NCERT Exemplar - Determinants(Part-1)

# 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 – 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), 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, …, ar are in G.P., then prove that the determinant
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

Taking ARr, ARr+6 and ARr+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.

C→ 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.

Q.18. Iffind A–1Using 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
∵ (AT)–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 → R- R3

Expanding along C1

⇒ (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 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 + 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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## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

## 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.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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