Class 12th Chapter -4 Determinants | NCERT Maths Solution | NCERT Solution

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

Table of Contents

NCERT Solutions for Class 12 Maths Chapter : 4 Determinants

Class 12 Maths Chapter 4 Determinants Ex 4.1

Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

Solution:
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
= 2x(-1)-(-5)x(4)
=-2+20
=18

Ex 4.1 Class 12 Maths Question 2.
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}$
(ii) $\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}$
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q2.1

Ex 4.1 Class 12 Maths Question 3.
If $A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ then show that |2A|=|4A|
Solution:
$A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$
=> $2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
L.H.S = |2A|
= $2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
= – 24
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q3.1

Ex 4.1 Class 12 Maths Question 4.
$A=\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right]$ , then show that |3A| = 27|A|
Solution:
3A = $3\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right]$
= $3\left[ \begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q4.1

Ex 4.1 Class 12 Maths Question 5.
Evaluate the following determinant:
(i) $\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
(ii) $\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} \right|$
(iii) $\left| \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right|$
(iv) $\left| \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
Solution:
(i) $\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q5.1 >

Ex 4.1 Class 12 Maths Question 6.
If $\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right]$ , find |A|
Solution:
|A| = $\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right]$
= 1(-9+12)-1(-18+15)-2(8-5)
= 0

Ex 4.1 Class 12 Maths Question 7.
Find the values of x, if
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2

Ex 4.1 Class 12 Maths Question 8.
If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$ , then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6

Class 12 Maths Chapter 4 Determinants Ex 4.2

Ex 4.2 Class 12 Maths Question 1.
$\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} x & a & x \\ y & b & y \\ z & c & z \end{matrix} \right| +\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right|$
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Ex 4.2 Class 12 Maths Question 2.
$\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q2.1

Ex 4.2 Class 12 Maths Question 3.
$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0$
Solution:
$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{matrix} \right|$
${ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }-{ 9C }_{ 2 }=0$

Ex 4.2 Class 12 Maths Question 4.
$\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q4.1

Ex 4.2 Class 12 Maths Question 5.
$\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| =2\left| \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} \right|$
Solution:
L.H.S = ∆ = $\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q5.1

By using properties of determinants in Q 6 to 14, show that

Ex 4.2 Class 12 Maths Question 6.
$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$
Solution:
L.H.S = ∆ = $\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right|$ …(i)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q6.1

Ex 4.2 Class 12 Maths Question 7.
$\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$
Solution:
L.H.S = $\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q7.1

Ex 4.2 Class 12 Maths Question 8.
(a) $\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$
(b) $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 3 } & { b }^{ 3 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(a+b+c)$
Solution:
(a) L.H.S = $\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q8.1

Ex 4.2 Class 12 Maths Question 9.
$\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| =(x-y)(y-z)(z-x)(xy+yz+zx)$
Solution:
Let ∆ = $\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right|$
Applying R1–>R1 – R2, R2–>R2 – R3
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q9.1

Ex 4.2 Class 12 Maths Question 10.
(a) $\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right| =(5x+4){ (4-x) }^{ 2 }$
(b) $\left| \begin{matrix} y+x & y & y \\ y & y+k & y \\ y & y & y+k \end{matrix} \right| ={ k }^{ 2 }(3y+k)$
Solution:
(a) L.H.S = $\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right|<br />$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q10.1

Ex 4.2 Class 12 Maths Question 11.
(a) $\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| ={ (a+b+c) }^{ 3 }$
(b) $\left| \begin{matrix} x+y+2z & \quad z & \quad z \\ x & \quad y+z+2x & \quad x \\ y & y & \quad z+x+2y \end{matrix} \right| ={ 2(x+y+z) }^{ 3 }$
Solution:
(a) L.H.S = $\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right|$
= $\left( a+b+c \right) \left| \begin{matrix} 1 & \quad 1 & \quad 1 \\ 2b & \quad b-c-a & \quad 2b \\ 2c & \quad 2c & \quad c-a-b \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q11.1

Ex 4.2 Class 12 Maths Question 12.
$\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| ={ { (1-x }^{ 3 }) }^{ 2 }$
Solution:
L.H.S = $\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q12.1

Ex 4.2 Class 12 Maths Question 13.
$\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| ={ (1+{ a }^{ 2 }+{ b }^{ 2 }) }^{ 3 }$
Solution:
L.H.S = $\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q13.1

