NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants 2021

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

Edited By Komal Miglani | Updated on Mar 30, 2025 10:55 PM IST | #CBSE Class 12th

Let’s say you are involved in a complicated project, perhaps designing a bridge, studying the movement of a car, or solving complicated puzzles. Before long, you may find that you have to solve them purposely for the sake of working through various complex variables—how do you do that? The answer is Determinants. This chapter is a good foundation for the students so that they not only have a theoretical understanding of determinants but can apply them to circumstances for engineering, physics and even computer science.

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NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants
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The NCERT Exemplar for Class 12 Determinants provides an easier way to solve systems of linear equations, find area and volume, and determine if a matrix is invertible for example. Students will learn the rules and properties of determinants in this chapter, including cofactor expansion, the determinant of a 3x3 matrix, and solving linear equations using Cramer's Rule. These are important concepts to learn to understand higher-level topics in math and science. Students will need to regularly practice the material to fully understand it. They should be able to apply the properties and procedures for determining determinants to solve problems and equations with ease. If they need further background or explanations, students can use the NCERT Class 12 Maths Solutions.

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

Class 12 Maths Chapter 4 Exemplar Solutions Exercise: 4.3
Page number: 77-85
Total questions: 58

Question:1

Using the properties of determinants in evaluation:
$| \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} |$

Answer:

$| \begin{array}{cc} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{array} |$
If $A = | \begin{array}{ll} a & b \\ c & d \end{array} |$
The value of the determinant of A can then be found by:

$\begin{aligned} | A | = | \begin{array}{ll} a & b \\ c & d \end{array} | = a d - b c \\ = (x^{2} - x + 1) \times (x + 1) - (x + 1) \times (x - 1) \\ = x (x^{2} - x + 1) + 1 (x^{2} - x + 1) - (x^{2} - 1) \\ (Since (a - b) (a + b) = (a^{2} - b^{2})) \\ = x^{3} - x^{2} + x + x^{2} - x + 1 - x^{2} + 1 \\ = x^{3} - x^{2} + 2 \end{aligned}$

Question:2

Using the properties of determinants in evaluation:
$| \begin{array}{ccc} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{array} |$

Answer:

$A = | \begin{array}{ccc} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{array} |$
$C_{1} \to C_{1} + C_{2} + C_{3}$
$\begin{array}{l} = | \begin{array}{ccc} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{array} | \\ = (a + x + y + z) | \begin{array}{ccc} 1 & y & z \\ 1 & a + y & z \\ y & a + z \end{array} | \end{array}$
$R_{2} \to R_{2} - R_{1}$
$R_{3} \to R_{3} - R_{1}$
$\begin{array}{l} = (a + x + y + z) | \begin{array}{ccc} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{array} | \\ = (a + x + y + z) | \begin{array}{ll} a & 0 \\ 0 & a \end{array} | \\ = a^{2} (a + z + x + y) \end{array}$

Question:3

Using the properties of determinants in evaluation:
$| \begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array} |$

Answer:

$Let A = | \begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array} |$
$= x^{2} y^{2} z^{2} | \begin{array}{lll} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{array} |$
$C_{2} \to C_{2} - C_{3}$
$= x^{2} y^{2} z^{2} | \begin{array}{ccc} 0 & 0 & x \\ y & - y & y \\ z & z & 0 \end{array} |$
Expand IAI along C3 to get:
$= x^{2} y^{2} z^{2} (x (y z + y z))$
$= x^{2} y^{2} z^{2} (2 x y z)$
$= 2 x^{3} y^{3} z^{3}$

Question:4

Using the properties of determinants in evaluation:
$| \begin{array}{ccc} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{array} |$

Answer:

$L e t A = | \begin{array}{ccc} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{array} |$

Apply - $C_{1} \to C_{1} + C_{2} + C_{3}$

$\begin{aligned} = | \begin{array}{ccc} 3 x - x + y - x + z & - x + y & - x + z \\ x - y + 3 y + z - y & 3 y & z - y \\ x - z + y - z + 3 z & y - z & 3 z \end{array} | \\ = | \begin{array}{ccc} x + y + z & - x + y & - x + z \\ x + y + z & 3 y & z - y \\ x + y + z & y - z & 3 z \end{array} | \\ Take (x + y + z) common from C_{1} \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 1 & 3 y & z - y \\ 1 & y - z & 3 z \end{array} | \end{aligned}$

$\begin{aligned} Apply R_{2} \to R_{2} - R_{1,} you will get \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 1 - 1 & 3 y - (- x + y) & z - y - (- x + z) \\ 1 & y - z & 3 z \end{array} | \end{aligned}$

$\begin{aligned} = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 0 & 3 y + x - y & z - y + x - z \\ 1 & y - z & 3 z \end{array} | \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 0 & x + 2 y & x - y \\ 1 & y - z & 3 z \end{array} | \\ Now apply, R_{3} \to R_{3} - R_{1} \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 0 & x + 2 y & x - y \\ 1 - 1 & y - z - (- x + y) & 3 z - (- x + z) \end{array} | \end{aligned}$

$\begin{aligned} = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 0 & x + 2 y & x - y \\ 0 & y - z + x - y & 3 z + x - z \end{array} | \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y & - x + z \\ 0 & x + 2 y & x - y \\ 0 & x - z & 2 z + x \end{array} | \\ Apply, C_{2} \to C_{2} - C_{3}, \\ = (x + y + z) | \begin{array}{ccc} 1 & - x + y - (- x + z) & - x + z \\ 0 & x + 2 y - (x - y) & x - y \\ 0 & x - z - (2 z + x) & 2 z + x \end{array} | \end{aligned}$

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

Now, expand the determinant along Column 1

$\begin{aligned} = (x + y + z) [1 \times {(3 y) (2 z + x) - (- 3 z) (x - y)}] \\ = (x + y + z) [6 y z + 3 y x + (3 z) (x - y)] \\ = (x + y + z) [6 y z + 3 y x + 3 z x - 3 z y] \\ = (x + y + z) [3 y z + 3 z x + 3 y x] \\ = 3 (x + y + z) (y z + z x + y x) \end{aligned}$

Question:5

Using the properties of determinants in evaluation:
$| \begin{array}{ccc} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{array} |$

Answer:

$L e t A = | \begin{array}{ccc} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{array} |$
$\begin{aligned} Using C_{1} \to C_{1} + C_{2} + C_{3}, \\ = | \begin{array}{ccc} x + 4 + x + x & x & x \\ x + x + 4 + x & x + 4 & x \\ x + x + x + 4 & x & x + 4 \end{array} | \\ = | \begin{array}{ccc} 3 x + 4 & x & x \\ 3 x + 4 & x + 4 & x \\ 3 x + 4 & x & x + 4 \end{array} | \\ Take (3 x + 4) common form C, \\ = (3 x + 4) | \begin{array}{ccc} 1 & x & x \\ 1 & x + 4 & x \\ 1 & x & x + 4 \end{array} | \\ Apply, R_{3} \to R_{3} - R_{1} \\ = (3 x + 4) | \begin{array}{ccc} 1 & x & x \\ 0 & 4 & 0 \\ 1 - 1 & x - x & x + 4 - x \end{array} | \end{aligned}$
$\begin{aligned} = (3 x + 4) | \begin{array}{lll} 1 & x & x \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array} | \\ Expand along C_{1} \\ = (3 x + 4) [1 \times {(16) - 0}] \\ = (3 x + 4) (16) \\ = 16 (3 x + 4) \end{aligned}$

Question:6

Using the properties of determinants in evaluation:
$| \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$

Answer:

$L e t A = | \begin{array}{ccc} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} |$
$\begin{aligned} Apply R_{1} \to R_{1} + R_{2} + R_{y} \\ = | \begin{array}{ccc} a - b - c + 2 b + 2 c & 2 a + b - c - a + 2 c & 2 a + 2 b + c - a - b \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ = | \begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ Take (a + b + c) common form the first row \\ = (a + b + c) | \begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ Apply C_{2} \to C_{2} - C_{1} \end{aligned}$
$\begin{array}{l} = (a + b + c) | \begin{array}{ccc} 1 & 1 - 1 & 1 \\ 2 b & b - c - a - 2 b & 2 b \\ 2 c & 2 c - 2 c & c - a - b \end{array} | \\ = (a + b + c) | \begin{array}{ccc} 1 & 0 & 1 \\ 2 b & - b - c - a & 2 b \\ 2 c & 0 & c - a - b \end{array} | \end{array}$
Now apply $C_{3} \to C_{3} - C_{1}$
$\begin{array}{l} = (a + b + c) | \begin{array}{ccc} 1 & 0 & 1 - 1 \\ 2 b & - (a + b + c) & 2 b - 2 b \\ 2 c & 0 & c - a - b - 2 c \end{array} | \\ = (a + b + c) | \begin{array}{ccc} 1 & 0 & 0 \\ 2 b & - (a + b + c) & 0 \\ 2 c & 0 & - (a + b + c) \end{array} | \end{array}$
Expand the determinant along Row 1

$\begin{aligned} = (a + b + c) [1 \times {- (a + b + c) \times {- (a + b + c)} - 0} \\ = (a + b + c) [(a + b + c)^{2}] \\ = (a + b + c)^{3} \end{aligned}$

Question:7

Using the properties of determinants to prove that:
$| \begin{array}{lll} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{array} | = 0$

Answer:

Taking LHS, $| \begin{array}{lll} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{array} |$
$\begin{aligned} Multiply and divide R_{1} R_{2}, R_{3} respectively by x, y, z \\ = \frac{1}{x y z} | \begin{array}{lll} x y^{2} z^{2} & x y z & x (y + z) \\ y z^{2} x^{2} & y z x & y (z + x) \\ z x^{2} y^{2} & z x y & z (x + y) \end{array} | \\ = \frac{1}{x y z} | \begin{array}{lll} x y^{2} z^{2} & x y z & x y + x z \\ x^{2} y z^{2} & x y z & y z + x y \\ x^{2} y^{2} z & x y z & x z + y z \end{array} | \\ Now, take xyz common from the first and second Column \\ = \frac{1}{xyz} \times xyz \times xyz | \begin{array}{lll} yz & 1 & xy + xz \\ xz & 1 & yz + xy \\ xy & 1 & xz + yz \end{array} | \\ = xyz | \begin{array}{lll} y z & 1 & x y + x z \\ x z & 1 & y z + x y \\ x y & 1 & x z + y z \end{array} | \end{aligned}$
$\begin{aligned} Apply, C_{3} \to C_{3} + C_{2} \\ = xyz | \begin{array}{lll} y z & 1 & x y + x z + y z \\ x z & 1 & y z + x y + x z \\ x y & 1 & x z + y z + x y \end{array} | \\ Take (xy + yz + xz) common from C_{3} \\ = (xyz) (xy + yz + xz) | \begin{array}{lll} yz & 1 & 1 \\ xz & 1 & 1 \\ xy & 1 & 1 \end{array} | \end{aligned}$

Whenever any of the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0

Hence, $| \begin{array}{lll} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{array} | = 0$

∴ LHS = RHS

Question:8

Using the properties of determinants to prove that:
$| \begin{array}{ccc} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{array} | = 4 x y z$

Answer:

LHS given,

$| \begin{array}{ccc} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{array} |$

$\begin{aligned} = | \begin{array}{ccc} y + z + z + y & z + z + x + x & y + x + x + y \\ z & z + x & x \\ y & x & x + y \end{array} | \\ = | \begin{array}{ccc} 2 z + 2 y & 2 z + 2 x & 2 x + 2 y \\ z & z + x & x \\ y & x & x + y \end{array} | \\ 2 can be taken common from R_{1} \\ = 2 | \begin{array}{ccc} z + y & z + x & x + y \\ z & z + x & x \\ y & x & x + y \end{array} | \end{aligned}$

$\begin{aligned} Apply, R_{1} \to R_{1} - R_{2} \\ = 2 | \begin{array}{ccc} z + y - z & z + x - (z + x) & x + y - x \\ z & z + x & x \\ y & x & x + y \end{array} | \end{aligned}$

$\begin{aligned} = 2 | \begin{array}{ccc} y & 0 & y \\ z & z + x & x \\ y & x & x + y \end{array} | \\ Apply, R_{3} \to R_{3} - R_{1}, \\ = 2 | \begin{array}{ccc} y & 0 & y \\ z & z + x & x \\ y - y & x - 0 & x + y - y \end{array} | \\ = 2 | \begin{array}{ccc} y & 0 & y \\ z & z + x & x \\ 0 & x & x \end{array} | \\ Now, Apply R_{2} \to R_{2} - R_{3}, \\ = 2 | \begin{array}{ccc} y & 0 & y \\ z - 0 & z + x - x & x - x \\ 0 & x & x \end{array} | \\ = 2 | \begin{array}{lll} y & 0 & y \\ z & z & 0 \\ 0 & x & x \end{array} | \end{aligned}$

Take y,z,x common from the R1, R2 and R3 respectively

$\begin{aligned} Expand along Column 1 \\ \begin{array}{r} | A | = a_{11} (- 1)^{1 + 1} | \begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | + a_{21} (- 1)^{2 + 1} | \begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} | \\ + a_{31} (- 1)^{3 + 1} | \begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} | \end{array} \end{aligned}$

$\begin{aligned} = 2 x y z [(1) {(1) - 0} - (1) {0 - 1} + 0}] \\ = 2 x y z [1 + 1] \\ = 4 x y z \\ = RHS \\ ∴ LHS = RHS \end{aligned}$

Hence Proved

Question:9

Using the properties of determinants to prove that:
$| \begin{array}{ccc} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} | = (a - 1)^{3}$

Take LHS

$| \begin{array}{ccc} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} |$
Apply, $R_{1} \to R_{1} - R_{2}$

$= | \begin{array}{ccc} a^{2} + 2 a - (2 a + 1) & 2 a + 1 - (a + 2) & 1 - 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} |$

$= | \begin{array}{ccc} a^{2} + 2 a - 2 a - 1 & 2 a + 1 - a - 2 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} |$

$= | \begin{array}{ccc} a^{2} - 1 & a - 1 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} |$

$= | \begin{array}{ccc} (a - 1) (a + 1) & a - 1 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} | [∵ (a^{2} - b^{2}) = (a - b) (a + b)]$

Take (a-1) common from $R$

$= (a - 1) | \begin{array}{ccc} a + 1 & 1 & 0 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{array} |$

Apply, $R_{2} \to R_{2} - R_{3}$

$= | \begin{array}{ccc} a + 1 & 1 & 0 \\ 2 a + 1 - 3 & a + 2 - 3 & 1 - 1 \\ 3 & 3 & 1 \end{array} |$

$\begin{aligned} = (a - 1) | \begin{array}{ccc} a + 1 & 1 & 0 \\ 2 a - 2 & a - 1 & 0 \\ 3 & 3 & 1 \end{array} | \\ = (a - 1) | \begin{array}{ccc} a + 1 & 1 & 0 \\ 2 (a - 1) & a - 1 & 0 \\ 3 & 3 & 1 \end{array} | \end{aligned}$

Take (a-1) common from $R_{2}$

$= (a - 1)^{2} | \begin{array}{ccc} a + 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array} |$

Expand along $C_{3}$

$\begin{aligned} = (a - 1)^{2} [1 {(a + 1) - 2}] \\ = (a - 1)^{2} [a + 1 - 2] \\ = (a - 1)^{3} \\ = RHS \end{aligned}$
Hence, proved.

Question:10

If A + B + C = 0, then prove that $\begin{array}{ccc} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array} = 0$

Answer:

Let the determinant be:

$D = | \begin{array}{ccc} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array} |$
Expand the determinant:

$D = 1 - \cos^{2} A - \cos^{2} B - \cos^{2} C + 2 \cos A \cos B \cos C$
Use identity:

$\cos^{2} A + \cos^{2} B + \cos^{2} C = 1 + 2 \cos A \cos B \cos C$
So:

$D = 1 - (1 + 2 \cos A \cos B \cos C) + 2 \cos A \cos B \cos C = 0$
Hence, proved.

$D = 0$

Question:11

If the coordinates of the vertices of an equilateral triangle with sides of length ‘a’ are $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ , then ${| \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |}^{2} = \frac{3 a^{4}}{4}$

Answer:

The area of a triangle with the given vertices will be:

$Δ = \frac{1}{2} | \begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array} |$

Given: Length of the sides of the equilateral triangle = a

Thus, the area

$= \frac{\sqrt{3}}{4} a^{2}$

$∴ \frac{1}{2} | \begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array} | = \frac{\sqrt{3}}{4} a^{2}$ $

Square both sides

$\Rightarrow {(\frac{1}{2} | \begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array} |)}^{2} = {(\frac{\sqrt{3}}{4} a^{2})}^{2}$

$\begin{aligned} \Rightarrow {| \begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array} |}^{2} = \frac{3}{4} a^{4} \\ Hence Proved \end{aligned}$

Question:12

Find the value of θ satisfying $| \begin{array}{ccc} 1 & 1 & \sin 3 θ \\ - 4 & 3 & \cos 2 θ \\ 7 & - 7 & - 2 \end{array} | = 0$

Answer:

Given:

$| \begin{array}{ccc} 1 & 1 & \sin 3 θ \\ - 4 & 3 & \cos 2 θ \\ 7 & - 7 & - 2 \end{array} | = 0$
Expand along Row 1

$\begin{aligned} | A | = a_{11} (- 1)^{1 + 1} | \begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | + a_{12} (- 1)^{1 + 2} | \begin{array}{ll} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} | \\ + a_{13} (- 1)^{1 + 3} | \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} | \\ = | \begin{array}{ccc} 3 & \cos 2 θ \\ - 7 & - 2 \end{array} | - 1 | \begin{array}{cc} - 4 & \cos 2 θ \\ 7 & - 2 \end{array} | + \sin 3 θ | \begin{array}{cc} - 4 & 3 \\ 7 & - 7 \end{array} | \end{aligned}$
$\begin{aligned} \Rightarrow (1) {- 6 - {(- 7) \cos 2 θ}} - 1 {8 - 7 \cos 2 θ} + \sin 3 θ {28 - 21} = 0 \\ \Rightarrow - 6 + 7 \cos 2 θ - 8 + 7 \cos 2 θ + 7 \sin 3 θ = 0 \\ \Rightarrow 14 \cos 2 θ + 7 \sin 3 θ - 14 = 0 \\ \Rightarrow 2 \cos 2 θ + \sin 3 θ - 2 = 0 \end{aligned}$
We know,

$\begin{aligned} \cos 2 θ = 1 - 2 \sin^{2} θ \\ \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ \\ \Rightarrow 2 (1 - 2 \sin^{2} θ) + (3 \sin θ - 4 \sin^{3} θ) - 2 = 0 \\ \Rightarrow 2 - 4 \sin^{2} θ + 3 \sin θ - 4 \sin^{3} θ - 2 = 0 \\ \Rightarrow - 2 + 4 \sin^{2} θ - 3 \sin θ + 4 \sin^{3} θ + 2 = 0 \\ \Rightarrow \sin θ (4 \sin θ - 3 + 4 \sin^{2} θ) = 0 \end{aligned}$

$\begin{aligned} \Rightarrow \sin θ (4 \sin^{2} θ - 6 \sin θ + 2 \sin θ - 3) = 0 \\ \Rightarrow \sin θ [2 \sin θ (2 \sin θ - 3) + 1 (2 \sin θ - 3)] = 0 \\ \Rightarrow \sin θ (2 \sin θ + 1) (2 \sin θ - 3) = 0 \\ \Rightarrow \sin θ = 0 or 2 \sin θ + 1 = 0 or 2 \sin θ - 3 = 0 \\ \Rightarrow θ = n π or 2 \sin θ = - 1 or 2 \sin θ = 3 \\ \Rightarrow or \sin θ = - \frac{1}{2} or \sin θ = \frac{3}{2} \\ \Rightarrow θ = n π or θ = m π + (- 1)^{n} (- \frac{π}{6}); m, n \in Z \end{aligned}$
But it is not possible to have $\sin θ = \frac{3}{2}$

Question:13

If $| \begin{array}{ccc} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{array} | = 0$ then find values of x.

