NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Mathematics (Maths) Class 12

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Using the properties of determinants in Exercises 1 to 6, evaluate:
Q.1. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 - C2
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= (x + 1) (x2 – 2x + 2) – 0
= x3 – 2x2 + 2x + x2 – 2x + 2 = x3 – x2 + 2

Q.2. NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
(Taking a + x + y + z common)
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1 = (a + x+y +z) NCERT Exemplar - Determinants(Part-1) Notes | EduRev= a2(a + x + y + z)

Q.3. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking x2, y2 and z2 common from C1, C2 and C3 respectively
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= x2y2z2 [- x(0 - yz) + x( yz - 0)]
= x2y2z2(xyz + xyz) = x2y2z2(2xyz) = 2x3y3z3

Q.4. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (x + y + z) common from C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= (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. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (3x + 4) common from C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Q.6. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (a + b + c) common from R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (b + c + a) from C1 and C2
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Using the properties of determinants in Exercises 7 to 9, prove that:
Q.7. NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Ans.
L.H.S. =NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → xR1, R2 → yR2, R3 → zR3 and dividing the determinant by xyz.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking xyz common from C1 and C2
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C3 → C3 + C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (xy + yz + zx) common from C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
[∵ C2 and C3 are identical]
L.H.S. = R.H.S. Hence proved.

Q.8. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
L.H.S. =NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 - (C2 + C3)
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking –2 common from C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
= - 2 [ x - zy- zy ] = – 2(– 2xyz) = 4xyz R.H.S.
L.H.S. = R.H.S.
Hence, proved.

Q.9. NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
L.H.S.=NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (a – 1) common from C1 and C2
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= (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 thatNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
L.H.S. = NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev

= 1(1 – cos2 A) – cos C(cos C – cos A cos B) + cos B (cos A cos C – cos B)
= sin2 A – cosC + 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), thenNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
Area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3)
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRevNCERT Exemplar - Determinants(Part-1) Notes | EduRev
But area of equilateral triangle whose side is ‘a‘ =NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence, proved.

Q.12. Find the value of θ satisfyingNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 - C2
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking 7 common from C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
⇒ –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
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev is not possible as – 1 ≤ x ≤ 1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence,NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Q.13. IfNCERT Exemplar - Determinants(Part-1) Notes | EduRev, then find values of x.
Ans.
Let NCERT Exemplar - Determinants(Part-1) Notes | EduRev  
NCERT Exemplar - Determinants(Part-1) Notes | EduRev 
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (12 + x) common from R1,
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
(12 + x) (4x2 – 0) = 0 ⇒ 12 + x = 0 or 4x2 = 0 
 x = -12 or x = 0

Q.14. If a1, a2, a3, …, ar are in G.P., then prove that the determinant 
NCERT Exemplar - Determinants(Part-1) Notes | EduRev is independent of r.
Ans.
If a1, a2, a3, … abe the terms of G.P., then 
an = ARn–1 
(where A is the first term and R is the common ratio of the G.P. )
∴ar+1 = ARr+1–1 = ARr; ar+5 = ARr+5–1 = ARr+4 
ar+9 = ARr +9-1 = ARr +8 ; ar+7 = ARr+7–1 = ARr+6 
ar+11 = ARr+11–1 = ARr+10; ar+15 = ARr+15–1 = ARr+14 
ar+17 = AR r +17-1 = ARr +16 ;  ar+21 = ARr+21–1 = ARr+20 
The determinant becomes
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking ARr, ARr+6 and ARr+10 common from R1, R2 and R3 respectively.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
[∵ 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.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
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
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
C→ C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking (cos A – cos B) and (cos B – cos C) common from C1 and C2 respectively.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
⇒ (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 ifNCERT Exemplar - Determinants(Part-1) Notes | EduRevand show thatNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
Here,NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= 0 – 1 (0 – 1) + 1 (1 – 0)
= 1 + 1 = 2 ≠ 0 (non-singular matrix.)
Now, co-factors,
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Now, A2 =NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence,  A2 = NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Now, we have to prove thatNCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence, proved.

Long Answer  (L.A.)
Q.18. IfNCERT Exemplar - Determinants(Part-1) Notes | EduRevfind A–1Using A–1, solve the system of linear equations x – 2y = 10 , 2x – y – z = 8 , –2y + z = 7.
Ans.
Given that
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= 1(–1 – 2) – 2(–2 – 0) + 0
= – 3 + 4 = 1 ≠ 0 (non-singular matrix.)
Now co-factors,
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
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.
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
whereNCERT Exemplar - Determinants(Part-1) Notes | EduRev
∵ (AT)–1 = (A–1)T
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
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
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
= 3(2 + 3) – 2(1 – 6) – 2(– 1 – 4)
= 15 + 10 + 10 = 35 ≠ 0 non-singular matrix
Now, co-factors,
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence, x = 1, y = 1 and z = 1.

Q.20. GivenNCERT Exemplar - Determinants(Part-1) Notes | EduRevfind BA an⇒d use this to solve the system of equations y + 2z = 7, x – y = 3,  2x + 3y + 4z = 17.
Ans.
We have,NCERT Exemplar - Determinants(Part-1) Notes | EduRev
BA =NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
The given equations can be re-write as,
x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Hence, x = 2, y = – 1 and z = 4

Q.21. If a + b + c ≠ 0 andNCERT Exemplar - Determinants(Part-1) Notes | EduRevthen prove that a = b = c.
Ans.
Given that: a + b + c ≠ 0 andNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev(Taking a + b + c common from C1)
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2 and R2 → R- R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
⇒ (b – c) (a – b) – (c – a)2 = 0
⇒ ab – b2 – ac + bc – c– 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 thatNCERT Exemplar - Determinants(Part-1) Notes | EduRevis divisible by a + b + c and find the quotient.
Ans.
LetNCERT Exemplar - Determinants(Part-1) Notes | EduRev
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Taking ab + bc + ac – a2 – b2 – c2 common from C1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 - R2 and R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along C1 
⇒ (a +b + c)2 (ab + bc + ac - a2 - b2 - c2 )NCERT Exemplar - Determinants(Part-1) Notes | EduRev
  (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)
NCERT Exemplar - Determinants(Part-1) Notes | EduRev

Q.23. If x + y + z = 0, prove thatNCERT Exemplar - Determinants(Part-1) Notes | EduRev
Ans.
L.H.S.
Let NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
⇒ 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)
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
(Taking a + b + c common from R1)
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
Expanding along R1
NCERT Exemplar - Determinants(Part-1) Notes | EduRev
⇒ xyz(a + b+ c) [(c - a)2 - (b- c) (a - b)]
⇒ xyz(a + b + c)(c + 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. 

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