 2 (1 – 2 sin² θ) + 3 sin θ – 4 sin³ θ = 0
⇒ – 2 + 2 – 4 sin² θ + 3 sin θ – 4 sin³ θ = 0
⇒ – 4 sin³ θ – 4 sin² θ + 3 sin θ = 0
⇒ – sin θ(4 sin² θ + 4 sin θ – 3) = 0
sin θ = 0 or 4 sin² θ + 4 sin θ – 3 = 0
∴ θ = nπ or  4 sin² θ + 6 sin θ – 2 sin θ – 3 = 0 when n ∈ I     
⇒ 2 sin θ(2 sin θ + 3) - 1 (2 sin θ + 3) = 0  
⇒ (2 sin θ + 3)(2 sin θ – 1) = 0
⇒ 2 sin θ + 3 = 0 or 2 sin θ - 1 = 0

 is not possible as – 1 ≤ x ≤ 1

Hence,

Q.13. If, then find values of x.
Ans.
Let   
 
R₁ → R₁ + R₂ + R₃
⇒
Taking (12 + x) common from R₁,
⇒
C₁ → C₁ - C₂, C₂ → C₂ - C₃

Expanding along R₁
⇒
(12 + x) (4x² – 0) = 0 ⇒ 12 + x = 0 or 4x² = 0 
⇒ x = -12 or x = 0

Q.14. If a₁, a₂, a₃, …, a_r are in G.P., then prove that the determinant 
 is independent of r.
Ans.
If a₁, a₂, a₃, … a_r be the terms of G.P., then 
a_n = AR^n–1 
(where A is the first term and R is the common ratio of the G.P. )
∴a_r+1 = AR^r+1–1 = AR^r; a_r+5 = AR^r+5–1 = AR^r+4 
a_r+9 = AR^{r +9-1} = AR^{r +8} ; a_r+7 = AR^r+7–1 = AR^r+6 
a_r+11 = AR^r+11–1 = AR^r+10; a_r+15 = AR^r+15–1 = AR^r+14 
a_r+17 = AR ^{r +17-1} = AR^{r +16} ;  a_r+21 = AR^r+21–1 = AR_r+20 
∴ The determinant becomes

Taking AR^r, AR^r+6 and AR^r+10 common from R₁, R₂ and R₃ respectively.


[∵ R₁ and R₂ are identical rows]
= 0
Hence, the given determinant is independent of r.

Q.15. Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.
Ans.
If the given points lie on a straight line, then the area of the triangle formed by joining the points pairwise is zero.

R₁ → R₁ - R₂, R₂ → R₂ - R₃
⇒
Expanding along C₃

As 7 ≠ 0. Hence,, the three points do not lie on a straight line for any value of a.

Q.16. Show that the ∆ABC is an isosceles triangle if the determinant

Ans.

C₁ → C₁ - C₂, C₂ → C₂ - C₃

Taking (cos A – cos B) and (cos B – cos C) common from C₁ and C₂ respectively.


⇒ (cos A - cos B) (cos B- cos C) [cos B + cos C + 1 - cos A - cos B - 1] = 0
⇒ (cos A – cos B) (cos B – cos C) (cos C – cos A) = 0
cos A – cos B = 0 or cos B – cos C = 0
or cos C – cos A = 0
⇒ cos A = cos B or cos B = cos C or cos C = cos A  
⇒ ∠A = ∠C or ∠B = ∠C ⇒ ∠A = ∠B
Hence, ΔABC is an isosceles triangle.

Q.17. Find A^–1 ifand show that
Ans.
Here,

= 0 – 1 (0 – 1) + 1 (1 – 0)
= 1 + 1 = 2 ≠ 0 (non-singular matrix.)
Now, co-factors,



∴
Now, A² =

Hence,  A² =
Now, we have to prove that

Hence, proved.

Long Answer  (L.A.)
Q.18. Iffind A^–1. Using A^–1, solve the system of linear equations x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7.
Ans.
Given that

= 1(–1 – 2) – 2(–2 – 0) + 0
= – 3 + 4 = 1 ≠ 0 (non-singular matrix.)
Now co-factors,


∴
⇒
Now, the system of linear equations is given by x – 2y = 10, 2x – y – z = 8 and – 2y + z = 7, which is in the form of CX = D.

where
∵ (A^T)^–1 = (A^–1)^T

Hence, x = 0, y = – 5 and z = – 3

Q.19. Using matrix method, solve the system of equation 
3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.
Ans.
Given that
3x + 2y – 2z = 3
x + 2y + 3z = 6
2x – y + z = 2
Let

= 3(2 + 3) – 2(1 – 6) – 2(– 1 – 4)
= 15 + 10 + 10 = 35 ≠ 0 non-singular matrix
Now, co-factors,





Hence, x = 1, y = 1 and z = 1.

