NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants 2021

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

Komal MiglaniUpdated on 30 Mar 2025, 10:55 PM IST

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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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:
$\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|$

Answer:

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

$
\begin{aligned}
& |A|=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c \\
& =\left(x^2-x+1\right) \times(x+1)-(x+1) \times(x-1) \\
& =x\left(x^2-x+1\right)+1\left(x^2-x+1\right)-\left(x^2-1\right) \\
& \quad\left(\text { Since }(a-b)(a+b)=\left(a^2-b^2\right)\right) \\
& =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:
$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$

Answer:

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

Question:3

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

Answer:

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

Question:4

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

Answer:

$Let \: \: A=\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|$

Apply - $C_1 \rightarrow C_1+C_2+C_3$

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

$\\\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1,} \text { you will get }\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 1-1 & 3 \mathrm{y}-(-\mathrm{x}+\mathrm{y}) & \mathrm{z}-\mathrm{y}-(-\mathrm{x}+\mathrm{z}) \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\\end{aligned}$

$\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y}+\mathrm{x}-\mathrm{y} & \mathrm{z}-\mathrm{y}+\mathrm{x}-\mathrm{z} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &\text { Now apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1-1 & \mathrm{y}-\mathrm{z}-(-\mathrm{x}+\mathrm{y}) & 3 \mathrm{z}-(-\mathrm{x}+\mathrm{z}) \end{array}\right| \end{aligned}$

$\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{y}-\mathrm{z}+\mathrm{x}-\mathrm{y} & 3 \mathrm{z}+\mathrm{x}-\mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3},\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}-(-\mathrm{x}+\mathrm{z}) & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-(\mathrm{x}-\mathrm{y}) & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-(2 \mathrm{z}+\mathrm{x}) & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\\end{aligned}$

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

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:
$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$

Answer:

$Let\: \: A=\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$
$\\\begin{aligned} &\text { Using } C_{1} \rightarrow C_{1}+C_{2}+C_{3},\\ &=\left|\begin{array}{ccc} x+4+x+x & x & x \\ x+x+4+x & x+4 & x \\ x+x+x+4 & x & x+4 \end{array}\right|\\ &=\left|\begin{array}{ccc} 3 x+4 & x & x \\ 3 x+4 & x+4 & x \\ 3 x+4 & x & x+4 \end{array}\right|\\ &\text { Take }(3 x+4) \text { common form } C \text { , }\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{array}\right|\\ &\mathrm{Apply}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 0 & 4 & 0 \\ 1-1 & x-x & x+4-x \end{array}\right| \end{aligned}$
$\\\begin{aligned} &=(3 \mathrm{x}+4)\left|\begin{array}{lll} 1 & \mathrm{x} & \mathrm{x} \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{1}\\ &=(3 x+4)[1 \times\{(16)-0\}]\\ &=(3 x+4)(16)\\ &=16(3 x+4) \end{aligned}$

Question:6

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

Answer:

$Let\: \: A=\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$
$\\\begin{aligned} &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{y}\\ &=\left|\begin{array}{ccc} \mathrm{a}-\mathrm{b}-\mathrm{c}+2 \mathrm{~b}+2 \mathrm{c} & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}-\mathrm{a}+2 \mathrm{c} & 2 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}-\mathrm{a}-\mathrm{b} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &=\left|\begin{array}{ccc} \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &\text { Take }(a+b+c) \text { common form the first row }\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\\ &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \end{aligned}$
$\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 2 b & b-c-a-2 b & 2 b \\ 2 c & 2 c-2 c & c-a-b \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 b & -b-c-a & 2 b \\ 2 c & 0 & c-a-b \end{array}\right| \end{array}$
Now apply $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$
$\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 2 b & -(a+b+c) & 2 b-2 b \\ 2 c & 0 & c-a-b-2 c \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{array}\right| \end{array}$
Expand the determinant along Row 1

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

Question:7

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

Answer:

Taking LHS, $\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|$
$\\\begin{aligned} &\text { Multiply and divide } \mathrm{R}_{1} \mathrm{R}_{2}, \mathrm{R}_{3} \text { respectively by } \mathrm{x}, \mathrm{y}, \mathrm{z}\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x(y+z) \\ y z^{2} x^{2} & y z x & y(z+x) \\ z x^{2} y^{2} & z x y & z(x+y) \end{array}\right|\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x y+x z \\ x^{2} y z^{2} & x y z & y z+x y \\ x^{2} y^{2} z & x y z & x z+y z \end{array}\right|\\ &\text { Now, take xyz common from the first and second Column }\\ &=\frac{1}{\mathrm{xyz}} \times \mathrm{xyz} \times \mathrm{xyz}\left|\begin{array}{lll} \mathrm{yz} & 1 & \mathrm{xy}+\mathrm{xz} \\ \mathrm{xz} & 1 & \mathrm{yz}+\mathrm{xy} \\ \mathrm{xy} & 1 & \mathrm{xz}+\mathrm{yz} \end{array}\right|\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right| \end{aligned}$
$\\\begin{aligned} &\text { Apply, } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z+y z \\ x z & 1 & y z+x y+x z \\ x y & 1 & x z+y z+x y \end{array}\right|\\ &\text { Take }(\mathrm{xy}+\mathrm{yz}+\mathrm{xz}) \text { common from } \mathrm{C}_{3}\\ &=(\text { xyz })(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\left|\begin{array}{lll} \mathrm{yz} & 1 & 1 \\ \mathrm{xz} & 1 & 1 \\ \mathrm{xy} & 1 & 1 \end{array}\right| \end{aligned}$

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

∴ LHS = RHS

Question:8

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

Answer:

LHS given,

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

$\\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z}+\mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{z}+\mathrm{x}+\mathrm{x} & \mathrm{y}+\mathrm{x}+\mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ &=\left|\begin{array}{ccc} 2 z+2 y & 2 z+2 x & 2 x+2 y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &2 \text { can be taken common from } \mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ \end{aligned}$

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

$\\\begin{aligned} &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &\text { Apply, } R_{3} \rightarrow R_{3}-R_{1},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y}-\mathrm{y} & \mathrm{x}-0 & \mathrm{x}+\mathrm{y}-\mathrm{y} \end{array}\right|\\ &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ 0 & x & x \end{array}\right|\\ &\text { Now, Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z}-0 & \mathrm{z}+\mathrm{x}-\mathrm{x} & \mathrm{x}-\mathrm{x} \\ 0 & \mathrm{x} & \mathrm{x} \end{array}\right|\\ &=2\left|\begin{array}{lll} y & 0 & y \\ z & z & 0 \\ 0 & x & x \end{array}\right| \end{aligned}$

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

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

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

Hence Proved

Question:9

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

Take LHS

$
\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|
$
Apply, $R_1 \rightarrow R_1-R_2$

$
=\left|\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}\right|
$

$
=\left|\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}\right|
$

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

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

Take (a-1) common from $R$

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

Apply, $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3$

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

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

Take (a-1) common from $R_2$

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

Expand along $\mathrm{C}_3$

$
\begin{aligned}
& =(a-1)^2[1\{(a+1)-2\}] \\
& =(a-1)^2[a+1-2] \\
& =(a-1)^3 \\
& =\text { 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=\left|\begin{array}{ccc}
1 & \cos C & \cos B \\
\cos C & 1 & \cos A \\
\cos B & \cos A & 1
\end{array}\right|
$
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 $\left (x_1, y_1 \right )$, $\left (x_2, y_2 \right )$, $\left (x_3, y_3 \right )$, then $\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|^{2}=\frac{3 a^{4}}{4}$

Answer:

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

$\\\Delta=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|$

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

Thus, the area

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

$\therefore \frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$$

Square both sides

$\Rightarrow\left(\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\right)^{2}=\left(\frac{\sqrt{3}}{4} \mathrm{a}^{2}\right)^{2}$

$\\\begin{aligned} &\Rightarrow\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3} \end{array}\right|^{2}=\frac{3}{4} \mathrm{a}^{4}\\ &\text { Hence Proved } \end{aligned}$

Question:12

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

Answer:

Given:

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

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

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

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

Question:13

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

Answer:

Given:
$\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$
$\\\begin{aligned} &\text { Apply, } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{ccc} 4-x+4+x+4+x & 4+x & 4+x \\ 4+x+4-x+4+x & 4-x & 4+x \\ 4+x+4+x+4-x & 4+x & 4-x \end{array}\right|=0\\ &\text { Take, }(12+\mathrm{x}) \text { common from Row } 1\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x+4+x & 4+x \\ 1 & 4-x+4+x & 4+x \\ 1 & 4+x+4-x & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1 & 8 & 4+x \\ 1 & 8 & 4-x \end{array}\right|=0 \end{aligned}$
$\\\begin{aligned} &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1-1 & 8-8 & 4+x-(4-x) \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1-1 & 8-8-2 x & 4-x-4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 0 & -2 x & -2 x \end{array}\right|=0 \end{aligned}$
Expand along Column 1

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

Question:14

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

Answer:

$a_1, a_2, \ldots, a_r$ are in G.P
We know that, $a_{r+1}=A R^{(r+1)-1}=A R^r \ldots(i)$
[ $\because a_n=a r^{n-1}$, where $a=$ first term and $r=$ common ratio]
A is the first term of G.P
R is the common ratio of G.P.

$
\therefore\left|\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}\right|=\left|\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}\right| \ldots[\text { from (i) }]
$
Taking $A R^r, A R^{r+6}$ and $A R^{r+10}$ common from $R_1, R_2$ and $R_3$

$
=\mathrm{AR}^{\mathrm{r}} \times \mathrm{AR}^{\mathrm{r}+6} \times \mathrm{AR}^{\mathrm{r}+10}\left|\begin{array}{ccc}
1 & \mathrm{AR}^4 & \mathrm{AR}^8 \\
1 & \mathrm{AR}^4 & \mathrm{AR}^8 \\
1 & \mathrm{AR}^6 & \mathrm{AR}^{10}
\end{array}\right|
$
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.

$
\therefore\left|\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}\right|=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:-
$\\\begin{array}{l} \Delta=\frac{1}{2}\left|\begin{array}{ccc} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array}\right| \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{array}\right| \\ \text { Apply } R_{2} \rightarrow R_{2}-R_{1} \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2-a-5 & a+3-a+4 & 1-1 \\ a & a & 1 \end{array}\right| \\\\ =\frac{1}{2}\left| \begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a & a & 1 \end{array}\right| \end{array}$
$\\\begin{aligned} &\text { Apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a-a-5 & a-a+4 & 1-1 \end{array}\right|\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ -5 & 4 & 0 \end{array}\right|\\ &\text { Expand along Column } 3\\ &=\frac{1}{2}[(1)(-28-(7)(-5))]\\ &=\frac{1}{2}(-28+35)=\frac{7}{2} \neq 0 \end{aligned}$
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

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

$\\\begin{aligned} &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\\ &=\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 1+\cos A & 1+\cos B-1-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)(\cos B+\cos A)+(\cos B-\cos A) & \cos ^{2} C+\cos C \end{array}\right|=0 \end{aligned}$
$\\\begin{aligned} &\begin{array}{|ccc|} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)[(\cos B+\cos A)+1] & \cos ^{2} C+\cos C \end{array} \mid=0\\ &\text { Take, cosB-cosA common from column } 2\\ &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & 1 & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B+\cos A)+1 & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\\ &\cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 1+\cos A & 1 & 1+\cos C-1-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C+\cos C-\cos ^{2} A-\cos A \end{array}\right|=0 \end{aligned}$
$\\\begin{aligned} &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & \cos C-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C-\cos ^{2} A+\cos C-\cos A \end{array}\right|=0\\ &\Rightarrow(\cos B-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & (\cos C-\cos A) \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & (\cos C-\cos A)(\cos C+\cos A+1) \end{array}\right|=0\\ &\text { Take cosC-cosA common from Column } 3\\ &\Rightarrow(\cos B-\cos A)(\cos C-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & 1 \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos C+\cos A+1 \end{array}\right|=0 \end{aligned}$

