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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. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 - C2
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= (x + 1) (x2 – 2x + 2) – 0
= x3 – 2x2 + 2x + x2 – 2x + 2 = x3 – x2 + 2

Q.2. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE

Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
(Taking a + x + y + z common)
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1 = (a + x+y +z) NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE= a2(a + x + y + z)

Q.3. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking x2, y2 and z2 common from C1, C2 and C3 respectively
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= x2y2z2 [- x(0 - yz) + x( yz - 0)]
= x2y2z2(xyz + xyz) = x2y2z2(2xyz) = 2x3y3z3

Q.4. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (x + y + z) common from C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= (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) | Mathematics (Maths) Class 12 - JEE
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (3x + 4) common from C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE

Q.6. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (a + b + c) common from R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (b + c + a) from C1 and C2
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE

Using the properties of determinants in Exercises 7 to 9, prove that:
Q.7. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE

Ans.
L.H.S. =NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → xR1, R2 → yR2, R3 → zR3 and dividing the determinant by xyz.
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking xyz common from C1 and C2
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C3 → C3 + C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (xy + yz + zx) common from C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
[∵ C2 and C3 are identical]
L.H.S. = R.H.S. Hence proved.

Q.8. NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
L.H.S. =NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 - (C2 + C3)
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking –2 common from C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
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) | Mathematics (Maths) Class 12 - JEE
Ans.
L.H.S.=NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (a – 1) common from C1 and C2
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= (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) | Mathematics (Maths) Class 12 - JEE
Ans.
L.H.S. = NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE

= 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) | Mathematics (Maths) Class 12 - JEE
Ans.
Area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3)
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEENCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
But area of equilateral triangle whose side is ‘a‘ =NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Hence, proved.

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

Q.13. IfNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE, then find values of x.
Ans.
Let NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE  
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE 
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (12 + x) common from R1,
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
(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) | Mathematics (Maths) Class 12 - JEE 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) | Mathematics (Maths) Class 12 - JEE
Taking ARr, ARr+6 and ARr+10 common from R1, R2 and R3 respectively.
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
[∵ 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) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2, R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
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) | Mathematics (Maths) Class 12 - JEE
Ans.
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C→ C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking (cos A – cos B) and (cos B – cos C) common from C1 and C2 respectively.
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
⇒ (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) | Mathematics (Maths) Class 12 - JEEand show thatNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
Here,NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= 0 – 1 (0 – 1) + 1 (1 – 0)
= 1 + 1 = 2 ≠ 0 (non-singular matrix.)
Now, co-factors,
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Now, A2 =NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Hence,  A2 = NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Now, we have to prove thatNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Hence, proved.

Long Answer  (L.A.)
Q.18. IfNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEEfind 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) | Mathematics (Maths) Class 12 - JEE
= 1(–1 – 2) – 2(–2 – 0) + 0
= – 3 + 4 = 1 ≠ 0 (non-singular matrix.)
Now co-factors,
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
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) | Mathematics (Maths) Class 12 - JEE
whereNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
∵ (AT)–1 = (A–1)T
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
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) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
= 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) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Hence, x = 1, y = 1 and z = 1.

Q.20. GivenNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEEfind 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) | Mathematics (Maths) Class 12 - JEE
BA =NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
The given equations can be re-write as,
x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Hence, x = 2, y = – 1 and z = 4

Q.21. If a + b + c ≠ 0 andNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEEthen prove that a = b = c.
Ans.
Given that: a + b + c ≠ 0 andNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE(Taking a + b + c common from C1)
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2 and R2 → R- R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
⇒ (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) | Mathematics (Maths) Class 12 - JEEis divisible by a + b + c and find the quotient.
Ans.
LetNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
C1 → C1 + C2 + C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Taking ab + bc + ac – a2 – b2 – c2 common from C1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
R1 → R1 - R2 and R2 → R2 - R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along C1 
⇒ (a +b + c)2 (ab + bc + ac - a2 - b2 - c2 )NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
  (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) | Mathematics (Maths) Class 12 - JEE

Q.23. If x + y + z = 0, prove thatNCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Ans.
L.H.S.
Let NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
⇒ 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) | Mathematics (Maths) Class 12 - JEE
R1 → R1 + R2 + R3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
(Taking a + b + c common from R1)
C1 → C1 - C2, C2 → C2 - C3
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
Expanding along R1
NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE
⇒ 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. 

The document NCERT Exemplar - Determinants(Part-1) | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Exemplar - Determinants(Part-1) - Mathematics (Maths) Class 12 - JEE

1. What is the definition of a determinant?
Ans. A determinant is a scalar value that is calculated from the elements of a square matrix. It provides important information about the matrix, such as whether it is invertible or singular.
2. How is the determinant of a matrix calculated?
Ans. The determinant of a square matrix can be calculated using various methods, such as cofactor expansion, row or column operations, or using special properties of certain types of matrices. The specific method used depends on the size and properties of the matrix.
3. Why is the determinant important in linear algebra?
Ans. The determinant plays a crucial role in linear algebra as it provides information about the invertibility of a matrix. If the determinant of a matrix is non-zero, then the matrix is invertible, and its inverse can be calculated. Additionally, the determinant is used in solving systems of linear equations, calculating the area/volume of geometric shapes, and finding eigenvalues of a matrix.
4. What are the properties of determinants?
Ans. Determinants have several key properties, including: - If a matrix has two identical rows or columns, its determinant is zero. - Swapping two rows (or columns) of a matrix changes the sign of its determinant. - Multiplying a row (or column) of a matrix by a scalar multiplies its determinant by the same scalar. - If a matrix has a row (or column) that is a linear combination of other rows (or columns), then its determinant is zero. - The determinant of the product of two matrices is equal to the product of their determinants.
5. Can determinants be negative?
Ans. Yes, determinants can be negative. The sign of a determinant depends on the number of row (or column) swaps required to convert the matrix to its row-echelon form. If an odd number of swaps is needed, the determinant is negative, and if an even number of swaps is needed, the determinant is positive.
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