Q.20. Givenfind BA an⇒d use this to solve the system of equations y + 2z = 7, x – y = 3,  2x + 3y + 4z = 17.
Ans.
We have,
BA =

∴
The given equations can be re-write as,
x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7
∴
⇒

Hence, x = 2, y = – 1 and z = 4

Q.21. If a + b + c ≠ 0 andthen prove that a = b = c.
Ans.
Given that: a + b + c ≠ 0 and
C₁ → C₁ + C₂ + C₃
⇒
⇒(Taking a + b + c common from C₁)
⇒
R₁ → R₁ - R₂ and R₂ → R₂ - R₃
⇒
Expanding along C₁
⇒
⇒ (b – c) (a – b) – (c – a)2 = 0
⇒ ab – b² – ac + bc – c² – a² + 2ac = 0
⇒ – a² – b² – c² + ab + bc + ac = 0
⇒ a² + b² + c² – ab – bc – ac = 0
⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ac = 0
(Multiplying both sides by 2)
⇒ (a² + b² – 2ab) + (b² + c² – 2bc) + (a² + c² – 2ac) = 0
⇒ (a – b)² + (b – c)² + (a – c)² = 0
It is only possible when (a – b)² = (b – c)² = (a – c)² = 0
∴ a = b = c Hence, proved.

Q.22. Prove thatis divisible by a + b + c and find the quotient.
Ans.
Let
C₁ → C₁ + C₂ + C₃
⇒
Taking ab + bc + ac – a² – b² – c² common from C₁

R₁ → R₁ - R₂ and R₂ → R₂ - R₃





Expanding along C₁ 
⇒ (a +b + c)² (ab + bc + ac - a² - b² - c² )
⇒  (a + b + c)²(ab + bc + ac - a² - b² - c²)[(c - b) (b - a)-(a- c)²]
⇒  (a + b + c)² (ab + bc + ac - a²- b²- c² )(bc - ca - b2+ ab - a²- c² + 2ac)
⇒  (a + b + c)² (ab + bc + ac - a²- b²- c² )(ab + bc + ca - a²- b²- c² )
⇒  (a + b + c)² (ab + bc + ac - a² - b² - c²)²
⇒  (a + b + c) (a + b + c) (a² + b² + c² - ab - bc - ac)²
Hence, the given determinant is divisible by a + b + c and the quotient is
(a + b + c) (a² + b² + c² - ab - bc - ac)²
⇒  (a + b + c) (a² + b² + c² - ab - bc - ac) (a² + b² + c² - ab - bc - ac)
⇒  (a³ + b³ + c³ - 3abc) (a² + b² + c² - ab - bc - ac)
⇒ (a³ + b³ + c³- 3abc) (2a² + 2b² + 2c² - 2ab - 2bc - 2ac)


Q.23. If x + y + z = 0, prove that
Ans.
L.H.S.
Let
Expanding along R₁

⇒ xa(yza² – x²bc) – yb(y²ac – xzb²) + zc(xyc² – z²ab)
⇒ xyza³ – x³abc – y³abc + xyzb³ + xyzc³ – z³abc
⇒ xyz(a³ + b³ + c³) – abc(x³ + y³ + z³)
⇒ xyz(a³ + b³ + c³) – abc(3xyz)
[(∵ x + y + z = 0) (∴ x³ + y³ + z³ = 3xyz)]
⇒ xyz(a³ + b³ + c³ – 3abc)

R₁ → R₁ + R₂ + R₃


(Taking a + b + c common from R₁)
C₁ → C₁ - C₂, C₂ → C₂ - C₃

Expanding along R₁

⇒ xyz(a + b+ c) [(c - a)² - (b- c) (a - b)]
⇒ xyz(a + b + c)(c ² + a²- 2ca - ab + b²+ ac - bc)
⇒ xyz(a + b+ c)( a² + b² + c² - ab - bc - ca)
⇒ xyz(a³ + b³ + c³ - 3abc)
[a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)]
L.H.S. = R.H.S.
Hence, proved.