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 \backslash \text { or } \cos C-\cos A=0 \backslash \text { or } \cos C-\cos B=0 \\
& \Rightarrow \cos B=\cos A \text { or } \cos C=\cos A \text { or } \cos C=\cos B \\
& \Rightarrow B=A \text { or } C=A \text { or } C=B
\end{aligned}
$
$
\text { Hence, } \triangle A B C \text { is an isosceles triangle. }
$

Question:17

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

Answer:

To find $\operatorname{adj} \mathrm{A}$

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

Now, we need to prove that.t

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

Question:18

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

Answer:

Find IAI Expand IAI along Column 1

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

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

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

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

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

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

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:

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

$
A^{-1}=\left[\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}\right]
$

Multiply:

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

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

Question:20

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

Answer:

$
\begin{aligned}
& A=\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right] \\
& \therefore B A=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right] \\
& =\left[\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}\right] \\
& =\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
& =6\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& B A=6 I \ldots(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,

$
\left[\begin{array}{ccc}
0 & 1 & 2 \\
1 & -1 & 0 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
7 \\
3 \\
17
\end{array}\right]
$
Apply, $\mathrm{R}_1 \leftrightarrow \mathrm{R}_2$

$
\begin{aligned}
& \mathrm{R}_2 \leftrightarrow \mathrm{R}_3 \\
& {\left[\begin{array}{lll}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
3 \\
17 \\
7
\end{array}\right]} \\
& {\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]^{-1}\left[\begin{array}{c}
3 \\
17 \\
7
\end{array}\right]}
\end{aligned}
$
So, $B A=6 I[$ from eg(i) $]$

$
\begin{aligned}
& {\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\frac{1}{6}\left[\begin{array}{l}
12 \\
-6 \\
24
\end{array}\right]} \\
& {\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1 \\
4
\end{array}\right]}
\end{aligned}
$

$\therefore x=2, y=-1$ and $z=4$

Question:21

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

Answer:

$
A=\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|
$
Apply $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$

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

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

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

Given that $\Delta=0$

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

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

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

$\text {but it is given that }(a+b+c) \neq 0 \therefore(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 $\left|\begin{array}{ccc} \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right|$ is divisible by a + b + c and find the quotient.

Answer:

$\left|\begin{array}{lll}b c-a^2 & c a-b^2 & a b-c^2 \\ c a-b^2 & a b-c^2 & b c-a^2 \\ a b-c^2 & b c-a^2 & c a-b^2\end{array}\right|$ is given.
Apply, $R_1 \rightarrow R_1-R_2$,

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

$
=(a+b+c)\left|\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}\right|
$

Apply $R_2 \rightarrow R_2-R_3$

$
\begin{aligned}
& =(a+b+c)\left|\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}\right| \\
& =(a+b+c)\left|\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}\right|
\end{aligned}
$

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

$
=(a+b+c)^2\left|\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}\right|
$

Expand along Column 1

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

Question:23

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

Answer:

Given LHS,

$
\left|\begin{array}{ccc}
x a & y b & z c \\
y c & z a & x b \\
z b & x c & y a
\end{array}\right|
$
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\left\{a^2 y z-x^2 b c\right\}-y b\left\{y^2 a c-b^2 x z\right\}+z c\left\{c^2 x y-z^2 a b\right\} \\
& =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\left(a^3+b^3+c^3\right)-a b c\left(x^3+y^3+z^3\right)
\end{aligned}
$
Given $x+y+z=0$

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

Question:24

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

Answer:

Given:

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

$
\begin{aligned}
& |\mathrm{A}|=\left|\begin{array}{ll}
\mathrm{a} & \mathrm{~b} \\
\mathrm{c} & \mathrm{~d}
\end{array}\right|=\mathrm{ad}-\mathrm{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 $\left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|$
A. $a^3 + b^3 + c^3$
B. 3 bc
C. $a^3 + b^3 + c^3-3abc$
D. none of these

Answer:

Given:

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

Apply C2 $\rightarrow \mathrm{C} 2+\mathrm{C} 3$

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

$
=(a+b+c)\left|\begin{array}{lll}
a-b & 1 & a \\
b-c & 1 & b \\
c-a & 1 & c
\end{array}\right|
$
Apply $\mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_3$

