16. |xx2yzyy2zxzz2xy|=(x−y)(y−z)(z−x)(xy+yz+zx)

LHS |xx2yzyy2zxzz2xy| =1xyz |x·xx·x2x·yzy·yy·y2y·zxz·zz·z2z·xy| Multiplying R1, R2& R3 by x, y&z. =1xyz|x2x3xyzy2y3xyzz2z3xyz|. =xyzxyz|x2x31y2y31z2z31| Taking xyz common from c3. =|x2x31y2−x2y3−x31−1z2−x223−x31−1| 2→2−13→3−1 . = |x2x31(y−x)(y+x)(y−x)(y2+x2+xy)0(z−x)(z+x)(2−x)(z2+x2+2x)0| Expanding along c3 we get, = 1 |(y−x)(y+x)(y−x)(y2+x2+xy)(z−x)(z+x)(z−x)(z2+x2+zx)| = (y-x)(z-x). |y+xy2+x2+xyz+xz2+x2+zx| Taking (yx) & (zx) common from R1 and R2. = (y-x)(z-x). Taking (y-x) & (z-x) common from R1 and R2. = (y-x)(z-x)[(y + x)(z2 + x2 + zx) - (z + x)(y2 + x2 + xy)] = (y-x)(z-x)[yz2 + yx2 + xyz + xz2 + x3 + x2z-zy2-zx2-xyz-xy2-x3-x2y] = y-x)(z-x)[yz2-zy2 + xz2-xy2] = (y-x)(z-x)[yz (z-y) + x(z2-y2)] = y-x)(z-x)(z-y)[yz + x (z + y)] = (- 1)(x-y)(z-x)(- 1)(y-z)[yz+ 1 xz + xy] = (x-y)(y-z)(z-x)(xy + yz + xz). = R.H.S

Kindly Consider the following 58. 2x−y=−23x+4y=3

The given system of eqn in matrix form is AK = B

NCERT Solutions for Class 12 Maths Chapter 4 – Determinants (Updated 2025)

Q: 20. |1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2|=(1+a2+b2)3

LHS = |1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2|. =|1+a2−b2−b(−2b)2ab+a(−2b)−2b2ab−b(2a)1−a2+b2+a(2a)2a2b−b(1−a2−b2)−2a+a(1−a2−b2)1−a2−b2| = | 1 + a 2 − b 2 + 2 b 2 2 a b − 2 a b − 2 b 2 a b − 2 a b 1 − a 2 + b 2 + 2 a 2 2 a 2 b − b + a 2 b + b 3 − 2 a + a − a 3 − a b 2 1 − a 2 − b 2 | = | 1 + a 2 + b 2 0 − 2 b 0 1 + a 2 + b 2 2 a b ( 1 + a 2 + b 2 ) − a ( 1 + a 2 + b 2 ) 1 − a 2 − b 2 | = (1 + a2 + b2)2 [(1 -a2 + b2) - 2a (-a)] = (1 + a2 + b2)2 (1 -a2 + b2 + 2a2) = (1 + a2 + b2)2 (1 + a2 + b2) = (1 + a2 + b2)3 = R.H.S.

Updated on Aug 1, 2025 15:40 IST

By Pallavi Pathak, Assistant Manager Content

Class 12 Determinants is an important chapter as it has a variety of applications in various fields including Social Science, Economics, Science, Engineering etc. Determinants Class 12 covers many properties of determinants, cofactors, minors and applications of determinants in finding inverse of a square matrix, adjoint, and area of a triangle, and inconsistency of system of linear equations. The chapter also covers finding the solution of linear equations in two or three variables using the inverse of a matrix.
The experts at Shiksha, with years of experience, have created the Class 12 determinants NCERT solutions to help students score well in their CBSE Board exam and other competitive exams like JEE Mains. It will help students to understand the concept better and know how to solve even the complex problems of the chapter. It offers step-by-step solutions to all the questions of this chapter in the NCERT textbook.
If you are looking for notes on all topics for Class 12 Maths with solved examples, PDF, and questions and answers, check Class 12 Maths Notes.

Table of content

Glance at Determinants Class 12
Class 12 Math Determinants NCERT Solution PDF: Free Download
Class 12 Math Determinants: Key Topics, Weightage
Important Formulas of Class 12 Determinants
Exercise-wise NCERT Solutions for Class 12 Math Chapter 4 Determinants
NCERT Class 12 Math Determinants Exercise 4.1 Solutions

Glance at Determinants Class 12

See below the quick summary of the Determinants Class 12 NCERT Solutions:

$Determinant of a matrix A = [a_{i j} 1 \times 1] is given by | a_{i j} | = a_{11}$
$Determinant of a matrix A = \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} is given by | \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} | = a_{11} a_{22} - a_{12} a_{21}$
$Determinant of a matrix A = \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} is given by (expanding along R) | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | = a_{1} | \begin{matrix} b_{2} & c_{2} \\ b_{3} & c_{3} \end{matrix} | - b_{1} | \begin{matrix} a_{2} & c_{2} \\ a_{3} & c_{3} \end{matrix} | + c_{1} | \begin{matrix} a_{2} & b_{2} \\ a_{3} & b_{3} \end{matrix} |$
$Area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is given by Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |$
The chapter includes many more such concepts and formulas.

Explore here the NCERT solutions of all three subjects - Chemistry, Physics, and Maths of Class 11 and Class 12. The subject matter experts at Shiksha have created all these solutions.

Class 12 Math Determinants NCERT Solution PDF: Free Download

Students can download the free Determinants Class 12 PDF from the link below. It will help them understand the concepts clearly and improve their exam preparation.
Class 12 Math Chapter 4 Determinants Solution: Free PDF Download

Class 12 Math Determinants: Key Topics, Weightage

In Class 12 Maths Chapter 4, the focus area of preparation for students should be determinants, their properties, and applications. To score well in the exams, the students should master the fundamental concepts and practice a variety of problems. The following topics are covered in this chapter:

Exercise	Topics Covered
4.1	Introduction
4.2	Determinant
4.3	Area of a Triangle
4.4	Minors and Cofactors
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices

Class 12 Determinants Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Main	2 questions	6.6%

Important Formulas of Class 12 Determinants

Determinants Important Formulae for CBSE and Competitive Exams

Students must learn important formulae and basic concepts of the Class 12 Determinants for a better grasp of the concepts. Students can also use these formulae to solve NCERT Determinants exercises.

Determinant of a $2 \times 2$ Matrix: Students can check the formula to find the value of determinant for $2 \times 2$ Matrix.

$∣ \begin{array}{cc} a & b \\ c & d \end{array} ∣ = a d - b c$

Determinant of a $3 \times 3$ Matrix: Students can check the formula to find the value of determinant for $3 \times3$ Matrix.

$∣ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} ∣ = a (e i - f h) - b (d i - f g) + c (d h - e g)$

Cramer's Rule:

Solution of $AX = B$ $x_i = \frac{\det(A_i)}{\det(A)} \quad (\det(A) \neq 0)$ where $A_i$ is the matrix obtained by replacing the $i^{\text{th}}$ column of $A$ with $B$ .