Answer:

Given:
$| \begin{array}{ccc} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{array} | = 0$
$\begin{aligned} Apply, C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{array}{ccc} 4 - x + 4 + x + 4 + x & 4 + x & 4 + x \\ 4 + x + 4 - x + 4 + x & 4 - x & 4 + x \\ 4 + x + 4 + x + 4 - x & 4 + x & 4 - x \end{array} | = 0 \\ Take, (12 + x) common from Row 1 \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 4 + x & 4 + x \\ 1 & 4 - x & 4 + x \\ 1 & 4 + x & 4 - x \end{array} | = 0 \\ Apply C_{2} \to C_{2} + C_{3} \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 4 + x + 4 + x & 4 + x \\ 1 & 4 - x + 4 + x & 4 + x \\ 1 & 4 + x + 4 - x & 4 - x \end{array} | = 0 \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 2 (4 + x) & 4 + x \\ 1 & 8 & 4 + x \\ 1 & 8 & 4 - x \end{array} | = 0 \end{aligned}$
$\begin{aligned} Apply, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 2 (4 + x) & 4 + x \\ 1 - 1 & 8 - 8 & 4 + x - (4 - x) \\ 1 & 8 & 4 - x \end{array} | = 0 \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 2 (4 + x) & 4 + x \\ 0 & 0 & 2 x \\ 1 & 8 & 4 - x \end{array} | = 0 \\ Apply R_{3} \to R_{3} - R_{1} \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 2 (4 + x) & 4 + x \\ 0 & 0 & 2 x \\ 1 - 1 & 8 - 8 - 2 x & 4 - x - 4 - x \end{array} | = 0 \\ \Rightarrow (12 + x) | \begin{array}{ccc} 1 & 2 (4 + x) & 4 + x \\ 0 & 0 & 2 x \\ 0 & - 2 x & - 2 x \end{array} | = 0 \end{aligned}$
Expand along Column 1

$\Rightarrow (12 + x) [(1) {0 - (2 x) (- 2 x)}] = 0$
$\Rightarrow (12 + x) (4 x^{2}) = 0$
$\Rightarrow 12 + x = 0 or 4 x^{2} = 0$
$\Rightarrow x = - 12 or x = 0$
Hence, the value of $x = - 12$ and 0

Question:14

If a1, a2, a3, ..., ar are in G.P., then prove that the determinant $| \begin{array}{ccc} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{array} |$ is independent of r.

Answer:

$a_{1}, a_{2}, \dots, a_{r}$ are in G.P
We know that, $a_{r + 1} = A R^{(r + 1) - 1} = A R^{r} \dots (i)$
[ $∵ a_{n} = a r^{n - 1}$ , where $a =$ first term and $r =$ common ratio]
A is the first term of G.P
R is the common ratio of G.P.

$∴ | \begin{array}{ccc} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{array} | = | \begin{array}{ccc} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{array} | \dots [from (i)]$
Taking $A R^{r}, A R^{r + 6}$ and $A R^{r + 10}$ common from $R_{1}, R_{2}$ and $R_{3}$

$= {AR}^{r} \times {AR}^{r + 6} \times {AR}^{r + 10} | \begin{array}{ccc} 1 & {AR}^{4} & {AR}^{8} \\ 1 & {AR}^{4} & {AR}^{8} \\ 1 & {AR}^{6} & {AR}^{10} \end{array} |$
Whenever any of the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0 Rows 1 and 2 are identical.

$∴ | \begin{array}{ccc} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{array} | = 0$
Hence Proved

Question: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.

Answer:

(a + 5, a – 4), (a – 2, a + 3) and (a, a) is given.

We need to prove that they don’t line in a straight line for any value of a
This can be done by proving the points to be vertices of the triangle.
Area of triangle:-
$\begin{array}{l} Δ = \frac{1}{2} | \begin{array}{ccc} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array} | \\ = \frac{1}{2} | \begin{array}{ccc} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{array} | \\ Apply R_{2} \to R_{2} - R_{1} \\ = \frac{1}{2} | \begin{array}{ccc} a + 5 & a - 4 & 1 \\ a - 2 - a - 5 & a + 3 - a + 4 & 1 - 1 \\ a & a & 1 \end{array} | \\ = \frac{1}{2} | \begin{array}{ccc} a + 5 & a - 4 & 1 \\ - 7 & 7 & 0 \\ a & a & 1 \end{array} | \end{array}$
$\begin{aligned} Apply, R_{3} \to R_{3} - R_{1} \\ = \frac{1}{2} | \begin{array}{ccc} a + 5 & a - 4 & 1 \\ - 7 & 7 & 0 \\ a - a - 5 & a - a + 4 & 1 - 1 \end{array} | \\ = \frac{1}{2} | \begin{array}{ccc} a + 5 & a - 4 & 1 \\ - 7 & 7 & 0 \\ - 5 & 4 & 0 \end{array} | \\ Expand along Column 3 \\ = \frac{1}{2} [(1) (- 28 - (7) (- 5))] \\ = \frac{1}{2} (- 28 + 35) = \frac{7}{2} \neq 0 \end{aligned}$
This proves that the given points form a triangle and therefore do not lie on a straight line.

Question:16

Show that the Δ ABC is an isosceles triangle if the determinant

$Δ = [\begin{array}{ccc} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{array}] = 0$
Answer:

$\begin{aligned} Δ = | \begin{array}{ccc} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{array} | = 0 \\ Apply, C_{2} \to C_{2} - C_{1} \\ = | \begin{array}{ccc} 1 & 1 - 1 & 1 \\ 1 + \cos A & 1 + \cos B - 1 - \cos A & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B - \cos^{2} A - \cos A & \cos^{2} C + \cos C \end{array} | = 0 \\ = | \begin{array}{ccc} 1 & 0 & 1 \\ 1 + \cos A & \cos B - \cos A & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B - \cos^{2} A - \cos A & \cos^{2} C + \cos C \end{array} | = 0 \\ = | \begin{array}{ccc} 1 & 0 & 1 \\ 1 + \cos A & \cos B - \cos A & 1 + \cos C \\ \cos^{2} A + \cos A & (\cos B - \cos A) (\cos B + \cos A) + (\cos B - \cos A) & \cos^{2} C + \cos C \end{array} | = 0 \end{aligned}$
$\begin{aligned} \begin{array}{ccc} 1 & 0 & 1 \\ 1 + \cos A & \cos B - \cos A & 1 + \cos C \\ \cos^{2} A + \cos A & (\cos B - \cos A) [(\cos B + \cos A) + 1] & \cos^{2} C + \cos C \end{array} ∣= 0 \\ Take, cosB-cosA common from column 2 \\ \Rightarrow \cos B - \cos A | \begin{array}{ccc} 1 & 0 & 1 \\ 1 + \cos A & 1 & 1 + \cos C \\ \cos^{2} A + \cos A & (\cos B + \cos A) + 1 & \cos^{2} C + \cos C \end{array} | = 0 \\ Apply C_{3} \to C_{3} - C_{1} \\ \cos B - \cos A | \begin{array}{ccc} 1 & 0 & 1 - 1 \\ 1 + \cos A & 1 & 1 + \cos C - 1 - \cos A \\ \cos^{2} A + \cos A & \cos B + \cos A + 1 & \cos^{2} C + \cos C - \cos^{2} A - \cos A \end{array} | = 0 \end{aligned}$
$\begin{aligned} \Rightarrow \cos B - \cos A | \begin{array}{ccc} 1 & 0 & 0 \\ 1 + \cos A & 1 & \cos C - \cos A \\ \cos^{2} A + \cos A & \cos B + \cos A + 1 & \cos^{2} C - \cos^{2} A + \cos C - \cos A \end{array} | = 0 \\ \Rightarrow (\cos B - \cos A) | \begin{array}{ccc} 1 & 0 & 0 \\ 1 + \cos A & 1 & (\cos C - \cos A) \\ \cos^{2} A + \cos A & \cos B + \cos A + 1 & (\cos C - \cos A) (\cos C + \cos A + 1) \end{array} | = 0 \\ Take cosC-cosA common from Column 3 \\ \Rightarrow (\cos B - \cos A) (\cos C - \cos A) | \begin{array}{ccc} 1 & 0 & 0 \\ 1 + \cos A & 1 & 1 \\ \cos^{2} A + \cos A & \cos B + \cos A + 1 & \cos C + \cos A + 1 \end{array} | = 0 \end{aligned}$

Expand along Row 1

$\begin{aligned} \Rightarrow (\cos B - \cos A) (\cos C - \cos A) [(1) {\cos C + \cos A + 1 - (\cos B + \cos A + 1)}] = 0 \\ \Rightarrow (\cos B - \cos A) (\cos C - \cos A) [\cos C + \cos A + 1 - \cos B - \cos A - 1] = 0 \\ \Rightarrow (\cos B - \cos A) (\cos C - \cos A) (\cos C - \cos B) = 0 \\ \Rightarrow \cos B - \cos A = 0 ∖ or \cos C - \cos A = 0 ∖ or \cos C - \cos B = 0 \\ \Rightarrow \cos B = \cos A or \cos C = \cos A or \cos C = \cos B \\ \Rightarrow B = A or C = A or C = B \end{aligned}$
$Hence, △ A B C is an isosceles triangle.$

Question:17

Find $A^{- 1}$ if $A = [\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}]$ and show that $A^{- 1} = \frac{A^{2} - 3 I}{2}$

Answer:

To find $adj A$

$\begin{aligned} a_{11} = | \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array} | = 0 - 1 = - 1 \\ a_{12} = - | \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} | = - (0 - 1) = 1 \\ a_{13} = | \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array} | = 1 - 0 = 1 \\ a_{21} = - | \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} | = - (0 - 1) = 1 \\ a_{22} = | \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array} | = 0 - 1 = - 1 \\ a_{23} = - | \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array} | = - (0 - 1) = 1 \\ a_{31} = | \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array} | = 1 - 0 = 1 \\ a_{32} = - | \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array} | = - (0 - 1) = 1 \\ ∴ adjA = {[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}]}^{T} = {[\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]}^{T} = [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] \end{aligned}$
$∴ A^{- 1} = \frac{adj A}{| A |} = \frac{[\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]}{2} = \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]$
$A^{- 1} = \frac{A^{2} - 31}{2}$

Now, we need to prove that.t

$\begin{aligned} A^{2} = [\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}] \times [\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}] = [\begin{array}{lll} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{array}] \\ A^{2} = [\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}] \\ ∴ \frac{A^{2} - 3 I}{2} = \frac{1}{2} {[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}] - 3 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]} \\ = \frac{1}{2} {[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}] - [\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}]} \\ 2 - 3 \\ = \frac{1}{2} {[\begin{array}{cc} 1 \\ 1 & 2 - 3 \\ 1 & 1 \\ 1 & 1 \end{array}]} \\ = \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]} \\ = A^{- 1} \end{aligned}$
Hence Proved

Question:18

If $A = [\begin{array}{ccc} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{array}]$ find $A^{- 1}$ . Using $A^{- 1}$ , solve the system of linear equations $x - 2 y = 10, 2 x - y - z = 8, - 2 y + z = 7$

Answer:

Find IAI Expand IAI along Column 1

$\begin{aligned} | A | = a_{11} (- 1)^{1 + 1} | \begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | + a_{21} (- 1)^{2 + 1} | \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} | + a_{31} (- 1)^{3 + 1} | \begin{array}{cc} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} | \\ | A | = (1) | \begin{array}{cc} 0 & 0 \\ - 1 & - 2 \\ - 1 & 1 \end{array} | - (- 2) | + 0 | \begin{array}{cc} 2 & 0 \\ - 1 & - 2 \end{array} | \\ = (- 1 + 2) + 2 (0) + 0 \\ = 1 \end{aligned}$
To find adj $A$

$\begin{aligned} a_{11} = | \begin{array}{cc} - 1 & - 2 \\ - 1 & 1 \end{array} | = - 1 - 2 = - 3 \\ a_{12} = - | \begin{array}{cc} - 2 & - 2 \\ 0 & 1 \end{array} | = - (- 2 + 0) = 2 \\ a_{13} = | \begin{array}{cc} - 2 & - 1 \\ 0 & - 1 \end{array} | = 2 + 0 = 2 \\ a_{21} = - | \begin{array}{cc} 2 & 0 \\ - 1 & 1 \end{array} | = - (2 + 0) = - 2 \\ a_{22} = | \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} | = 1 \\ a_{23} = - | \begin{array}{cc} 1 & 2 \\ 0 & - 1 \end{array} | = - (- 1) = 1 \\ a_{31} = | \begin{array}{cc} 2 & 0 \\ - 1 & - 2 \end{array} | = - 4 \\ a_{32} = - | \begin{array}{cc} 1 & 0 \\ - 2 & - 2 \end{array} | = - (- 2) = 2 \\ a_{33} = | \begin{array}{cc} 1 & 2 \\ - 2 & - 1 \end{array} | = - 1 + 4 = 3 \end{aligned}$

$\begin{aligned} ∴ adj A = [\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}] = [\begin{array}{lll} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{array}] = [\begin{array}{ccc} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}] \\ ∴ A^{- 1} = \frac{adj A}{| A |} = \frac{[\begin{array}{ccc} - 3 & - 2 & - 4 \\ 2 & 1 & 3 \end{array}]}{1} = [\begin{array}{ccc} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}] \end{aligned}$
According to the linear equation:

$x - 2 y = 10$
$2 x - y - z = 8$
$- 2 y + z = 7$
We know that, $AX = B$

$A = [\begin{array}{ccc} 1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1 \end{array}]$
Here,
So, transpose of $A^{- 1}$

$\begin{aligned} A^{- 1} = [\begin{array}{lll} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{array}] \\ \Rightarrow X = A^{- 1} B \end{aligned}$
$\begin{aligned} \Rightarrow [\begin{matrix} x \\ y \\ z \\ x \\ y \\ y \end{matrix}] = [\begin{array}{lll} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{array}] [\begin{matrix} 10 \\ 8 \\ - 30 + 16 + 14 \\ - 20 + 8 + 7 \\ - 40 + 16 + 21 \end{matrix}] \\ ∴ x = 0, y = - 5 and z = - 3 \end{aligned}$

Question:19

Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.

Answer:

Given system:

$\begin{array}{r} 3 x + 2 y - 2 z = 3 \\ x + 2 y + 3 z = 6 \\ 2 x - y + z = 2 \end{array}$
Matrix form:

$[\begin{array}{ccc} 3 & 2 & - 2 \\ 1 & 2 & 3 \\ 2 & - 1 & 1 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 3 \\ 6 \\ 2 \end{array}]$
The inverse of the coefficient matrix:

$A^{- 1} = [\begin{array}{ccc} \frac{7}{17} & \frac{8}{17} & - \frac{4}{17} \\ \frac{1}{17} & \frac{5}{17} & \frac{6}{17} \\ \frac{5}{17} & - \frac{2}{17} & \frac{3}{17} \end{array}]$

Multiply:

$X = A^{- 1} B = [\begin{array}{l} 1 \\ 2 \\ 1 \end{array}]$
Final Answer:

$x = 1, y = 2, z = 1$

Question:20

Given $A = [\begin{array}{ccc} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{array}], B = [\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}]$ find BA and use this to solve the system of equations $y + 2 z = 7, x - y = 3, 2 x + 3 y + 4 z = 17.$

Answer:

$\begin{aligned} A = [\begin{array}{ccc} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{array}] and B = [\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}] \\ ∴ B A = [\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}] [\begin{array}{ccc} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{array}] \\ = [\begin{array}{ccc} 2 + 4 & 2 - 2 & - 4 + 4 \\ 4 - 12 + 8 & 4 + 6 - 4 & - 8 - 12 + 20 \\ - 4 + 4 & 2 - 2 & - 4 + 10 \end{array}] \\ = [\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] \\ = 6 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ B A = 6 I \dots (i) \end{aligned}$
Now, the given system of equations is:

$y + 2 z = 7,$
$x - y = 3$
$2 x + 3 y + 4 z = 17$

So,

$[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & - 1 & 0 \\ 2 & 3 & 4 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 7 \\ 3 \\ 17 \end{matrix}]$
Apply, $R_{1} \leftrightarrow R_{2}$

$\begin{aligned} R_{2} \leftrightarrow R_{3} \\ [\begin{array}{lll} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{array}{c} 3 \\ 17 \\ 7 \end{array}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] = {[\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}]}^{- 1} [\begin{array}{c} 3 \\ 17 \\ 7 \end{array}] \end{aligned}$
So, $B A = 6 I [$ from eg(i) $]$

$\begin{aligned} [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 12 \\ - 6 \\ 24 \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{array}{c} 2 \\ - 1 \\ 4 \end{array}] \end{aligned}$

$∴ x = 2, y = - 1$ and $z = 4$

Question:21

If a + b + c ≠ 0 and $| \begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array} | = 0$ then prove that a = b = c.

Answer:

$A = | \begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array} |$
Apply $C_{1} \to C_{1} + C_{2} + C_{3}$

$= | \begin{array}{lll} a + b + c & b & c \\ b + a + c & c & a \\ c + a + b & a & b \end{array} |$
Take $(a + b + c)$ common from Column 1

$= (a + b + c) | \begin{array}{ccc} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array} |$
Expand along Column 1

$\begin{aligned} = (a + b + c) [(1) (b c - a^{2}) - (1) (b^{2} - a c) + (1) (b a - c^{2})] \\ = (a + b + c) [b c - a^{2} - b^{2} + a c + a b - c^{2}] \\ = (a + b + c) [- (a^{2} + b^{2} + c^{2} - a b - b c - a c)] \\ = - \frac{1}{2} (a + b + c) (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 a c) \\ = - \frac{1}{2} (a + b + c) [(a^{2} + b^{2} - 2 a b) + (b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 a c)] \\ = - \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \\ [∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b] \end{aligned}$

Given that $Δ = 0$

$\Rightarrow - \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0$

$\Rightarrow (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0$

$Either (a + b + c) = 0 or (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$

$but it is given that (a + b + c) \neq 0 ∴ (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$

$\Rightarrow a - b = b - c = c - a = 0 \Rightarrow a = b = c$

Hence Proved

Question:22

Prove that $| \begin{array}{ccc} bc - a^{2} & ca - b^{2} & ab - c^{2} \\ ca - b^{2} & ab - c^{2} & bc - a^{2} \\ ab - c^{2} & bc - a^{2} & ca - b^{2} \end{array} |$ is divisible by a + b + c and find the quotient.