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

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

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

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

Question:27

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

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

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

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

Apply

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

Question:28

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

A. 0
B. –1
C. 1
D. None of these

Answer:

C)
Given
$\left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{lll} \sin x+\cos x+\cos x & \cos x & \cos x \\ \cos x+\sin x+\cos x & \sin x & \cos x \\ \cos x+\cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\Rightarrow\left|\begin{array}{ccc} 2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\text { Take }(2 \cos x+\sin x) \text { common from Column } 1\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1-1 & \sin x-\cos x & \cos x-\cos x \\ 1 & \cos x & \sin x \end{array}\right|=0 \end{aligned}$
$\\\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1 & \cos x & \sin x \end{array}\right|=0 \\ \text { Apply } R_{3} \rightarrow R_{3}-R_{1} \\ \Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1-1 & \cos x-\cos x & \sin x-\cos x \end{array}\right|=0 \\$

$\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x \end{array}\right|=0$
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 \operatorname{or}(\sin X-\cos X)^2=0 \\
& \Rightarrow 2=-\frac{\sin \mathrm{x}}{\cos \mathrm{x}} \text { or } \sin \mathrm{x}=\cos \mathrm{x} \\
& \Rightarrow \tan x=-2 \text { ortan } x=1\left[\because \tan x=\frac{\sin x}{\cos x}\right] \\
& \text { buttan } x=-2 \text { isnotpossibleasfor }-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}
\end{aligned}
$
$
\text { So, } \tan x=1
$
$
\therefore x=\frac{\pi}{4}
$
Only one real and distinct root occurs.

Question:29

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

B. –1
C. 1
D. None of these

Answer:

Given:

$
\left.\left|\begin{array}{ccc}
-1 & \cos \mathrm{C} & \cos \mathrm{~B} \\
\cos \mathrm{C} & -1 & \cos \mathrm{~A} \\
\cos \mathrm{~B} & \cos \mathrm{~A} & -1
\end{array}\right| \right\rvert\,
$
Expand along Column 1

$
\begin{aligned}
& |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc}
\mathrm{a}_{22} & \mathrm{a}_{23} \\
\mathrm{a}_{32} & \mathrm{a}_{33}
\end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc}
\mathrm{a}_{12} & \mathrm{a}_{13} \\
\mathrm{a}_{32} & \mathrm{a}_{33}
\end{array}\right|+\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{ll}
\mathrm{a}_{12} & \mathrm{a}_{13} \\
\mathrm{a}_{22} & \mathrm{a}_{23}
\end{array}\right| \\
& \Delta=(-1)\left|\begin{array}{cc}
-1 & \cos A \\
\cos A & -1
\end{array}\right|-\cos C\left|\begin{array}{cc}
\cos C & \cos B \\
\cos A & -1
\end{array}\right|+\cos B\left|\begin{array}{cc}
\cos C & \cos B \\
-1 & \cos A
\end{array}\right| \\
& =\left[(-1)\left\{1-\cos ^2 A\right\}-\cos C\{-\cos C-\cos A \cos B\}+\cos B\{\cos A \cos C+\right. \\
& \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 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}+2 \cos \mathrm{~A} \cos \mathrm{~B} \cos \mathrm{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} \\
& \text { Now use: } \cos (A+B) \cos (A-B)=2 \cos A \cos B \\
& \text { 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 \mathrm{C}+[2 \cos (\mathrm{~A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})\}+4 \cos \mathrm{~A} \cos \mathrm{~B} \cos C}{2} \ldots \text { (i) }
\end{aligned}
$

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

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

Question:30

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

Answer:

Given:

$
f(t)=\left|\begin{array}{ccc}
\cos t & t & 1 \\
2 \sin t & t & 2 t \\
\sin t & t & t
\end{array}\right|
$
Divide $\mathrm{R}_2$ and $\mathrm{R}_3$ by $t$