Properties of Determinants:

- $\det(A^T) = \det(A)$
- $\det(AB) = \det(A) \cdot \det(B)$
- $\det(kA) = k^n \cdot \det(A)$ for an $n \times n$ matrix.
Adjoint and Inverse:
$Adj (A) = [C_{i j}]^{T}, A^{- 1} = \frac{Adj (A)}{\det (A)}$

Exercise-wise NCERT Solutions for Class 12 Math Chapter 4 Determinants

Class 12 Determinants chapter deals with important topics such as finding determinants of square Matrices, findingarea of triangle with help of determinant, co-factors, minors, adjoint and inverse of a Matrix. consistency of linear systems. All exercise deals with different concepts, Class 12 Determinants exeercies 4.1 deals with the basics of determinants and finding the value. Class 12 Determinants Exercise 4.2 deals with finding the area of triangle. Class Determinants Exercises 4.3 deals with finding Minors and Co-factors. Class 12 Determinants Exercise 4.4 focuses on adjoints and inverse of Matrix. Class Determinants Exercise 4.5 deals with checking consistency of system of equations. Students can check Exercise-wise Class 12 Chapter 4 Determinants NCERT Solutions below;

NCERT Class 12 Math Determinants Exercise 4.1 Solutions

Ex 4.1 Class 12 Maths NCERT solutions discusses basic concept of determinants, starting with the calculation of determinants for $2 \times 2$ and $3 \times 3$ matrices. This exercise helps in building foundational blocks of the determinants. The Determinants Exercise 4.1 Solutions consist of 8 Questions including 2 Long, 5 Short Answers, 1 MCQ. Students can check complete solution of Exercuse 4.1 below;

Determinants Exercise 4.1 Solutions

Q1.Evaluate the determinants in Exercises 1 and 2.

$| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} |$

A.1. $| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} | =$ 2×(−1) – 4×(−5) = −2+20 = 18

Q2.(i) $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$ (ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

A.2. (i) $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$

$= c o s θ \times c o s θ - (- s i n θ) \times s i n θ$

= cos²θ + sin²θ

(ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

$= (x^{2} - x + 1) (x + 1) - (x - 1) (x + 1)$

= $x^{3}$ + $x^{2}$ − $x^{2}$ − $x$ + $x$ +1 – $x^{2}$ + 1

= $x^{3}$ - $x^{2}$ + 2

Q3.If $[\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ , then show that | 2A | = 4 | A |

A.3.Given, A= $[\begin{matrix} 1 & 3 \\ 4 & 2 \end{matrix}]$

So, 2A= 2 $[\begin{matrix} 1 & 4 \\ 4 & 2 \end{matrix}]$

= $[\begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix}]$ .

L.H.S. = $| 2 |$ = $| \begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix} |$ = 2×4 – 4×8 = 8−32 = −24

R.H.S. = 4 $| |$ = 4 $| \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} |$ = 4(1×2 – 2×4)

= 4[2 − 8]

= 4[−6] = 24.

LHS = RHS

Q4. If A= ,then show that | 3 A | = 27 | A |

Commonly asked questions

16. $| \begin{matrix} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{matrix} | = (x - y) (y - z) (z - x) (x y + y z + z x)$

LHS $| \begin{matrix} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{matrix} |$

$= \frac{1}{x y z}$ $| \begin{matrix} x \cdot x & x \cdot x^{2} & x \cdot y z \\ y \cdot y & y \cdot y^{2} & y \cdot z x \\ z \cdot z & z \cdot z^{2} & z \cdot x y \end{matrix} |$ Multiplying R1, R2& R3 by x, y&z.

$= \frac{1}{x y z} | \begin{matrix} x^{2} & x^{3} & x y z \\ y^{2} & y^{3} & x y z \\ z^{2} & z^{3} & x y z \end{matrix} | .$

$= \frac{x y z}{x y z} | \begin{matrix} x^{2} & x^{3} & 1 \\ y^{2} & y^{3} & 1 \\ z^{2} & z^{3} & 1 \end{matrix} |$ Taking xyz common from c3.

$= | \begin{matrix} x^{2} & x^{3} & 1 \\ y^{2} - x^{2} & y^{3} - x^{3} & 1 - 1 \\ z^{2} - x^{2} & 2^{3} - x^{3} & 1 - 1 \end{matrix} | \begin{matrix} _{2} \to_{2} -_{1} & _{3} \to_{3} -_{1} . \end{matrix}$

= $| \begin{matrix} x^{2} & x^{3} & 1 \\ (y - x) (y + x) & (y - x) (y^{2} + x^{2} + x y) & 0 \\ (z - x) (z + x) & (2 - x) (z^{2} + x^{2} + 2 x) & 0 \end{matrix} |$

Expanding along c3 we get,

= 1 $| \begin{matrix} (y - x) (y + x) & (y - x) (y^{2} + x^{2} + x y) \\ (z - x) (z + x) & (z - x) (z^{2} + x^{2} + z x) \end{matrix} |$

= (y-x)(z-x). $| \begin{matrix} y + x & y^{2} + x^{2} + x y \\ z + x & z^{2} + x^{2} + z x \end{matrix} |$ Taking (yx) & (zx) common from R₁ and R₂.

= (y-x)(z-x). Taking (y-x) & (z-x) common from R₁ and R₂.

= (y-x)(z-x)[(y + x)(z² + x² + zx) - (z + x)(y² + x² + xy)]

= (y-x)(z-x)[yz² + yx² + xyz + xz² + x³ + x²z-zy²-zx²-xyz-xy²-x³-x²y]

= y-x)(z-x)[yz²-zy² + xz²-xy²]

= (y-x)(z-x)[yz (z-y) + x(z²-y²)]

= y-x)(z-x)(z-y)[yz + x (z + y)]

= (- 1)(x-y)(z-x)(- 1)(y-z)[yz+ 1 xz + xy]

= (x-y)(y-z)(z-x)(xy + yz + xz). = R.H.S

Kindly Consider the following

58. $\begin{array}{l} 2 x - y = - 2 \\ 3 x + 4 y = 3 \end{array}$

The given system of eqⁿ in matrix form is AK = B

10. Kindly Consider the following

Kindly go through the solution

=0=RHS

20. $| \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b \\ 2 a b & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} | = {(1 + a^{2} + b^{2})}^{3}$

LHS = ${| \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b \\ 2 a b & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} |}_{.}$

$= | \begin{matrix} 1 + a^{2} - b^{2} - b (- 2 b) & 2 a b + a (- 2 b) & - 2 b \\ 2 a b - b (2 a) & 1 - a^{2} + b^{2} + a (2 a) & 2 a \\ 2 b - b (1 - a^{2} - b^{2}) & - 2 a + a (1 - a^{2} - b^{2}) & 1 - a^{2} - b^{2} \end{matrix} |$

$= | \begin{matrix} 1 + a^{2} - b^{2} + 2 b^{2} & 2 a b - 2 a b & - 2 b \\ 2 a b - 2 a b & 1 - a^{2} + b^{2} + 2 a^{2} & 2 a \\ 2 b - b + a^{2} b + b^{3} & - 2 a + a - a^{3} - a b^{2} & 1 - a^{2} - b^{2} \end{matrix} |$

$= | \begin{matrix} 1 + a^{2} + b^{2} & 0 & - 2 b \\ 0 & 1 + a^{2} + b^{2} & 2 a \\ b (1 + a^{2} + b^{2}) & - a (1 + a^{2} + b^{2}) & 1 - a^{2} - b^{2} \end{matrix} |$

= (1 + a² + b²)² [(1 -a² + b²) - 2a (-a)]

= (1 + a² + b²)² (1 -a² + b² + 2a²)

= (1 + a² + b²)² (1 + a² + b²)

= (1 + a² + b²)³ = R.H.S.