Answer:

$| \begin{array}{lll} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{array} |$ is given.
Apply, $R_{1} \to R_{1} - R_{2}$ ,

$\begin{aligned} = | \begin{array}{ccc} b c - a^{2} - ca + b^{2} & ca - b^{2} - ab + c^{2} & ab - c^{2} - bc + a^{2} \\ ca - b^{2} & ab - c^{2} & bc - a^{2} \\ ab - c^{2} & bc - a^{2} & ca - b^{2} \end{array} | \\ = | \begin{array}{ccc} (b c - c a) + (b^{2} - a^{2}) & (c a - a b) + (c^{2} - b^{2}) & (a b - b c) + (a^{2} - c^{2}) \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \\ c (b - a) + (b - a) (b + a) & a (c - b) + (c - b) (c + b) & b (a - c) + (a - c) (a + c) \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \\ = | \begin{array}{ccc} (b - a) (c + b + a) & (c - b) (a + c + b) & (a - c) (b + a + c) \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{array} | \end{array} | \end{aligned}$
Take (a+b+c) common from Column 1

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

Apply $R_{2} \to R_{2} - R_{3}$

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

Take (a+b+c) common from Column 2

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

Expand along Column 1

$\begin{aligned} = (a + b + c)^{2} {[a b + b c + c a - (a^{2} + b^{2} + c^{2})]}^{2} \\ = (a + b + c) (a + b + c) {[a b + b c + c a - (a^{2} + b^{2} + c^{2})]}^{2} \end{aligned}$
The determinant is divisible by $(a + b + c)$ and the quotient is $(a + b + c) {[a b + b c + c a - (a^{2} + b^{2} + c^{2})]}^{2}$

Question:23

If x + y + z = 0, prove that $| \begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array} | = x y z | \begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array} |$

Answer:

Given LHS,

$| \begin{array}{ccc} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array} |$
Expand along Row 1

$\begin{aligned} = x a {(z a) (y a) - (x c) (x b)} - (y b) {(y c) (y a) - (z b) (x b)} + (z c) {(y c) (x c) - \\ (z b) (z a)} \\ = x a {a^{2} y z - x^{2} b c} - y b {y^{2} a c - b^{2} x z} + z c {c^{2} x y - z^{2} a b} \\ = a^{3} x y z - x^{3} a b c - y^{3} a b c + b^{3} x y z + c^{3} x y z - z^{3} a b c \\ = x y z (a^{3} + b^{3} + c^{3}) - a b c (x^{3} + y^{3} + z^{3}) \end{aligned}$
Given $x + y + z = 0$

$\begin{aligned} \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z = x y z (a^{3} + b^{3} + c^{3}) - a b c (3 x y z) = x y z (a^{3} + b^{3} + c^{3} - 3 a b c) \\ = x y z | \begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array} | \end{aligned}$

Question:24

If $| \begin{array}{cc} 2 x & 5 \\ 8 & x \end{array} | = | \begin{array}{cc} 6 & - 2 \\ 7 & 3 \end{array} |$ then value of x is
A. 3
B. ± 3
C. ± 6
D. 6

Answer:

Given:

$\begin{aligned} | \begin{array}{cc} 2 x & 5 \\ 8 & x \end{array} | = | \begin{array}{cc} 6 & - 2 \\ 7 & 3 \end{array} | \\ A = | \begin{array}{ll} a & b \\ c & d \end{array} | \end{aligned}$
Then the determinant of A is

$\begin{aligned} | A | = | \begin{array}{ll} a & b \\ c & d \end{array} | = ad - bc \\ \Rightarrow (2 x) (x) - (5) (8) = (6) (3) - (7) (- 2) \Rightarrow 2 x^{2} - 40 = 18 - (- 14) \Rightarrow 2 x^{2} - 40 = 18 + 14 \Rightarrow x^{2} - 20 = 9 + 7 \Rightarrow x^{2} - 20 = 16 \Rightarrow x^{2} = 16 + 20 \Rightarrow x^{2} = 36 \Rightarrow x = \sqrt{36} \Rightarrow x = \pm 6 \end{aligned}$

Question:25

The value of determinant $| \begin{array}{lll} a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c \end{array} |$
A. $a^{3} + b^{3} + c^{3}$
B. 3 bc
C. $a^{3} + b^{3} + c^{3} - 3 a b c$
D. none of these

Answer:

Given:

$| \begin{array}{ccc} a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c \end{array} |$

Apply C2 $\to C 2 + C 3$

$= | \begin{array}{lll} a - b & a + b + c & a \\ b - c & c + a + b & b \\ c - a & a + b + c & c \end{array} |$
Take $(a + b + c)$ common from Column 2

$= (a + b + c) | \begin{array}{lll} a - b & 1 & a \\ b - c & 1 & b \\ c - a & 1 & c \end{array} |$
Apply $C_{1} \to C_{1} - C_{3}$

$\begin{aligned} = (a + b + c) | \begin{array}{ccc} a - b - a & 1 & a \\ b - c - b & 1 & b \\ c - a - c & 1 & c \end{array} | \\ = (a + b + c) | \begin{array}{lll} - b & 1 & a \\ - c & 1 & b \\ - a & 1 & c \end{array} | \end{aligned}$
Expand along Row 1

$\begin{aligned} = (a + b + c) [(- b) {c - b} - (1) {- c^{2} - (- a b)} + a {- c - (- a)}] \\ = (a + b + c) (- b c + b^{2} + c^{2} - a b - a c + a^{2}) \end{aligned}$
$\begin{aligned} = a (- b c + b^{2} + c^{2} - a b - a c + a^{2}) + b (- b c + b^{2} + c^{2} - a b - a c + a^{2}) + c (- b c + \\ b^{2} + c^{2} - a b - a c + a^{2}) \\ = - a b c + a b^{2} + a c^{2} - a^{2} b - a^{2} c + a^{3} - b^{2} c + b^{3} + b c^{2} - a b^{2} - a b c + a^{2} b - b c^{2} + \\ b^{2} c + c^{3} - a b c - a c^{2} + a^{2} c \\ = a^{3} + b^{3} + c^{3} - 3 a b c \end{aligned}$

Question:26

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
A. 9
B. 3
C. – 9
D. 6

Answer:

B)
$\begin{aligned} The area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) is given by \\ Δ = \frac{1}{2} | \begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | \\ Δ = \frac{1}{2} | \begin{array}{ccc} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array} | = 9 \\ Δ = | \begin{array}{ccc} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array} | = 18 \end{aligned}$
Expand along Column 2

$\Rightarrow - (k) {- 3 - 3} = 18$
$\Rightarrow - k (- 6) = 18$
$\Rightarrow 6 k = 18$
$\Rightarrow k = 3$

Question:27

The determinant $| \begin{array}{ccc} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{array} |$ equals

A. abc (b–c) (c – a) (a – b)
B. (b–c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. None of these

Answer:

D)
Given:

$| \begin{array}{ccc} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{array} |$
$= | \begin{array}{lll} b (b - a) & b - c & c (b - a) \\ a (b - a) & a - b & b (b - a) \\ c (b - a) & c - a & a (b - a) \end{array} |$

Take (b-a) common from both Columns 1 and 3

$= (b - a) (b - a) | \begin{array}{lll} b & b - c & c \\ a & a - b & b \\ c & c - a & a \end{array} |$

Apply

$C_{1} \to C_{1} - C_{3}$ $= (b - a) (b - a) | \begin{array}{lll} b - c & b - c & c \\ a - b & a - b & b \\ c - a & c - a & a \end{array} |$
Whenever any two columns or rows in any determinant are equal, its value becomes = 0
Here Columns 1 and 2 are identical $∴ | \begin{array}{lll} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{array} | = 0$

Question:28

The number of distinct real roots of $| \begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array} | = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

A. 0
B. –1
C. 1
D. None of these

Answer:

C)
Given
$| \begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array} | = 0$
$\begin{aligned} Apply C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{array}{lll} \sin x + \cos x + \cos x & \cos x & \cos x \\ \cos x + \sin x + \cos x & \sin x & \cos x \\ \cos x + \cos x + \sin x & \cos x & \sin x \end{array} | = 0 \\ \Rightarrow | \begin{array}{ccc} 2 \cos x + \sin x & \cos x & \cos x \\ 2 \cos x + \sin x & \sin x & \cos x \\ 2 \cos x + \sin x & \cos x & \sin x \end{array} | = 0 \\ Take (2 \cos x + \sin x) common from Column 1 \\ \Rightarrow 2 \cos x + \sin x | \begin{array}{ccc} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{array} | = 0 \\ Apply R_{2} \to R_{2} - R_{1} \\ \Rightarrow 2 \cos x + \sin x | \begin{array}{ccc} 1 & \cos x & \cos x \\ 1 - 1 & \sin x - \cos x & \cos x - \cos x \\ 1 & \cos x & \sin x \end{array} | = 0 \end{aligned}$
$\Rightarrow 2 \cos x + \sin x | \begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x - \cos x & 0 \\ 1 & \cos x & \sin x \end{array} | = 0 Apply R_{3} \to R_{3} - R_{1} \Rightarrow 2 \cos x + \sin x | \begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x - \cos x & 0 \\ 1 - 1 & \cos x - \cos x & \sin x - \cos x \end{array} | = 0$

$\Rightarrow 2 \cos x + \sin x | \begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x - \cos x & 0 \\ 0 & 0 & \sin x - \cos x \end{array} | = 0$
Expand along Column 1

$\begin{aligned} (2 \cos X + \sin X) [(1) {(\sin X - \cos X) (\sin X - \cos X)}] \\ \Rightarrow (2 \cos X + \sin X) (\sin X - \cos X)^{2} = 0 \\ \Rightarrow 2 \cos X = - \sin X or (\sin X - \cos X)^{2} = 0 \\ \Rightarrow 2 = - \frac{\sin x}{\cos x} or \sin x = \cos x \\ \Rightarrow \tan x = - 2 ortan x = 1 [∵ \tan x = \frac{\sin x}{\cos x}] \\ buttan x = - 2 isnotpossibleasfor - \frac{π}{4} \leq x \leq \frac{π}{4} \end{aligned}$
$So, \tan x = 1$
$∴ x = \frac{π}{4}$
Only one real and distinct root occurs.