$
f(t)=t^2\left|\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}\right|
$
$
\Rightarrow \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^2}=\frac{\mathrm{t}^2}{\mathrm{t}^2}\left|\begin{array}{ccc}
\cos t & t & 1 \\
\frac{2 \sin t}{t} & 1 & 2 \\
\frac{\sin t}{t} & 1 & 1
\end{array}\right|
$
$
\Rightarrow \lim _{t \rightarrow 0} \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^2}=\left|\begin{array}{lll}
\lim _{t \rightarrow 0} \cos t & \lim _{t \rightarrow 0} t & \lim _{t \rightarrow 0} 1 \\
\lim _{t \rightarrow 0} \frac{2 \sin t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 2 \\
\lim _{t \rightarrow 0} \frac{\sin ^t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 1
\end{array}\right|
$
$
=\left|\begin{array}{lll}
1 & 0 & 1 \\
2 & 1 & 2 \\
1 & 1 & 1
\end{array}\right|\left(\because \lim _{t \rightarrow 0} \frac{\sin t}{t}=1\right)
$

Expand along Row 1

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

Question:31

The maximum value of $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right|_{\text {is }(\theta \text { is a real number })}$

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

Answer:

We have:

$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1+\cos \theta & 1 & 1
\end{array}\right|
$
Apply, $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3$ and $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3$

$
\Rightarrow \Delta=\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & \sin \theta & 1 \\
\cos \theta & 0 & 1
\end{array}\right|
$

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

$
\begin{aligned}
& \therefore \Delta=\frac{1}{2}\left(\sin 2\left(45^{\circ}\right)\right)=\frac{1}{2} \sin 90^{\circ}=\frac{1}{2}(1) \\
& \therefore \Delta=\frac{1}{2}
\end{aligned}
$

Question:32

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

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

Answer:

We have:

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

If we put $x=a$

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

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

$
\begin{aligned}
& =0\left|\begin{array}{cc}
0 & -c \\
c & 0
\end{array}\right|-(-a)\left|\begin{array}{cc}
a & -c \\
b & 0
\end{array}\right|+(-b)\left|\begin{array}{ll}
a & 0 \\
b & c
\end{array}\right| \\
& =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 $A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]$

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

Answer:

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

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

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

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

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

Question:34

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

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

Answer:

We know, that A and B are invertible matrices

$
\begin{aligned}
& \text { Consider }(A B) B^{-1} A^{-1} \Rightarrow(A B) B^{-1} A^{-1}=A\left(B B^{-1}\right) A^{-1}=A I A^{-1}=(A I) A^{-1}=A A^{-1}=I \Rightarrow(A B)^{-1}=B^{-1} A^{-1} \ldots \text { option }(C) \\
& \text { Also } A A^{-1}=I \Rightarrow\left|A A^{-1}\right|=|I| \Rightarrow|A|\left|A^{-1}\right|=1 \\
& \Rightarrow|\mathrm{~A}|^{-1}=\frac{1}{|\mathrm{~A}|} \\
& \therefore \operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1} \ldots(B)
\end{aligned}
$
We know that $\frac{|\mathrm{A}|^{-1}}{\operatorname{adj} \mathrm{~A}} \frac{|\vec{A}|}{\text { We }}$
$\Rightarrow \operatorname{adj} A=|A| \cdot A^{-1} \ldots$ option $(A)$

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

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

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

Question:35

If x, y, z are all different from zero and ,$\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0$, 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
$\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0$
$\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$
$\left|\begin{array}{ccc} x & 0 & 1 \\ 0 & y & 1 \\ -z & -z & 1+z \end{array}\right|=0$
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
$\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\ \therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x^{-1}+y^{-1}+z^{-1}=-1$

Question:36

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

Answer:

Matrix given:

$
\begin{aligned}
& \left|\begin{array}{ccc}
x & x+y & x+2 y \\
x+2 y & x & x+y \\
x+y & x+2 y & x
\end{array}\right| \\
& =\mathrm{x}\left|\begin{array}{cc}
\mathrm{x} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+2 \mathrm{y} & \mathrm{x}
\end{array}\right|-(\mathrm{x}+\mathrm{y})\left|\begin{array}{cc}
\mathrm{x}+2 \mathrm{y} & \mathrm{x}+\mathrm{y} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}
\end{array}\right|+(\mathrm{x}+2 \mathrm{y})\left|\begin{array}{cc}
\mathrm{x}+2 \mathrm{y} & \mathrm{x} \\
\mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y}
\end{array}\right| \\
& =x\left[x^2-(x+y)(x+2 y)\right]-(x+y)\left[(x+2 y)(x)-(x+y)^2\right]+(x+2 y)[(x+ \\
& \left.2 y)^2-x(x+y)\right] \\
& =x\left[x^2-x^2-3 x y-2 y^2\right]-(x+y)\left[x^2+2 x y-x^2-2 x y-y^2\right]+(x+2 y)\left[x^2+\right. \\
& \left.4 x y+4 y^2-x^2-x y\right] \\
& =x\left[-3 x y-2 y^2\right]-(x+y)\left[-y^2\right]+(x+2 y)\left[3 x y+4 y^2\right] \\
& =-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, $\Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86$ ,then sum of these number is
A. 4
B. 5
C. -4
D. 9

Answer:

We have:

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

$
1\left[2 a^2+4\right]+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\left(a^2+4 a\right)=42
$

$
\left(a^2+4 a\right)=21 \Rightarrow a^2+4 a-21=0 \Rightarrow(a+7)(a-3)=0 \therefore a=-7 \text { 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}
& \text { if } A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|k \cdot A|=k^3|A| \\
& \therefore|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 $\left|A^{-1}\right|=|A|^{-1}$
Given
$\mathrm{A}=$ invertible matrix $3 \times 3$

$
\begin{aligned}
& A A^{-1}=I \Rightarrow|A|\left|A^{-1}\right|=1 \\
& \therefore\left|A^{-1}\right|=\frac{1}{|A|}
\end{aligned}
$

Question:40

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

Answer:

$\\ \begin{aligned} &\text { Given }\\ &\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\\ &=\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply Formula: }(a+b)^{2}-(a-b)^{2}=4 a b \text { . } \end{aligned}$
$\\ \begin{aligned} &=\left|\begin{array}{lll} 4\left(2^{x}\right)\left(2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4\left(3^{x}\right)\left(3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4\left(4^{x}\right)\left(4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} 4 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Column } 1 \text { and } 3 \text { thus become proportional }\\ &\therefore\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0 \end{aligned}$

Question:41

Fill in the blanks

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

Answer:

We know

$
\cos 2 \theta=0 \Rightarrow \cos 2 \theta=\cos \pi / 2 \Rightarrow 2 \theta=\pi / 2 \therefore \theta=\pi / 4
$

$\\ \begin{aligned} &\therefore \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\\ &\text { Then }\\ &\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\left|\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right|^{2}\\ &\Rightarrow\left[0-\frac{1}{\sqrt{2}}\left(\frac{1}{2}\right)+\frac{1}{\sqrt{2}}\left(-\frac{1}{2}\right)\right]^{2}=\left[\frac{-2}{2 \sqrt{2}}\right]^{2}=\left[-\frac{1}{\sqrt{2}}\right]^{2}\\ &\therefore\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\frac{1}{2} \end{aligned}$

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
$\begin{aligned} \left(A^{2}\right)^{-1} &=(A \cdot A)^{-1} \\ &=A^{-1} \cdot A^{-1} \\ &=\left(A^{-1}\right)^{2} \end{aligned}$

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, $A=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|$
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 $\left|\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|=0$ , then other two roots are ___.