Using the property of determinants and without expanding in Exercises 1 to 7, prove that:

Kindly go through the solution

Kindly Consider the following

60. $\begin{array}{l} 5 x + 2 y = 3 \\ 3 x + 2 y = 5 \end{array}$

The given system of equation in matrix form is

AK = B

67. Prove that the determinant is independent of θ.

Kindly go through the solution

By using properties of determinants, in Exercises 8 to 14, show that:

=(b-a)(c-a)[(c+a)-(b+a)]

=(b-a)(c-a)[(c+a – b – a)]

=(b-a)(c-a)(c-b)

=(-1)(a-b) ×(-1)(b-c)(c-a)

=(a-b)(b-c)(c-a).

= R.H.S.

$= | \begin{matrix} 1 & 1 - 1 & 1 - 1 \\ a & b - a & c - a \\ a^{3} & b^{3} - a^{3} & c^{3} - a^{3} \end{matrix} | \begin{matrix} c_{2} \to c_{2} - c_{1} & c_{3} \to c_{3} - c_{1} \end{matrix}$

= $| \begin{matrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^{3} & (b - a) (b^{2} + a^{2} + a b) & (c - a) (c^{2} + a^{2} + a b) \end{matrix} |$ ${\begin{cases} ? x^{3} - y^{3} \\ = (x - y) (x^{2} + y^{2} + x y) \end{cases}$

Expanding along R1

= $| \begin{matrix} b - a & c - a \\ (b - a (b^{2} + a^{2} + a b) & (c - a) (c^{2} + a^{2} + a b) \end{matrix} |$

= (ba)(ca) $| \begin{matrix} 1 & 1 \\ b^{2} + a^{2} + a b & c^{2} + a^{2} + a b \end{matrix} |$ ${\begin{cases} (b - a) & (c - a) \\ c_{1} & c_{3} \end{cases}$

= (b-a)(c-a) {(c² + a² + ab)- (b² + a² + ab)}

= (b-a) (c -a) {c² + a² + ab-b²-a²-ab}

= (b-a) (c-a)(c²-b² + ac-ab)

= (b-a)(c-a)(c-b)(c + b) + (c-b) a}

= (b-a)(c-a)(c-b)(a + c + b).

= (- 1)(a-b)(c-a)(- 1)(b-c)(a + c + b)

= (a-b)(b-c)(c-a)(a + c + b)

= R.H.S.

32. Using Cofactors of elements of third column, evaluate $Δ = | \begin{matrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{matrix} |$

Given, $Δ = | \begin{matrix} 1 & x & y z \\ 1 & y & z x \\ 1 & z & x y \end{matrix} |$

Co-factor of elements of third column

∴ Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

= yz (z-y) + zx [- (z-x)] + xy (y-x)

= y z²-y²z-z²x+ zx² + xy²-x²y.

= yz²-y²z+ (xy²-xz²) + (zx²-x²y)

= yz (z-y) + x (y² – z²) -x² (y-z)

= -yz (y-z) + x (y + z) (y-z) -x² (y-z)

= (y-z) [-yz + x (y + z) -x²]

= (y-z) [-yz + xy + xz-x²]

= (y-z) [-y (z-x) + x (z-x)]

= (y-z) (z-x) (x -y)

17. (i) $| \begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} | = (5 x + 4) {(4 - x)}^{2}$

(i) LHS = $| \begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$

$= | \begin{matrix} x + 4 + 2 x + 2 x & 2 x + x + 4 + 2 x & 2 x + 2 x + x + 4 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$ R_1→R₁ + R₂ + R₃

$= | \begin{matrix} 5 x + 4 & 5 x + 4 & 5 x + 4 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$

Taking (5x + 4) common from R₁.

$= (5 x + 4) | \begin{matrix} 1 & 1 & 1 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} | .$

= (5x + 4)[0 - (4 -x)(x- 4)]

= (5x + 4)(4 -x)(4 -x)

= (5x + 4)(4 -x)² = R.H.S.

= $| \begin{matrix} y + k + y + y & y + y + k + y & y + y + y + k \\ y & y + k & y \\ y & y & y + k \end{matrix} |$ R_1→R₁ + R₂+ R₃

$= | \begin{matrix} 3 y + k & 3 y + k & 3 y + k \\ y & y + k & y \\ y & y & y + k \end{matrix} |$

= (3y + k) $| \begin{matrix} 1 & 1 & 1 \\ y & y + k & y \\ y & y & y + k \end{matrix} |$

65. $[\begin{array}{l} 2 - 3 5 \\ 3 2 - 5 \\ 1 1 - 5 \end{array}],find A^{- 1} .Using A^{- 1}$ solve the system of equations.

$\begin{array}{l} 2 x - 3 y + 5 z = 11 \\ 3 x + 2 y - 4 z = - 5 \\ x + y - 2 z = - 3 \end{array}$

Kindly go through the solution

61. $\begin{array}{l} 2 x + y + z = 1 \\ x - 2 y - z = \frac{3}{2} \\ 3 y - 5 z = 9 \end{array}$

The given system of solution in matrix form is AK = B.

62. $\begin{array}{l} x - y + z = 4 \\ 2 x + y - 3 z = 0 \\ x + y + z = 2 \end{array}$

The given system of eqⁿ can be written in matrix form as AK = B.

13. Kindly Consider the following

Kindly go through the solution

70. If a, b and c are real numbers and

$△ = | \begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0$ Show that either a+b+c=0 or a=b=c=0

$? = | \begin{matrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0$

$\begin{array}{l} R_{1} \to R_{1} + R_{2} + R_{3} \\ | \begin{matrix} 2 (a + b + c) & 2 (a + b + c) & 2 (a + b + c) \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0 \end{array}$

Taking 2(a + b + c) common from R₁

$\Rightarrow 2 (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0$

Either $2 (a + b + c) = 0$ i.e. a+b+c=0 or

$| \begin{matrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0$

$c_{2} \to c_{2} - c_{1}$ and $c_{3} \to c_{3} - c_{1}$

$\Rightarrow | \begin{matrix} 1 & 0 & 0 \\ c + a & b - c & b - a \\ a + b & c - a & c - b \end{matrix} | = 0$

Expanding along $R_{1}$

$\begin{array}{l} \Rightarrow | \begin{matrix} b - c & b - a \\ c - a & c - b \end{matrix} | = 0 \\ \Rightarrow (b - c) (c - b) - (b - a) (c - a) = 0 \\ \Rightarrow b c - b^{2} - c^{2} + c b - b c + a b + a c - a^{2} = 0 \\ \Rightarrow - a^{2} - b^{2} - c^{2} + a b + b c + c a = 0 \end{array}$