Question:29

If A, B and C are angles of a triangle, then the determinant $\begin{array}{ccc} - 1 & \cos C & \cos B \\ \cos C & - 1 & \cos A \\ \cos B & \cos A & - 1 \end{array} ∣$ is equal to
A. 0

B. –1
C. 1
D. None of these

Answer:

Given:

$| \begin{array}{ccc} - 1 & \cos C & \cos B \\ \cos C & - 1 & \cos A \\ \cos B & \cos A & - 1 \end{array} | |$
Expand along Column 1

$\begin{aligned} | A | = a_{11} (- 1)^{1 + 1} | \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | + a_{21} (- 1)^{2 + 1} | \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} | + a_{31} (- 1)^{3 + 1} | \begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} | \\ Δ = (- 1) | \begin{array}{cc} - 1 & \cos A \\ \cos A & - 1 \end{array} | - \cos C | \begin{array}{cc} \cos C & \cos B \\ \cos A & - 1 \end{array} | + \cos B | \begin{array}{cc} \cos C & \cos B \\ - 1 & \cos A \end{array} | \\ = [(- 1) {1 - \cos^{2} A} - \cos C {- \cos C - \cos A \cos B} + \cos B {\cos A \cos C + \\ \cos B}] \end{aligned}$

$= - 1 + \cos^{2} A + \cos^{2} C + \cos A \cos B \cos C + \cos A \cos B \cos C + \cos^{2} B$
$= - 1 + \cos^{2} A + \cos^{2} B + \cos^{2} C + 2 \cos A \cos B \cos C$
Using the formula

$\begin{aligned} 1 + \cos 2 A = 2 \cos^{2} A \\ = - 1 + \frac{1 + \cos 2 A}{2} + \frac{1 + \cos 2 B}{2} + \frac{1 + \cos 2 C}{2} + 2 \cos A \cos B \cos C \end{aligned}$
Taking L.C.M, we get

$= \frac{- 2 + 1 + \cos 2 A + 1 + \cos 2 B + 1 + \cos 2 C + 4 \cos A \cos B \cos C}{2}$

$\begin{aligned} = \frac{1 + (\cos 2 A + \cos 2 B) + \cos 2 C + 4 \cos C \cos A \cos B}{2} \\ Now use: \cos (A + B) \cos (A - B) = 2 \cos A \cos B \\ so, \cos 2 A + \cos 2 B = 2 \cos (A + B) \cos (A - B) \\ = \frac{1 + \cos 2 C + {2 \cos (A + B) \cos (A - B)} + 4 \cos A \cos B \cos C}{2} \\ = \frac{1 + 2 \cos^{2} C - 1 + {2 \cos (A + B) \cos (A - B)} + 4 \cos A \cos B \cos C}{2} \\ = \frac{2 \cos^{2} C + [2 \cos (A + B) \cos (A - B)} + 4 \cos A \cos B \cos C}{2} \dots (i) \end{aligned}$

We know that $A, B, C$ are angles of triangle

$\begin{aligned} \Rightarrow A + B + C = π \\ \Rightarrow A + B = π - C \\ = \frac{2 \cos^{2} C + {2 \cos (π - C) \cos (A - B)} + 4 \cos A \cos B \cos C}{2} \\ = \frac{2 \cos^{2} C + {- 2 \cos C \cos (A - B)} + 4 \cos A \cos B \cos C}{2} [∵ \cos (π - x) = - \cos x] \\ = \frac{- 2 \cos C {\cos (A - B) - \cos C} + 4 \cos A \cos B \cos C}{2} \end{aligned}$
$= - \cos C {\cos (A - B) - \cos C} + 2 \cos A \cos B \cos C = - \cos C [\cos (A - B) - \cos {π - (A + B)}] + 2 \cos A \cos B \cos C = - \cos C [\cos (A - B) + \cos (A + B)] + 2 \cos A \cos B \cos C = - \cos C [2 \cos A \cos B] + 2 \cos A \cos B \cos C = 0$

Question:30

Let $f (t) = | \begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{array} |$ then $lim_{t \to 0} \frac{f (t)}{t^{2}}$ is equal to
A. 0
B. –1
C. 2
D. 3

Answer:

Given:

$f (t) = | \begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{array} |$
Divide $R_{2}$ and $R_{3}$ by $t$

$f (t) = t^{2} | \begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{t} & \frac{t}{t} & \frac{2 t}{t} \\ \frac{\sin t}{t} & \frac{t}{t} & \frac{t}{t} \end{array} |$
$\Rightarrow \frac{f (t)}{t^{2}} = \frac{t^{2}}{t^{2}} | \begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{t} & 1 & 2 \\ \frac{\sin t}{t} & 1 & 1 \end{array} |$
$\Rightarrow lim_{t \to 0} \frac{f (t)}{t^{2}} = | \begin{array}{lll} lim_{t \to 0} \cos t & lim_{t \to 0} t & lim_{t \to 0} 1 \\ lim_{t \to 0} \frac{2 \sin t}{t} & lim_{t \to 0} 1 & lim_{t \to 0} 2 \\ lim_{t \to 0} \frac{\sin^{t}}{t} & lim_{t \to 0} 1 & lim_{t \to 0} 1 \end{array} |$
$= | \begin{array}{lll} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{array} | (∵ lim_{t \to 0} \frac{\sin t}{t} = 1)$

Expand along Row 1

$\begin{aligned} = (1) (1 - 2) + (1) (2 - 1) \\ = - 1 + 1 \end{aligned}$
$= 0$

Question:31

The maximum value of $Δ = {| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin θ & 1 \\ 1 + \cos θ & 1 & 1 \end{array} |}_{is (θ is a real number)}$

A. $\frac{1}{2}$
B. $\frac{\sqrt{3}}{2}$
C. $\sqrt{2}$
D. $\frac{2 \sqrt{3}}{4}$

Answer:

We have:

$Δ = | \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin θ & 1 \\ 1 + \cos θ & 1 & 1 \end{array} |$
Apply, $C_{2} \to C_{2} - C_{3}$ and $C_{2} \to C_{2} - C_{3}$

$\Rightarrow Δ = | \begin{array}{ccc} 0 & 0 & 1 \\ 0 & \sin θ & 1 \\ \cos θ & 0 & 1 \end{array} |$

$= 0 - 0 + 1 (\sin θ \cdot \cos θ)$ Multiply and divide by $2, = \frac{1}{2} (2 \sin θ \cos θ)$ We already know that $\underset{―}{2 \sin θ \cos θ = \sin 2 θ} = \frac{1}{2} (\sin 2 θ)$
The maximum value of $: \sin 2 θ$ is $1, θ = 45^{\circ}$ .

$\begin{aligned} ∴ Δ = \frac{1}{2} (\sin 2 (45^{\circ})) = \frac{1}{2} \sin 90^{\circ} = \frac{1}{2} (1) \\ ∴ Δ = \frac{1}{2} \end{aligned}$

Question:32

If $f (x) = | \begin{array}{ccc} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{array} |$

A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0

Answer:

We have:

$f (x) = | \begin{array}{ccc} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{array} |$

If we put $x = a$

$\begin{aligned} f (a) = | \begin{array}{ccc} 0 & a - a & a - b \\ a + a & 0 & a - c \\ a + b & a + c & 0 \end{array} | \\ = 0 | \begin{array}{cc} 0 & a - c \\ a + c & 0 \end{array} | - 0 | \begin{array}{cc} 2 a & a - c \\ a + b & 0 \end{array} | + (a - b) | \begin{array}{cc} 2 a & 0 \\ a + b & a + c \end{array} | \\ = 0 - 0 + (a - b) [2 a (a + c) - 0 (a + b)] = (a - b) [2 a^{2} + 2 a c - 0] = (a - b) (2 a^{2} + 2 a c) \neq 0 If x = b \\ f (b) = | \begin{array}{ccc} 0 & b - a & b - b \\ b + a & 0 & b - c \\ b + b & b + c & 0 \end{array} | \\ = 0 | \begin{array}{cc} 0 & b - c \\ b + c & 0 \end{array} | - (b - a) | \begin{array}{cc} b + a & b - c \\ 2 b & 0 \end{array} | + 0 | \begin{array}{cc} b + a & 0 \\ 2 b & b + c \end{array} | \\ = 0 - (b - a) [(b + a) (0) - (b - c) (2 b)] + 0 = - (b - a) [0 - 2 b^{2} + 2 b c] = (a - b) (2 b^{2} - 2 b c) \neq 0 \end{aligned}$
If $x = 0$ according to the given question:

$f (0) = | \begin{array}{ccc} 0 & 0 - a & 0 - b \\ 0 + a & 0 & 0 - c \\ 0 + b & 0 + c & 0 \end{array} |$

$\begin{aligned} = 0 | \begin{array}{cc} 0 & - c \\ c & 0 \end{array} | - (- a) | \begin{array}{cc} a & - c \\ b & 0 \end{array} | + (- b) | \begin{array}{ll} a & 0 \\ b & c \end{array} | \\ = 0 + a [a (0) - (- b c)] - b [a c - b (0)] \end{aligned}$
$= a [b c] - b [a c]$
$= a b c - a b c = 0$
Then the condition is satisfied.

Question:33

If $A = [\begin{array}{ccc} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}]$

then $A^{- 1}$ exists if
A. λ = 2
B. λ ≠ 2
C. λ ≠ -2
D. None of these

Answer:

D)
We have
$A = [\begin{array}{ccc} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}]$

$\begin{aligned} \Rightarrow | A | = 2 (6 - 5) - λ (0 - 5) + (- 3) (0 - 2) \\ = 2 + 5 λ + 6 \\ = 5 λ + 8 \end{aligned}$

ithe inverse of A exists only if A is non-singular.e. $| A | \neq 0$ .