Ex 4.2 Class 12 Maths Question 14.
$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$
Solution:
Let ∆ = $\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right|$
$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab+0 & \quad ac+0 \\ ab+0\quad & \quad b^{ 2 }+1 & \quad bc+0 \\ ca+0\quad & \quad cb+0 & \quad { c }^{ 2 }+1 \end{matrix} \right|$
This may be expressed as the sum of 8 determinants
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q14.1

Ex 4.2 Class 12 Maths Question 15.
If A be a square matrix of order 3×3, then | kA | is equal to
(a) k|A|
(b) k² |A|
(c) k³ |A|
(d) 3k|A|
Solution:
Option (c) is correct.

Ex 4.2 Class 12 Maths Question 16.
Which of the following is correct:
(a) Determinant is a square matrix
(b) Determinant is a number associated to a matrix
(c) Determinant is a number associated to a square matrix
(d) None of these
Solution:
Option (c) is correct

Class 12 Maths Chapter 4 Determinants Ex 4.3

Ex 4.3 Class 12 Maths Question 1.
Find the area of the triangle with vertices at the point given in each of the following:
(i) (1,0), (6,0) (4,3)
(ii) (2,7), (1,1), (10,8)
(iii) (-2,-3), (3,2), (-1,-8)
Solution:
(i) Area of triangle = $\frac { 1 }{ 2 } \left| \begin{matrix} 1\quad & 0 & \quad 1 \\ 6\quad & 0 & \quad 1 \\ 4\quad & 3 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [1(0-3)+1(18-0)]
= 7.5 sq units
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q1.1

Ex 4.3 Class 12 Maths Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:
The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q2.1

Ex 4.3 Class 12 Maths Question 3.
Find the value of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4,0), (0,2)
(ii) (-2,0), (0,4), (0, k).
Solution:
(i) Area of ∆ = 4 (Given)
$\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \\ 4\quad & 0 & \quad 1 \\ 0\quad & 2 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ==> k = 0
Case(b): -k + 4 = -4 ==> k = 8
Hence, k = 0,8
(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q3.1

Ex 4.3 Class 12 Maths Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Given: Points (1,2), (3,6)
Equation of the line is
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q4.1

Ex 4.3 Class 12 Maths Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ = $\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \\ 5\quad & 4 & \quad 1 \\ k\quad & 4 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

Class 12 Maths Chapter4 Determinants Ex 4.4

Ex 4.4 Class 12 Maths Question 1.
Write the minors and cofactors of the elements of following determinants:
(i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$
Solution:
(i) Let A = $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
M₁₁ = 3, M₁₂ = 0, M₂₁ = – 4, M₂₂= 2
For cofactors

Ex 4.4 Class 12 Maths Question 2.
Write Minors and Cofactor of elements of following determinant
(i) $\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right|$
(ii) $\left| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{matrix} \right|$
Solution:
(i) Minors M₁₁ = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q2.1

Ex 4.4 Class 12 Maths Question 3.
Using cofactors of elements of second row, evaluate
$\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|$
Solution:
Given
$\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q3.1

Ex 4.4 Class 12 Maths Question 4.
Using Cofactors of elements of third column, evaluate
$\Delta =\left| \begin{matrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{matrix} \right|$
Solution:
Elements of third column are yz, zx, xy
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q4.1

Ex 4.4 Class 12 Maths Question 5.
If $\Delta =\left| \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right|$ and Aij is the cofactors of aij? then value of ∆ is given by
(a) a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃
(b) a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁
(c) a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃
(d) a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁
Solution:
Option (d) is correct.

Class 12 Maths Chapter 4 Determinants Ex 4.5

Find the adjoint of each of the matrices in Questions 1 and 2.

Ex 4.5 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=A(say)$
Solution:
Let C_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
C₁₁ = (-1)¹⁺¹ (4) = 4; C₁₂ = (-1)¹⁺²(3) = -3
C₂₁ = (-1)²⁺¹ (2)= – 2; C₂₂ = (-1)²⁺² (1) = -1
Adj A = $\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$
= $\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Ex 4.5 Class 12 Maths Question 2.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{matrix} \right] =A(say)$
Solution:
${ A }_{ 11 }={ (-1) }^{ 1+1 }M_{ 11 }=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3$
Similarly,
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q2.1

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Ex 4.5 Class 12 Maths Question 3.
$\begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}=A(say)$
Solution:
|A| = 24
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q3.1