Multiplying by -2

$\begin{array}{l} \Rightarrow 2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a = 0 \\ \Rightarrow a^{2} + a^{2} + b^{2} + b^{2} + c^{2} + c^{2} - - 2 a b - 2 b c - 2 c a = 0 \\ \Rightarrow (a^{2} + b^{2} - 2 a b) + (a^{2} + c^{2} - 2 a c) + (b^{2} + c^{2} - 2 b c) = 0 \\ \Rightarrow {(a - b)}^{2} + {(a - c)}^{2} + {(b - c)}^{2} = 0 \\ \Rightarrow a - b = 0, b - c = 0, c - a = 0 \\ [? x^{2} + y^{2} + z^{2} = 0 \Rightarrow x = 0, y = 0, z = 0] \\ \Rightarrow a = b, b = c, c = a \\ \Rightarrow a = b = c \end{array}$

$∴$ Either a+b+c=0 or a=b=c

28. 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

Given,

Area of triangle = 35 sq. Units

$\Rightarrow \frac{1}{2} | \begin{matrix} 2 & - 6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} | = 35 .$

∴ Option D is correct.

Kindly Consider the following

42.

Kindly go through the solution

14. Kindly Consider the following

Kindly go through the solution

=a²b²c²× (−2) $(1 \times (- 1) - 1 \times 1)$

= a²b²c²× (−2)× (−2)

= 4a²b²c²

= R.H.S.

85. Let A = $[\begin{matrix} 1 & s i n θ & 1 \\ - s i n θ & 1 & s i n θ \\ - 1 & - s i n θ & 1 \end{matrix}]$ , where 0 ≤θ≤ 2. Then

(A) Det (A) = 0 (B) Det (A) (2, ∞)

(C) Det (A) (2, 4) (D) Det (A) [2, 4]

Kindly go through the solution

5. Evaluate the determinants

$= 3 [0 (- 5) (- 1)] + 1 (0 - 3 (- 1) - 2 [0 \times (- 5) - 0 \times 3]$

= 15 + 3

= 12

=3 [1×1− (−2)×3]+4 [1×1− (−2)×2]+5 [3×1−2×1]

= 3 [1+6]+4 [1+4]+5 (3−2)

= 3×7+4×5+5×1

= 21+20+5

= 46

= 0 –1 [ − 1×0 – (−2)× (−3)] + 2 [−1×3 – (−2)×0]

= (− 1)× ( − 6)+2 (− 3)

= 6 – 6

= 0

= 2 [2×0− ( − 5)× (−1)]+3 [ (−1)× (− 1) – (− 2)×2]

= 2 (−5)+3 (1+4)

= 2 (−5) + 3× (5)

= −10+15

= 5

33. If $Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$ and $_{i j}$ is Cofactors of $a_{i j}$ then value of $Δ$ is given by

Option D is correct.

Kindly Consider the following

37.

Kindly go through the solution

4. If A= ,then show that | 3 A | = 27 | A |

=3 [36 – 6×0]

= 108

RHS = 27. |A| = 27 4 = 108

∴ LHS = RHS

8. If , then x is equal to

(A) 6 (B) ± 6 (C) – 6 (D) 0

Kindly go through the solution

27. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants

(i) Let P (x, y) be any point on line joining A (1, 2) & B (3, 6)

Then, area of triangle (ABP) = 0 {the point are collinear

$\frac{1}{2} | \begin{matrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{matrix} | = 0$

31. Using Cofactors of elements of second row, evaluate $Δ = | \begin{array}{l} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array} |$

Given, $? = | \begin{array}{l} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array} |$

Co-factors of elements of second row,

? ? = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

= 2 * 7 + 0 * 7 + 1 * (-7) = 14 + 0 - 7 = 7.

Kindly Consider the following

69. Evaluate

$= - s i n α (- s i n^{2} β s i n α - s i n α c o s^{2} β) - 0 (c o s α c o s β s i n α s i n β - c o s α s i n β s i n α c o s β) + c o s α (c o s α c o s^{2} β + c o s α s i n^{2} β)$

$= s i n^{2} α (s i n^{2} β + c o s^{2} β) + c o s^{2} α (c o s^{2} β + s i n^{2} β)$

$= s i n^{2} α + c o s^{2} α$

76. Evaluate $|\begin{matrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{matrix}|$

Kindly go through the solution

21. ${| \begin{matrix} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ c a & c b & c^{2} + 1 \end{matrix} |}_{.} = 1 + a^{2} + b^{2} + c^{2}$

LHS = ${| \begin{matrix} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ c a & c b & c^{2} + 1 \end{matrix} |}_{.}$

$= \frac{1}{a b c} | \begin{matrix} a (a^{2} + 1) & a b^{2} & a c^{2} \\ a^{2} b & b (b^{2} + 1) & b c^{2} \\ a^{2} c & b^{2} c & c (c^{2} + 1) \end{matrix} |$

$= \frac{a b c}{a b c} [\begin{matrix} a^{2} + 1 & b^{2} & c^{2} \\ a^{2} & b^{2} + 1 & c^{2} . \\ a^{2} & b^{2} & c^{2} + 1 \end{matrix}]$ Taking a, b&c common from R₁, R₂&R₃

$[\begin{matrix} 1 + a^{2} + b^{2} + c^{2} & b^{2} & c^{2} \\ a^{2} + b^{2} + 1 + c^{2} & b^{2} + 1 & c^{2} \\ a^{2} + b^{2} + c^{2} + 1 & b^{2} & c^{2} + 1 \end{matrix}]$ C_1→ C₂ + C₃.

Choose the correct answer in Exercises 15 and 16.

22. Let A be a square matrix of order 3 × 3, then | kA| is equal to

(A) k| A| (B) | A | (C) | A | (D) 3k | A |

Then, KA = $k [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}]$

$= [\begin{matrix} k a_{11} & k a_{12} & k a_{13} \\ k a_{4} & k a_{22} & k a_{23} \\ k a_{31} & k a_{32} & k a_{33} \end{matrix}]$

$∴ |KA| = | \begin{matrix} _{a 11} & k a_{12} & _{13} \\ _{a_{21}} & _{22} & k a_{23} \\ _{31} & _{a_{32}} & k_{a_{33}} \end{matrix} |$

$=k^{3} | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{22} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$

$= k^{3} |A|$

So, option c is correct.

Kindly Consider the following

38.

Kindly go through the solution

48. For the matrix A = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] .$

Given, A= $[\begin{matrix} 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}]$

56. $\begin{array}{l} 5 x - y + 4 z = 5 \\ 2 x + 3 y + 5 z = 2 \\ 5 x - 2 y + 6 z = - 1 \end{array}$

The given system of eqⁿ in matrix form is AK = B

= 5 x 28 -13 + 4 x (-9)

= 140 - 13 - 76

= 51 ≠ 0 .

The given system of eqⁿ are consistent

66. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is `60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is `90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is `70. Find cost of each item per kg by matrix method.