$\begin{aligned} .5 λ + 8 \neq 0 \\ \Rightarrow 5 λ \neq - 8 \end{aligned}$

$∴ λ \neq - \frac{8}{5}$
So, $A^{- 1}$ exists if and only if $λ \neq - \frac{8}{5}$

Question:34

If A and B are invertible matrices, then which of the following is not correct?
A. $a d j A = | A | . A^{- 1}$
B. $d e t (A)^{- 1} = [d e t (A)]^{- 1}$

C. $(A B)^{- 1} = B^{- 1} A^{- 1}$
D. $(A + B)^{- 1} = B^{- 1} + A^{- 1}$

Answer:

We know, that A and B are invertible matrices

$\begin{aligned} Consider (A B) B^{- 1} A^{- 1} \Rightarrow (A B) B^{- 1} A^{- 1} = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = (A I) A^{- 1} = A A^{- 1} = I \Rightarrow (A B)^{- 1} = B^{- 1} A^{- 1} \dots option (C) \\ Also A A^{- 1} = I \Rightarrow | A A^{- 1} | = | I | \Rightarrow | A | | A^{- 1} | = 1 \\ \Rightarrow | A |^{- 1} = \frac{1}{| A |} \\ ∴ \det (A)^{- 1} = [\det (A)]^{- 1} \dots (B) \end{aligned}$
We know that $\frac{| A |^{- 1}}{adj A} \frac{| \vec{A} |}{We}$
$\Rightarrow adj A = | A | \cdot A^{- 1} \dots$ option $(A)$

$\Rightarrow (A + B)^{- 1} = \frac{1}{| A + B |} adj (A + B)$

But $B^{- 1} + A^{- 1} = \frac{1}{| B |} adj B + \frac{1}{| A |}$ adj A

$∴ (A + B)^{- 1} \neq B^{- 1} + A^{- 1}$

Question:35

If x, y, z are all different from zero and , $| \begin{array}{ccc} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{array} | = 0$ , then value of $x^{- 1} + y^{- 1} + z^{- 1}$ is
$A . x y z B . x^{- 1} y^{- 1} z^{- 1} C . - x - y - z D . - 1$

Answer:

We have
$| \begin{array}{ccc} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{array} | = 0$
$Apply C_{1} \to C_{1} - C_{3} and C_{2} \to C_{2} - C_{3}$
$| \begin{array}{ccc} x & 0 & 1 \\ 0 & y & 1 \\ - z & - z & 1 + z \end{array} | = 0$
Expand along Row 1

$\Rightarrow x [y (1 + z) + z] - 0 + 1 (y z) = 0$

$x y + x y z + x z + y z = 0$

Divide both sides by XYZ
$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0 ∴ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = x^{- 1} + y^{- 1} + z^{- 1} = - 1$

Question:36

The value of the determinant $| \begin{array}{ccc} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{array} |$ is
A. $9 x^{2} (x + y)$
B. $9 y^{2} (x + y)$
C. $3 y^{2} (x + y)$
D. $7 x^{2} (x + y)$

Answer:

Matrix given:

$\begin{aligned} | \begin{array}{ccc} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{array} | \\ = x | \begin{array}{cc} x & x + y \\ x + 2 y & x \end{array} | - (x + y) | \begin{array}{cc} x + 2 y & x + y \\ x + y & x \end{array} | + (x + 2 y) | \begin{array}{cc} x + 2 y & x \\ x + y & x + 2 y \end{array} | \\ = x [x^{2} - (x + y) (x + 2 y)] - (x + y) [(x + 2 y) (x) - (x + y)^{2}] + (x + 2 y) [(x + \\ 2 y)^{2} - x (x + y)] \\ = x [x^{2} - x^{2} - 3 x y - 2 y^{2}] - (x + y) [x^{2} + 2 x y - x^{2} - 2 x y - y^{2}] + (x + 2 y) [x^{2} + \\ 4 x y + 4 y^{2} - x^{2} - x y] \\ = x [- 3 x y - 2 y^{2}] - (x + y) [- y^{2}] + (x + 2 y) [3 x y + 4 y^{2}] \\ = - 3 x^{2} y - 2 x y^{2} + x y^{2} + y^{3} + 3 x^{2} y + 4 x y^{2} + 6 x y^{2} + 8 y^{3} \\ = 9 y^{3} + 9 x y^{2} \\ = 9 y^{2} (x + y) \end{aligned}$

Question:37

There are two values of a which makes determinant, $Δ = | \begin{array}{ccc} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{array} | = 86$ ,then sum of these number is
A. 4
B. 5
C. -4
D. 9

Answer:

We have:

$\begin{aligned} Δ = | \begin{array}{ccc} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{array} | = 86 \\ 1 | \begin{array}{cc} a & - 1 \\ 4 & 2 a \end{array} | - (- 2) | \begin{array}{cc} 2 & - 1 \\ 0 & 2 a \end{array} | + 5 | \begin{array}{cc} 2 & a \\ 0 & 4 \end{array} | = 86 \\ 1 [2 a^{2} - (- 4)] + 2 [4 a - 0] + 5 [8 - 0] = 86 \end{aligned}$

$1 [2 a^{2} + 4] + 2 [4 a] + 5 [8] = 86$

$2 a^{2} + 4 + 8 a + 40 = 86$

$2 a^{2} + 8 a + 44 = 86$

$2 a^{2} + 8 a = 42$

$2 (a^{2} + 4 a) = 42$

$(a^{2} + 4 a) = 21 \Rightarrow a^{2} + 4 a - 21 = 0 \Rightarrow (a + 7) (a - 3) = 0 ∴ a = - 7 or 3$

The sum of -7 and $3 = - 4$

Question:38

Fill in the blanks
If A is a matrix of order 3 × 3, then |3A| = ___.

Answer:

If A is a matrix of order $3 \times 3$ , then $| 3 A | = 27 | A |$ .
We know:

$\begin{aligned} if A = {[a_{i j}]}_{3 \times 3}, then | k \cdot A | = k^{3} | A | \\ ∴ | 3 A | = 3^{3} | A | = 27 | A | \end{aligned}$

Question:39

Fill in the blanks
If A is an invertible matrix of order 3 × 3, then | $A^{- 1}$ |= ____.

Answer:

If A is an invertible matrix of order $3 \times 3$ , then $| A^{- 1} | = | A |^{- 1}$
Given
$A =$ invertible matrix $3 \times 3$

$\begin{aligned} A A^{- 1} = I \Rightarrow | A | | A^{- 1} | = 1 \\ ∴ | A^{- 1} | = \frac{1}{| A |} \end{aligned}$

Question:40

Fill in the blanks
If x, y, z ∈ R, then the value of determinant $| \begin{array}{lll} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} |$ is equal to ___.

Answer:

$\begin{aligned} Given \\ | \begin{array}{lll} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} | \\ Apply C_{1} \to C_{1} - C_{2} \\ = | \begin{array}{lll} {(2^{x} + 2^{- x})}^{2} - {(2^{x} - 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} - {(3^{x} - 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} - {(4^{x} - 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} | \\ Apply Formula: (a + b)^{2} - (a - b)^{2} = 4 a b . \end{aligned}$
$\begin{aligned} = | \begin{array}{lll} 4 (2^{x}) (2^{- x}) & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 (3^{x}) (3^{- x}) & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 (4^{x}) (4^{- x}) & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} | \\ = | \begin{array}{lll} 4 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} | \\ Column 1 and 3 thus become proportional \\ ∴ | \begin{array}{lll} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{array} | = 0 \end{aligned}$

Question:41

Fill in the blanks

If cos 2θ = 0, then ${| \begin{array}{ccc} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{array} |}^{2} =$

Answer:

We know

$\cos 2 θ = 0 \Rightarrow \cos 2 θ = \cos π / 2 \Rightarrow 2 θ = π / 2 ∴ θ = π / 4$

$\begin{aligned} ∴ \sin \frac{π}{4} = \frac{1}{\sqrt{2}} and \cos \frac{π}{4} = \frac{1}{\sqrt{2}} \\ Then \\ {| \begin{array}{ccc} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{array} |}^{2} = {| \begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array} |}^{2} \\ \Rightarrow {[0 - \frac{1}{\sqrt{2}} (\frac{1}{2}) + \frac{1}{\sqrt{2}} (- \frac{1}{2})]}^{2} = {[\frac{- 2}{2 \sqrt{2}}]}^{2} = {[- \frac{1}{\sqrt{2}}]}^{2} \\ ∴ {| \begin{array}{ccc} 0 & \cos θ & \sin θ \\ \cos θ & \sin θ & 0 \\ \sin θ & 0 & \cos θ \end{array} |}^{2} = \frac{1}{2} \end{aligned}$

Question:42

Fill in the blanks
If A is a matrix of order 3 × 3, then $(A^{2})^{- 1}$ = ____.

Answer:

For matrix A is of order 3X3
$\begin{aligned} {(A^{2})}^{- 1} & = (A \cdot A)^{- 1} \\ = A^{- 1} \cdot A^{- 1} \\ = {(A^{- 1})}^{2} \end{aligned}$

Question:43

If A is a matrix of order 3 × 3, then the number of minors in the determinant of A is

Answer:

If matrix A is of order 3X3 then
Number of Minors of IAI = 9 as there are 9 elements in a 3x3 matrix

Question:44

Fill in the blanks
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ___.

Answer:

If, $A = | \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} |$
then $| A | = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}$

We know that the determinant is equal to the sum of corresponding co-factors of any row or column.

Question:45

Fill in the blanks
If x = -9 is a root of $| \begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} | = 0$ , then other two roots are ___.

Answer:

We know that

$\begin{aligned} x = - 9 is a root of | \begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array} | = 0 \\ \Rightarrow x | \begin{array}{ll} x & 2 \\ 6 & x \end{array} | - 3 | \begin{array}{ll} 2 & 2 \\ 7 & x \end{array} | + 7 | \begin{array}{ll} 2 & x \\ 7 & 6 \end{array} | = 0 \end{aligned}$
$\Rightarrow x [x^{2} - 12] - 3 [2 x - 14] + 7 [12 - 7 x] = 0 \Rightarrow x^{3} - 12 x - 6 x + 42 + 84 - 49 x = 0 \Rightarrow x^{3} - 67 x + 126 = 0 (x + 9) (2 - x) (7 - x) = 0 H$ ere, $126 \times 1 = 9 \times 2 \times 7$ Forx $= 2, \Rightarrow 2^{3} - 67 \cdot 2 + 126 = 134 - 134 = 0 ∴ x = 2$ is one root. For $= 7, \Rightarrow 7^{3} - 677 - 7 + 126 = 469 - 469 = 0; x = 7$ will be another root.