Answer:

We know that

$
\begin{aligned}
& x=-9 \text { is a root of }\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|=0 \\
& \Rightarrow \mathrm{x}\left|\begin{array}{ll}
\mathrm{x} & 2 \\
6 & \mathrm{x}
\end{array}\right|-3\left|\begin{array}{ll}
2 & 2 \\
7 & \mathrm{x}
\end{array}\right|+7\left|\begin{array}{ll}
2 & \mathrm{x} \\
7 & 6
\end{array}\right|=0
\end{aligned}
$
$\Rightarrow x\left[x^2-12\right]-3[2 x-14]+7[12-7 x]=0 \Rightarrow x^3-12 x-6 x+42+84-49 x=0 \Rightarrow x^3-67 x+126=0(x+9)(2-x)(7-x)=0 H$ ere, $126 \times 1=9 \times 2 \times 7$ Forx $=2, \Rightarrow 2^3-67 \cdot 2+126=134-134=0 \therefore x=2$ 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

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

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

Answer:

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

$
\therefore 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:

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

Question:49

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

Answer:

For a non-singular matrix, aA is invertible such that
$\\ (\mathrm{aA})\left(\frac{1}{\mathrm{a}} \mathrm{A}^{-1}\right)=\left(\mathrm{a} \cdot \frac{1}{\mathrm{a}}\right)\left(\mathrm{AA}^{-1}\right) \\ \text {i.e. } \quad(\mathrm{aA})^{-1}=\frac{1}{\mathrm{a}} \mathrm{A}^{-1} \\$
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|\left|A^{-1}\right|=1 \therefore\left|A^{-1}\right|=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| \text { andif } A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|k . A|=k^3|A| . \\
& \therefore|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 $\therefore|A|=12$
For any square matrix of order $\mathrm{n}, \operatorname{adj} A\left|=|A|^{n-1}\right.$

$
\text { Forn }=3,|\operatorname{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
$\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0$, where a, b, c are in A.P.

Answer:

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

Thus determinant = 0

Question:54

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

Answer:

For any square matrix of order $\mathrm{n}, \operatorname{adj} A\left|=|A|^{n-1}\right.$
Here $\mathrm{n}=2$,

$
\Rightarrow|\operatorname{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 $\left|\begin{array}{ccc} \sin A & \cos A & \sin A+\cos A \\ \sin B & \cos B & \sin B+\cos B \\ \sin C & \cos C & \sin C+\cos C \end{array}\right|$ is equal to zero.
Answer:

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

$
\left|\begin{array}{ccc}
x & p & 1 \\
y & q & m \\
z+c & r+w & n+h
\end{array}\right|+\left|\begin{array}{ccc}
a & u & f \\
y & q & m \\
z+c & r+w & n+h
\end{array}\right|+\left|\begin{array}{ccc}
x & p & l \\
b & v & g \\
z+c & r+w & n+h
\end{array}\right|
$
$
+\left|\begin{array}{ccc}
a & u & f \\
b & v & g \\
z+c & r+w & n+h
\end{array}\right|
$
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 $\Delta=\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|=16$ ,then $\Delta_{1}=\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=32$

Answer:

$
\begin{aligned}
& \text { Wehave } \Delta=\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|=16 \text { Weneedtoprove } \Delta_1=\left|\begin{array}{lll}
p+x & a+x & a+p \\
q+y & b+y & b+q \\
r+z & c+z & c+r
\end{array}\right|=32 \text {. } \\
& \mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \\
& \left|\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}\right|=32
\end{aligned}
$
2 can be taken common from Column 1

$
2\left|\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}\right|=32
$
After that apply $\mathrm{C} 1 \rightarrow \mathrm{C} 1-\mathrm{C} 2$ and $\mathrm{C} 2 \rightarrow \mathrm{C} 2-\mathrm{C} 3$

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

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

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

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

Question:58

State True or False for the statements

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

Answer:

$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|
$
Apply, $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$

$
\Rightarrow \Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
0 & \sin \theta & 0 \\
0 & 0 & \cos \theta
\end{array}\right|
$

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

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

$
\therefore \Delta=1 / 2\left(\sin 2\left(45^{\circ}\right)\right)=1 / 2 \sin 90^{\circ}=1 / 2(1) \therefore \Delta=1 / 2
$
Thus the given statement is true.

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Frequently Asked Questions (FAQs)

Q: 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.

Q: 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.

Q: 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.

Q: 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

On Question asked by student community

Have a question related to CBSE Class 12th ?

Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

Read Complete Answer

Rahul Mandal

8 Sep'25

how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer

gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer

Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

Read Complete Answer

gopi

3 Sep'25

12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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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
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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

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

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