Let k, y and z be the cost per kg of onion, wheat and rice. Then, we form a system of equations as follows

4k + 3y + 2z = 60

2k + 4y + 6z = 90

6k + 2y + 3z = 70

In matrix form we can write,

AX = B

= 0 - 3 (-30) + 2 (-20)

= 90 - 40

= 50 ≠ 0.

∴ The system has a unique solution.

The cost of onion, wheat and rice per kg are? 5? 8 and? 8 respectively.

68. Without expanding the determinants, prove that

$| \begin{matrix} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{matrix} | = | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} |$

L.H.S.= $| \begin{matrix} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{matrix} |$

Multiplying R₁, by a, R₂ by b and R₃ by c

$= \frac{1}{a b c} | \begin{matrix} a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c \end{matrix} |$

Taking abc common from c_{3 = $= \frac{a b c}{a b c} | \begin{matrix} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{matrix} |$}

Inter changing c₁ and c_{3 = $= - | \begin{matrix} 1 & a^{3} & a^{2} \\ 1 & b^{3} & b^{2} \\ 1 & c^{3} & c^{2} \end{matrix} |$}

Inter changing c_{2 and}c_{3 = $= | \begin{matrix} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{matrix} | = R . H . S$}

Kindly Consider the following

41.

Kindly go through the solution

24. Find 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)

(i) Area of triangle is given by,

Δ = $\frac{1}{2}$ $| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | = \frac{1}{2} | \begin{matrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{matrix} |$

$= \frac{1}{2} | [- 3 + 18] | = \frac{15}{2}$ =7.5sq. units.

(ii) Area of the triangle is given by,

Δ = $\frac{1}{2}$ $| \begin{matrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{matrix} |$

$= \frac{1}{2} | [2 (1 - 8) - 7 (1 - 10) + 1 (8 - 10)] |$

$= \frac{1}{2} | (2 \times (- 7) - 7 \times (- 9) + 21 \times (- 2)] |$

$= \frac{1}{2} | [- 14 + 63 - 2] | = \frac{1}{2} | 47 |$

= $\frac{47}{2}$ =23.5 sq. units

(iii) Area of triangle is given by,

Δ = $\frac{1}{2}$ ${| \begin{matrix} - 2 & - 3 & 1 \\ 3 & 2 & 1 \\ - 1 & - 8 & 1 \end{matrix} |}_{.}$

$= \frac{1}{2} | [- 2 \times 10 + 3 (4) + (- 24 + 2)] |$

$= \frac{1}{2} | [- 20 + 12 - 22] |$

$= \frac{1}{2} | - 30 | = \frac{30}{2}$ = 15 sq. units.

74. Let A = . Verify that

(i) [adj A]^-1 = adj (A^-1) (ii) (A^-1)^-1 = A

Kindly go through the solution

6. If A= ,find | A |

= 1 [1× (−9) – 4× (−3)]–1 [2× (–9)–5× (– 3)]– 2 [2×4 −5×1]

= 1 [− 9 + 12] −1 [− 18+15] − 2 [8 − 5]

= 3 + 3 −6

= 0

11. Kindly Consider the following

Kindly go through the solution

35. $[\begin{array}{l} 1 - 1 2 \\ 2 3 5 \\ - 2 0 1 \end{array}]$

We have,

$A_{11} = {(- 1)}^{1 + 1} \times | \begin{matrix} 3 & 5 \\ 0 & 1 \end{matrix} | = (3 \times 1 - 5 \times 0) = 3$

$A_{12} = {(- 1)}^{1 + 2} \times | \begin{matrix} 2 & 5 \\ - 2 & 1 \end{matrix} | = (- 1) (2 \times 1 - 5 \times (- 2) = - 1 (2 + 10) = - 12$

$A_{13} = {(- 1)}^{1 + 3} \times | \begin{matrix} 2 & 3 \\ - 2 & 0 \end{matrix} | = [2 \times 0 - (- 2) \times 3] = 6$

Kindly Consider the following

40.

Kindly go through the solution

Kindly Consider the following

43. $[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}]$

Let A = $[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}]$ Then |A| = $| \begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix} |$

$\Rightarrow | | = 1 \times | \begin{matrix} 2 & - 3 \\ - 2 & 4 \end{matrix} | - (- 1) | \begin{matrix} 0 & - 3 \\ 3 & 4 \end{matrix} | + 2 | \begin{matrix} 0 & 2 \\ 3 & - 2 \end{matrix} |$

= (8 - 6) + (0 - (- 9) + 2 (0 - 6)

= 2 + 9 - 12 = - 1 ¹ 0

So, A^-¹ exist

3. If $[\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ , then show that | 2A | = 4 | A |

Given, A= $[\begin{matrix} 1 & 3 \\ 4 & 2 \end{matrix}]$

So, 2A= 2 $[\begin{matrix} 1 & 4 \\ 4 & 2 \end{matrix}]$

= $[\begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix}]$ .

L.H.S. = $| 2 |$ = $| \begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix} |$ = 2×4 – 4×8 = 8−32 = −24

R.H.S. = 4 $| |$ = 4 $| \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} |$ = 4 (1×2 – 2×4)

= 4 [2 − 8]

= 4 [−6] = 24.

LHS = RHS

1.Evaluate the determinants in Exercises 1 and 2.

$| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} |$

$| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} | =$ 2× (−1) – 4× (−5) = −2+20 = 18

7. Find values of x, if

Kindly go through the solution

26. Find values of k if area of triangle is 4 sq. units and vertices are

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

(i) Area of the triangle = 4 sq. units (given)

$\Rightarrow \frac{1}{2} | \begin{matrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{matrix} | = 4$

(ii) Area of the triangle = 4 sq units

34. $[\begin{array}{l} 1 2 \\ 3 4 \end{array}]$

We have,

A_{11} = {(- 1)}^{1 + 1} \times 4 = 4

A_{12} = {(- 1)}^{1 + 2} \times 3 = - 3

A_{21} = {(- 1)}^{2 + 1} \times 2 = - 2

A_{22} = {(- 1)}^{2 + 2} \times 1 = 1 .

Kindly Consider the following

36. $[\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}]$

Let $A = [\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}] | A | = (2 \times (- 6) - 3 \times (- 4)) = - 12 + 12 = 0$

$\begin{array}{l} A_{11} = {(- 1)}^{1 + 1} \times (- 6) = - 6 \\ A_{12} = {(- 1)}^{1 + 2} \times (- 4) = 4 \end{array}$

$\begin{array}{l} A_{21} = {(- 1)}^{2 + 1} \times (3) = - 3 \\ A_{22} = {(- 1)}^{2 + 2} \times 2 = 2 \end{array}$

Kindly Consider the following

39.

Kindly go through the solution

44. $[\begin{matrix} 1 & 0 & 0 \\ 0 & c o s \propto & s i n \propto \\ 0 & s i n \propto & - c o s \propto \end{matrix}]$

Kindly go through the solution

Kindly Consider the following

49.