Question:46

Fill in the blanks

$| \begin{array}{ccc} 0 & x y z & x - z \\ y - x & 0 & y - z \\ z - x & z - y & 0 \end{array} | =$

Answer:

$\begin{aligned} = (z - x) [1 [0 - (y - z) (z - y)] - (x y z) [0 - (y - z)] + (x - z) [(z - y) - 0]] \\ = (z - x) (z - y) (- y + z - x y z + x - z) \\ = (z - x) (z - y) (x - y - x y z) \\ = (z - x) (y - z) (y - x + x y z) \end{aligned}$

Question:47

If $f (x) = | \begin{array}{lll} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{array} | = A + B x + C x^{2} + \dots$ , then A = ____.

Answer:

Given $f (x) = | \begin{array}{ccc} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{array} | \Rightarrow f (x) = (1 + x)^{17} (1 + x)^{23} (1 + x)^{41} | \begin{array}{ccc} 1 & (1 + x)^{2} & (1 + x)^{6} \\ 1 & (1 + x)^{6} & (1 + x)^{11} \\ 1 & (1 + x)^{2} & (1 + x)^{6} \end{array} |$ Wecanseethatrow1 androw3areidentical $∴ f (x) = (1 + x)^{17} (1 + x)^{23}$

$∴ A = 0$

Question:48

State True or False for the statements
$(A^{3})^{- 1} = (A^{- 1})^{3}$ , where A is a square matrix and |A| ≠ 0.

Answer:

${(A^{3})}^{- 1} = {(A^{- 1})}^{3}$ Because, ${(A^{n})}^{- 1} = {(A^{- 1})}^{n}$ , wheren $\in N$ .

Question:49

State True or False for the statements
$(a A)^{- 1} = (1 / a) A^{- 1}$ , where a is any real number and A is a square matrix.

Answer:

For a non-singular matrix, aA is invertible such that
$(aA) (\frac{1}{a} A^{- 1}) = (a \cdot \frac{1}{a}) ({AA}^{- 1}) i.e. (aA)^{- 1} = \frac{1}{a} A^{- 1}$
here a = any non-zero scalar. Here A should be a non-singular matrix which is not given in the statement, thus the statement given in the question is false.

Question:50

State True or False for the statements
$| A^{- 1} | \neq | A |^{- 1}$ , where A is non-singular matrix.

Answer:

We know A is a non-singular Matrix

In that case: $A A^{- 1} = I$ .

$\Rightarrow | A | | A^{- 1} | = 1 ∴ | A^{- 1} | = 1 / | A | = | A |^{- 1}$
Thus the statement is false.

Question:51

State True or False for the statements
If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.

Answer:

We know that:

$\begin{aligned} | A B | = | A | \cdot | B | andif A = {[a_{i j}]}_{3 \times 3}, then | k . A | = k^{3} | A | . \\ ∴ | 3 A | = 27 | A B | = 27 | A | | B | = 27 \cdot 5 \cdot 3 = 405 \end{aligned}$
Hence the statement given in question is true.

Question:52

State True or False for the statements
If the value of a third-order determinant is 12, then the value of the determinant formed by replacing each element with its co-factor will be 144.

Answer:

Given $∴ | A | = 12$
For any square matrix of order $n, adj A | = | A |^{n - 1}$

$Forn = 3, | adj A | = | A |^{3 - 1} = | A |^{2} = 12^{2} = 144$
Thus the given statement is true.

Question:53

State True or False for the statements
$| \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | = 0$ , where a, b, c are in A.P.

Answer:

$\begin{aligned} since a, b, c are in A P, 2 b = a + c . \\ ∴ | \begin{array}{lll} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | = 0 \\ A p p l y_{i}, R_{1} \to R_{1} + R_{3} \\ \Rightarrow | \begin{array}{ccc} 2 x + 4 & 2 x + 6 & 2 x + a + c \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | = 0 \end{aligned}$
Since 2b = a + c,
$\Rightarrow | \begin{array}{ccc} 2 (x + 2) & 2 (x + 3) & 2 (x + b) \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} | = 0$
We can see that Row 1 and 3 are proportional

Thus determinant = 0

Question:54

State True or False for the statements
$| a d j . A | = | A |^{2}$ , where A is a square matrix of order two.

Answer:

For any square matrix of order $n, adj A | = | A |^{n - 1}$
Here $n = 2$ ,

$\Rightarrow | adj A | = | A |^{n - 1} = | A |$
Thus the statement given in question is false

Question:55

State True or False for the statements

The determinant $| \begin{array}{ccc} \sin A & \cos A & \sin A + \cos A \\ \sin B & \cos B & \sin B + \cos B \\ \sin C & \cos C & \sin C + \cos C \end{array} |$ is equal to zero.
Answer:

$| \begin{array}{ccc} \sin A & \cos A & \sin A + \cos A \\ \sin B & \cos B & \sin B + \cos B \\ \sin C & \cos C & \sin C + \cos C \end{array} |$
$= | \begin{array}{ccc} \sin A & \cos A & \sin A \\ \sin B & \cos B & \sin B \\ \sin C & \cos C & \sin C \end{array} | + | \begin{array}{ccc} \sin A & \cos A & \cos A \\ \sin B & \cos B & \cos B \\ \sin C & \cos C & \cos C \end{array} |$
We can see that columns are identical in both the matrix on the right-hand side

Thus Determinant = 0

The statement in question is therefore true.

Question:56

State True or False for the statements

If the determinant $| \begin{array}{ccc} x + a & p + u & 1 + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{array} |$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8

Answer:

Given $| \begin{array}{ccc} x + a & p + u & 1 + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{array} |$ Splitrow $1 \Rightarrow | \begin{array}{ccc} x + a & p + u & 1 + f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{array} | = | \begin{array}{ccc} x & p & 1 \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{array} | + | \begin{array}{ccc} a & u & f \\ y + b & q + v & m + g \\ z + c & r + w & n + h \end{array} |$
Split row 2

$| \begin{array}{ccc} x & p & 1 \\ y & q & m \\ z + c & r + w & n + h \end{array} | + | \begin{array}{ccc} a & u & f \\ y & q & m \\ z + c & r + w & n + h \end{array} | + | \begin{array}{ccc} x & p & l \\ b & v & g \\ z + c & r + w & n + h \end{array} |$
$+ | \begin{array}{ccc} a & u & f \\ b & v & g \\ z + c & r + w & n + h \end{array} |$
We can split all the rows in the same way. Thus the statement given in the question is true.

Question:57

State True or False for the statements

Let $Δ = | \begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array} | = 16$ ,then $Δ_{1} = | \begin{array}{lll} p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r \end{array} | = 32$

Answer:

$\begin{aligned} Wehave Δ = | \begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array} | = 16 Weneedtoprove Δ_{1} = | \begin{array}{lll} p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r \end{array} | = 32 . \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ | \begin{array}{lll} 2 (p + x + a) & a + x & a + p \\ 2 (q + y + b) & b + y & b + q \\ 2 (r + z + c) & c + z & c + r \end{array} | = 32 \end{aligned}$
2 can be taken common from Column 1

$2 | \begin{array}{ccc} (p + x + a) & a + x & a + p \\ (q + y + b) & b + y & b + q \\ (r + z + c) & c + z & c + r \end{array} | = 32$
After that apply $C 1 \to C 1 - C 2$ and $C 2 \to C 2 - C 3$

$\begin{aligned} | \begin{array}{lll} p & x - p & a + p \\ q & y - q & b + q \\ r & z - r & c + r \end{array} | = 16 \\ | \begin{array}{ccc} p & x & a + p \\ q & y & b + q \\ r & z & c + r \end{array} | - | \begin{array}{lll} p & p & a + p \\ q & q & b + q \\ r & r & c + r \end{array} | = 16 \end{aligned}$
The second determinant of columns 2 and 3 are identical.l

$| \begin{array}{lll} p & x & a + p \\ q & y & b + q \\ r & z & c + r \end{array} | - 0 = 16$

$| \begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array} | + | \begin{array}{lll} p & x & p \\ q & y & q \\ r & z & r \end{array} | = 16$
Again, the second determinant of columns 1 and 3 is identic. al

$| \begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array} | = 16$
Hence the statement given in question is true

Question:58

State True or False for the statements

The maximum value of $| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin θ & 1 \\ 1 & 1 & 1 + \cos θ \end{array} |$ is 1/2.

Answer:

$Δ = | \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin θ & 1 \\ 1 & 1 & 1 + \cos θ \end{array} |$
Apply, $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$\Rightarrow Δ = | \begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin θ & 0 \\ 0 & 0 & \cos θ \end{array} |$

$= \cos θ \cdot \sin θ$ Multiply and divide by $2, = 1 / 2 (2 \sin θ \cos θ)$ We know, $2 \sin θ \cos θ = \sin 2 θ = 1 / 2 (\sin 2 θ)$

Since the maximum value of $\sin 2 θ$ is $1, θ = 45^{\circ}$ .

$∴ Δ = 1 / 2 (\sin 2 (45^{\circ})) = 1 / 2 \sin 90^{\circ} = 1 / 2 (1) ∴ Δ = 1 / 2$
Thus the given statement is true.

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Frequently Asked Questions (FAQs)

1. Can I download the solutions for this chapter?

Yes, you download NCERT exemplar Class 12 Maths solutions chapter 4 pdf by using the webpage to pdf tool available online.

2. What are the important topics of this chapter?

The Properties of Determinants, Adjoint and Inverse of a Matrix and Application of Determinants and Matrices are the more important topics among others as per their weightage.

3. How to study well for boards?

Practice, Practice and Practice. Once you have read the chapters well and made notes, you must practice being fast and precise with answers.

4. How many questions are there in this chapter?

The NCERT exemplar solutions for Class 12 Maths chapter 4 has one exercise with 58 questions for practice.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

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