Ex 4.5 Class 12 Maths Question 4.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] =A(say)$
Solution:
A₁₁ = 0, A₁₂= – 11, A₁₃= 0,
A₂₁= – 3, A₂₂= 1, A₂₃= 1, A₃₁ = – 2
A₃₂= 8, A₃₃= – 3
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q4.1

Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:

Ex 4.5 Class 12 Maths Question 5.
$\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix}=6+8=14\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q5.1

Ex 4.5 Class 12 Maths Question 6.
$\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix}=-2+15=13\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q6.1

Ex 4.5 Class 12 Maths Question 7.
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
Solution:
|A| = 10
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q7.1

Ex 4.5 Class 12 Maths Question 8.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right] =A$
Solution:
$\left| A \right| =\left| \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right| =1\left| \begin{matrix} 3 & 0 \\ 2 & -1 \end{matrix} \right| =-3\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ijin A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q8.1

Ex 4.5 Class 12 Maths Question 9.
$\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q9.1

Ex 4.5 Class 12 Maths Question 10.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and
${ A }^{ -1 }=\frac { Adj\quad A }{ |A| }$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q10.1

Ex 4.5 Class 12 Maths Question 11.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
Solution:
A = $\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
adj A = $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{matrix} \right]$
First find |A| = -cos²α-sin²α
=-1≠0
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q11.1

Ex 4.5 Class 12 Maths Question 12.
Let $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},B=\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$ , verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| = $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$
= 15-14
= 1≠0
$Adj A=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q12.1

Ex 4.5 Class 12 Maths Question 13.
If $A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A = $\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
A² = $\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q13.1

Ex 4.5 Class 12 Maths Question 14.
For the matrix A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$ find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$
A²+aA+bI²=0
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q14.1

Ex 4.5 Class 12 Maths Question 15.
For the matrix $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right]$ Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² = $\left[ \begin{matrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q15.1

Ex 4.5 Class 12 Maths Question 16.
If $A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$ show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have
$A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q16.1

Ex 4.5 Class 12 Maths Question 17.
Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:
Let A = $\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q17.1
Dividing by | A |, |Adj. A| = | A |²
Hence, Part (b) is the correct answer.

Ex 4.5 Class 12 Maths Question 18.
If A is an invertible matrix of order 2, then det. (A^-1) is equal to:
(a) det. (A)
(b) $\\ \frac { 1 }{ det.(A) }$
(c) 1
(d) 0
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I
$|{ A }^{ -1 }|=\frac { 1 }{ |A| }$
Hence option (b) is correct.

Class 12 Maths Chapter 4 Determinantsc Ex 4.6

Examine the consistency of the system of equations in Questions 1 to 6Ex 4.6 Class 12 Maths Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=> $\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}$ = 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=> $\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=> $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix}$
= 6 – 6
= 0.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q3.1
Hence, equations are consistent with no solution

Ex 4.6 Class 12 Maths Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z = $\\ \frac { 4 }{ a }$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q4.1 Ex 4.6 Class 12 Maths Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
$\left[ \begin{matrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]$
=> AX = B
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q5.1

Ex 4.6 Class 12 Maths Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
$\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 2 \\ -1 \end{matrix} \right]$
$AX=B|A|=\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right]$
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.

Solve system of linear equations using matrix method in Questions 7 to 14:

Ex 4.6 Class 12 Maths Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q7.1

Ex 4.6 Class 12 Maths Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q8.1

Ex 4.6 Class 12 Maths Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as
$\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 7 \end{matrix} \right] i.e,,AX=B$
where $A=\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q9.1

Ex 4.6 Class 12 Maths Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as
$\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] i.e,,AX=B$
where $A=\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q10.1

Ex 4.6 Class 12 Maths Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A^-1B
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q11.1

Ex 4.6 Class 12 Maths Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written
$\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 2 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q12.1

Ex 4.6 Class 12 Maths Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:
$\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q13.1

Ex 4.6 Class 12 Maths Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written
$\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q14.1

Ex 4.6 Class 12 Maths Question 15.
If A = $\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right]$ Find A^-1. Using A^-1. Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
We have AX = B
where $A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q15.1

Ex 4.6 Class 12 Maths Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q16.1

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NCERT Solutions for Class 12 Maths Chapter : 4 Determinants

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Class 12th Chapter -4 Determinants | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 4 Determinants

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