Kindly go through the solution

Kindly Consider the following

75. Evaluate

Expanding along C₁

$= 2 (x + y) {(- x) (x - y) - y . y}$

$= 2 (x + y) (- x^{2} + x y - y^{2})$

$= - 2 (x + y) (x^{2} - x y + y^{2})$

$= - 2 {x^{3} - x^{2} y + x y^{2} + x^{2} y - x y^{2} + y^{3}}$

= $- 2 (x^{3} + y^{3})$

81. $| \begin{matrix} s i n α & c o s α & c o s (α + δ) \\ s i n β & c o s β & c o s (β + δ) \\ s i n γ & c o s γ & c o s (γ + δ) \end{matrix} | = 0$

$\begin{array}{l} L . H . S . = | \begin{matrix} s i n α & c o s α & c o s (α + δ) \\ s i n β & c o s β & c o s (β + δ) \\ s i n γ & c o s γ & c o s (γ + δ) \end{matrix} | \\ = \frac{1}{s i n δ c o s δ} | \begin{matrix} s i n α s i n δ & c o s α c o s δ & c o s (α + δ) \\ s i n β s i n δ & c o s β c o s δ & c o s (β + δ) \\ s i n γ s i n δ & c o s γ c o s δ & c o s (γ + δ) \end{matrix} | \end{array}$

Applying $c o s (A + B) = c o s A c o s B - s i n A s i n B$ in $c_{3}$

$\begin{array}{l} c_{1} \to c_{1} + c_{3} \\ ? = \frac{1}{s i n δ c o s δ} | \begin{matrix} c o s α c o s δ & c o s α c o s δ & c o s α c o s δ - s i n α s i n δ \\ c o s β c o s δ & c o s β c o s δ & c o s β c o s δ - s i n β s i n δ \\ c o s γ c o s δ & c o s γ c o s δ & c o s γ c o s δ - s i n γ s i n δ \end{matrix} | \\ c_{1} = c_{2} \\ ∴ ? = 0 = R . H . S . \end{array}$

82. Solve the system of equations

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z}$ = 4

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z}$ = 1

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z}$ = 2

Given equations are :-

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$

$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$

$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

2. (i) $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$ (ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

(i) $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$

$= c o s θ \times c o s θ - (- s i n θ) \times s i n θ$

= cos²θ + sin²θ

(ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

$= (x^{2} - x + 1) (x + 1) - (x - 1) (x + 1)$

= $x^{3}$ + $x^{2}$ − $x^{2}$ − $x$ + $x$ +1 – $x^{2}$ + 1

= $x^{3}$ - $x^{2}$ + 2

12. Kindly Consider the following

Kindly go through the solution

18. (i) $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | = {(a + b + c)}^{3}$

${| \begin{matrix} x + y + 2 z & x & y \\ z & y + z + 2 x & y \\ z & x & z + x + 2 y \end{matrix} |}_{\cdot} = 2 {(x + y + z)}^{3}$

(i) LHS = $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

$= | \begin{matrix} a - b - c + 2 b + 2 c & 2 a + b - c - a + 2 c & 2 a + 2 b + c - a b \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$ R_1→ R₁ + R₂ + R₃

$= | \begin{matrix} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

= (a + b + c) $| \begin{matrix} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$ Taking (a + b + c) common from R₁

= (a + b+ c) $| \begin{matrix} 1 & 1 - 1 & 1 - 1 \\ 2 b & b - c - a - 2 b & 2 b - 2 b \\ 2 c & 2 c - 2 c & c - a - b - 2 c \end{matrix} | \begin{matrix} _{} \to_{2} -_{1} & _{} \to_{3} -_{1} \end{matrix}$

= (a + b + c) $| \begin{matrix} 1 & 0 & 0 \\ 2 b & - b - c - a & 0 \\ 2 c & 0 & - c - a - b \end{matrix} | .$

= (a + b + c) × 1. $| \begin{matrix} - (a + b + c) & 0 \\ 0 & - (a + b + c) \end{matrix} |$ Expand along R₁

= (a + b + c){(a + b + c)²- 0}

= (a + b +c)³ = R.H.S

(ii) LHS = ${| \begin{matrix} x + y + 2 z & x & y \\ z & y + z + 2 x & y \\ z & x & z + x + 2 y \end{matrix} |}_{\cdot}$

$= | \begin{matrix} x + y + 2 z + x + y & x & y \\ z + y + z + 2 x + y & y + z + 2 x & y \\ z + x + z + x + 2 y & x & z + x + 2 y \end{matrix}$
C_1→ C₁ + C₂ + C₃.

$= {| \begin{matrix} 2 (x + y + z) & x & y \\ 2 (x + y + z) & y + z + 2 x & y \\ 2 (x + y + z) & x & z + x + 2 y \end{matrix} |}_{.}$

= 2 (k + y + z) $| \begin{matrix} 1 & x & y \\ 1 & y + z + 2 x & y \\ 1 & x & z + x + 2 y \end{matrix} |$ Taking 2

19. $| \begin{matrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix} | = {(1 - x^{3})}^{2}$

LHS = $| \begin{matrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix} |$

$= | \begin{matrix} 1 + x^{2} + x & x + 1 + x^{2} & x^{2} + x + 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix} |_{}$ R_1→ R₁ + R₂ + R₃

= (1 + x² + x) (1 -x)² [ (1 + x)× 1 - (-x) x].

= (1 + x² + x) (1 -x)² (1 + x + x²).

= { (1 + x² + x) (1 -x)}²

= {1 -x + x²-x³ + x-x²}²

= (1 -x³)² = R.H.S.

23. 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

Option 'C' is correct as determinant is a number associated to a square matrix.

25. Show that points

A (a, b + c), B (b, c + a), C (c, a + b) are collinear.

The area of triangle from by the given points is area ( ΔABC) = $\frac{1}{2}$ $| \begin{matrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{matrix} |$

$= \frac{1}{2} | \begin{matrix} a + b + c + 1 & b + c & 1 \\ b + c + a + 1 & c + a & 1 \\ c + a + b + 1 & a + b & 1 \end{matrix} |$ C_1→ C₁ + C₂ + C₃

$= \frac{1}{2} (a + b + c + 1) | \begin{matrix} 1 & b + c & 1 \\ 1 & c + a & 1 \\ 1 & a + b & 1 \end{matrix} |$ Taking (a + b + c + 1) common from C₁

$= \frac{(a + b + c + 1)}{2} \times 0 {? c = c_{3}}$

= 0

Hence the points A, B C are collinear.

29. (i) $| \begin{matrix} 2 & - 4 \\ 0 & 3 \end{matrix} |$ (ii) $| \begin{matrix} a & c \\ b & d \end{matrix} |$

(i) We know that,

Minor of element a_ij is m_ij and its co-factor is A_ij = (–1)^i+j M_ij

So,

M₁₁ = 3 and A₁₁ = (- 1)^{1 + 1} M₁₁ = 1 * 3 = 3

M₁₂ = 0 and A₁₂ = (-1)¹⁺² M₁₂ = -1 * 0 = 0

M₂₁ = -4 and A₂₁ = (-1)²⁺¹ M₂₁ = (-1) * (-4) = 4

M₂₂ = 2 and A₂₂ = (-1)²⁺² M₂₂ = 1 * 2 = 2

(ii) Given A = $| \begin{matrix} a & c \\ b & d \end{matrix} |$

So,

M₁₁ = d and A₁₁ = (-1)¹⁺¹ M₁₁ = 1 * d = d

M₁₂ = b and A₁₂ = (-1)¹⁺² M₁₂ = (-1) * b = -b

M₂₁ = c and A₂₁ = (-1)^{2+ 1} M₂₁ = (–1) * c = –c

M₂₂ = a and A₂₂ = (-1)²⁺² M₂₂ = 1 * a = a

30. (i) $| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} |$ (ii) $| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & - 1 \\ 0 & 1 & 2 \end{matrix} |$

(i) Given, A = $| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 \end{matrix} |$

So,

(ii)Given, $Δ = | \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & - 1 \\ 0 & 1 & 2 \end{matrix} |$

Kindly Consider the following

45.

Kindly go through the solution

Kindly Consider the following

46.

Given, A= $[\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$

we have, A² = A?A = $[\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}]$

$= [\begin{matrix} 3 \times 3 + 1 \times (- 1) & 3 \times 1 + 1 \times 2 \\ (- 1) \times 3 + (- 1) \times 2 & (- 1) \times 1 + 2 \times 2 \end{matrix}]$

$= [\begin{matrix} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{matrix}] = [\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}]$

Hence, A² – 5A+71= $[\begin{matrix} 8 & 5 \\ - 5 & 3 \end{matrix}] - 5 [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] + 7 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$= [\begin{matrix} 8 - 5 \times 3 + 7 \times 1 & 5 - 5 \times 1 + 7 \times 0 \\ - 5 - 5 \times (1) + 7 \times 0 & 3 - 5 \times 2 + 7 \times 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}] = 0$

Now, A² – 5A+7I=0.

A?A – 5A= –7I.

A A(A^–1)–5(AA^–1)= –7I(A^–1) (Multiplying by A^-1on both sides)

AI – 5I= –7A^–1

7A^–1= –(AI – 5I)

A^–1= $\frac{1}{7} [- A + 5 I]$

= $\frac{1}{7} {(- 1) [\begin{matrix} 3 & 1 \\ - 1 & 2 \end{matrix}] + 5 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]}$

= $\frac{1}{7} {[\begin{matrix} (- 1) \times 3 + 5 \times 1 & - 1 + 5 \times 0 \\ (- 1) \times (- 1) + 5 \times 0 & (- 1) \times 2 + 5 \times 1 \end{matrix}]}$

A^–1= $\frac{1}{7} [\begin{matrix} 2 & - 1 \\ 1 & 3 \end{matrix}]$

Kindly Consider the following

47.

Kindly go through the solution

⇒ b = 1

Kindly Consider the following

50.

Given, A is an invertible matrix of order 2.

∴ Option (B) is correct.

51. $\begin{array}{l} x + 2 y = 2 \\ 2 x + 3 y = 3 \end{array}$

The given system of equation can be written in the form of AK = B

Now, = 3 × 1 - 2 × 2 = 3 - 4 = -1 ≠ 0.

∴ The system has unique solⁿ and hence equation are consistent

Kindly Consider the following

52. $\begin{array}{l} 2 x - y = 5 \\ x + y = 4 \end{array}$

The given system of eqⁿ can be written in the form AK = B where

Now, = 2 – (-1) = 2 + 1 = 3 ≠ 0.

∴ The system of eqⁿis consistent.

Kindly Consider the following

53. $\begin{array}{l} 2 x + 3 y = 5 \\ 2 x + 6 y = 8 \end{array}$

The given system of equation can be written in the form AK = B

Hence the given system of eqⁿ are inconsistent.

54. $\begin{array}{l} x + y + z = 1 \\ 2 x + 3 y + 2 z = 2 \\ a x + a y = 2 a z = 4 \end{array}$

The given system of eqⁿ using matrix method can be expressed as

AK = B

= (6 a - 2 a) - (4 a - 2 a) + (2 a - 3 a)

= 4a – 2a – a

= a ≠ 0.

Hence, the given system of eqⁿ is consistent

55. $\begin{array}{l} 3 x - y - 2 z = 2 \\ 2 y - z = - 1 \\ 3 x - 5 y = 3 \end{array}$

The given system of eqⁿ using matrix form can be written as AK = B

= 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 6)

= -15 + 3 + 12

= -15 + 15

= 0

Kindly Consider the following

57. $\begin{array}{l} 5 x + 2 y = 4 \\ 7 x + 3 y = 5 \end{array}$

The given system of eqⁿ in matrix form is AK = B.

Then, |A| = 5 x 3 - 7 x 2 = 15 - 14 = 1 ≠ 0

∴ The system has a unique solution.

∴ k = 2 and y = -3

Kindly Consider the following

59. $\begin{array}{l} 4 x - 3 y = 3 \\ 3 x - 5 y = 7 \end{array}$

The given system of eqⁿ can be written in matrix form as

AK = B

63. $\begin{array}{l} 2 x + 3 y + 3 z = 5 \\ x - 2 y + z = - 4 \\ 3 x - y - 2 z = 3 \end{array}$

The given system of equation can be represented in matrix form as AK = B

64. $\begin{array}{l} x - y + 2 z = 7 \\ 3 x + 4 y - 5 z = - 5 \\ 2 x - y + 3 z = 12 \end{array}$

The given system of eqⁿ can be written in matrix form as

AK = B

∴ x= 2, y = 1 and z = 3 .

71. Solve the equation $| \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0, a \neq 0$

$\begin{array}{l} | \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ | \begin{matrix} 3 x + a & 3 x + a & 3 x + a \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \end{array}$

Taking 3x+a common from $R_{1}$

$(3 x + a) | \begin{matrix} 1 & 1 & 1 \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0$

Either $3 x + a = 0 \Rightarrow x = - \frac{a}{3}$

$\begin{array}{l} | \begin{matrix} 1 & 1 & 1 \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0 \\ c_{2} \to c_{2} - c_{1}, c_{3} \to c_{3} - c_{1} \\ | \begin{matrix} 1 & 0 & 0 \\ x & a & 0 \\ x & 0 & a \end{matrix} | = 0 \end{array}$

Expanding along $R_{1}$ , $a^{2} = 0$

$\Rightarrow a = 0$

$∴$ $x = - \frac{a}{3}$ is the only solution.

72. Prove that $| \begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix} | = 4 a^{2} b^{2} c^{2}$

$L . H . S = | \begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix} |$

$= | \begin{matrix} a^{2} & b c & c (a + c) \\ a (a + b) & b^{2} & a c \\ a b & b (b + c) & c^{2} \end{matrix} |$

Taking a, b, c common from c₁, c₂and c₃ respectively

$\begin{array}{l} = a b c | \begin{matrix} a & c & a + c \\ a + b & b & a \\ b & b + c & c \end{matrix} | \\ R_{1} \to R_{1} - R_{2} - R_{3} \\ = a b c | \begin{matrix} - 2 b & - 2 b & 0 \\ a + b & b & a \\ b & b + c & c \end{matrix} | \\ c_{2} \to c_{2} - c_{1} \\ = a b c | \begin{matrix} - 2 b & 0 & 0 \\ a + b & - a & a \\ b & c & c \end{matrix} | \end{array}$

Expanding along $R_{1}$

$a b c (- 2 b) (- 2 a c) = 4 a^{2} + b^{2} + c^{2} = R . H . S .$

73. If A^-1 = $[\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ and B = $[\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$ find (AB)^-1

Kindly go through the solution

77. $| \begin{matrix} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{matrix} | = (β - γ) (γ - α) (α - β) (α + β + γ)$

$\begin{array}{l} L . H . S . = | \begin{matrix} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{matrix} | \\ c_{3} \to c_{3} + c_{1} \\ = | \begin{matrix} α & α^{2} & α + β + γ \\ β & β^{2} & α + β + γ \\ γ & γ^{2} & α + β + γ \end{matrix} | \\ = (α + β + γ) | \begin{matrix} α & α^{2} & 1 \\ β & β^{2} & 1 \\ γ & γ^{2} & 1 \end{matrix} | \end{array}$

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$= (α + β + γ) | \begin{matrix} α & α^{2} & 1 \\ β - α & β^{2} - α^{2} & 0 \\ γ - α & γ^{2} - α^{2} & 0 \end{matrix} |$

Expanding along $c_{3}$

$\begin{array}{l} = (α + β + γ) | \begin{matrix} β - α & β^{2} - α^{2} \\ γ - α & γ^{2} - α^{2} \end{matrix} | \\ = (α + β + γ) | \begin{matrix} β - α & (β - α) (β + α) \\ γ - α & (γ - α) (γ + α) \end{matrix} | \\ \Rightarrow (α + β + γ) (β - α) (γ - α) | \begin{matrix} 1 & β + γ \\ 1 & γ + α \end{matrix} | \\ \Rightarrow (α + β + γ) (β - α) (γ - α) {γ + 2 - (β - γ)} \\ \Rightarrow (α + β + γ) (β - α) (γ - α) (γ - β) \\ \Rightarrow (α + β + γ) (α - β) (β - γ) (γ - α) = R . H . S . \end{array}$

78. $| \begin{matrix} x & x^{2} & 1 + p x^{3} \\ y & y^{2} & 1 + p y^{3} \\ z & z^{2} & 1 + p z^{3} \end{matrix} | = (1 + p x y z) (x - y) (y - z) (z - x)$

$\begin{array}{l} L . H . S . = | \begin{matrix} x & x^{2} & 1 + p x^{3} \\ y & y^{2} & 1 + p y^{3} \\ z & z^{2} & 1 + p z^{3} \end{matrix} | \\ = | \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} | + | \begin{matrix} x & x^{2} & p x^{3} \\ y & y^{2} & p y^{3} \\ z & z^{2} & p z^{3} \end{matrix} | \\ = ?_{1} + ?_{2} - - - - - (1) \\ N o w ?_{2} = | \begin{matrix} x & x^{2} & p x^{3} \\ y & y^{2} & p y^{3} \\ z & z^{2} & p z^{3} \end{matrix} | \\ = p x y z | \begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix} | \\ c_{1} \leftrightarrow c_{3} \\ = - p x y z | \begin{matrix} x^{2} & x & 1 \\ y^{2} & y & 1 \\ z^{2} & z & 1 \end{matrix} | \\ c_{1} \leftrightarrow c_{2} \\ = p x y z | \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} | = p x y z ?_{1} \end{array}$

Putting value in (1)

$\begin{array}{l} \Rightarrow ?_{1} + p x y z ?_{1} \\ = (1 + p x y z) ?_{1} - - - - - (2) \\ N o w ?_{1} = | \begin{matrix} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{matrix} | \end{array}$

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$= | \begin{matrix} x & x^{2} & 1 \\ y - x & y^{2} - x^{2} & 0 \\ z - x & z^{2} - x^{2} & 0 \end{matrix} |$

Expanding along $c_{3}$

$\begin{array}{l} ?_{1} = | \begin{matrix} (y - x) & (y - x) (y + x) \\ (z - x) & (z - x) (z + x) \end{matrix} | \\ = (y - x) (z - x) | \begin{matrix} 1 & y + x \\ 1 & z + x \end{matrix} | \\ = (y - x) (z - x) (z + x - y - x) \\ = (x - y) (y - z) (z - x) \end{array}$

Putting $?_{1}$ in (2)

$L . H . S . = (1 + p x y z) (x - y) (y - z) (z - x) = R . H . S$

79. $| \begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix} | = 3 (a + b + c) (a b + b c + c a)$

$\begin{array}{l} L . H . S . = | \begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix} | \\ c_{1} \to c_{1} + c_{2} + c_{3} \\ = | \begin{matrix} a + b + c & - a + b & - a + c \\ a + b + c & 3 b & - b + c \\ a + b + c & - c + b & 3 c \end{matrix} | \\ (a + b + c) | \begin{matrix} 1 & - a + b & - a + c \\ 1 & 3 b & - b + c \\ 1 & - c + b & 3 c \end{matrix} | \end{array}$

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$= (a + b + c) | \begin{matrix} 1 & - a + b & - a + c \\ 0 & 3 b + a - b & - b + a \\ 0 & - a + b + a - b & 3 c + a - c \end{matrix} |$

Expanding along $c_{1}$

$\begin{array}{l} = (a + b + c) | \begin{matrix} 2 b + a & a - b \\ a - c & 2 c + a \end{matrix} | \\ = (a + b + c) [4 b c + 2 a b + 2 a c + a^{2} + a c + a b - b c] \\ = (a + b + c) (3 a b + 3 b c + 3 a c) \\ = 3 (a + b + c) (a b + b c + a c) \\ = R . H . S . \end{array}$

80. Using properties of determinants,

Prove that: $| \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} | = 1$

$\begin{array}{l} L . H . S . = | \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} | \\ R_{2} \to R_{2} - 2 R_{1} \\ R_{3} \to R_{3} - 3 R_{1} \\ = | \begin{matrix} 1 & 1 + p & 1 + p + q \\ 0 & 1 & 2 + p \\ 0 & 3 & 7 + 3 p \end{matrix} | \\ R_{3} \to R_{3} - 3 R_{2} \\ = | \begin{matrix} 1 & 1 + p & 1 + p + q \\ 0 & 1 & 2 + p \\ 0 & 0 & 1 \end{matrix} | \end{array}$

Expanding along $c_{1}$

$? = | \begin{matrix} 1 & 2 + p \\ 0 & 1 \end{matrix} | = 1 = R . H . S .$

83. Choose the correct answer

If a, b, c are in AP, then the determinant $| \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} |$

A. 0

B. 1

C. Z

D. 2x

A $| \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} | = 0$

84. If x, y, z are nonzero real numbers, then the inverse of matrix A = $[\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ is

(A) $[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$ (B) xyz $[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

(C) $\frac{1}{x y z} [\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}]$ (D) $\frac{1}{x y z} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Kindly go through the solution

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