NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 12 - Matrices

# NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 12 - Matrices

Edited By Ramraj Saini | Updated on Dec 04, 2023 12:16 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Chapter 3 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 3 class 12 Matrices are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In this article, you will get NCERT Solutions for Class 12 Maths Chapter 3 Miscellaneous Exercise. These miscellaneous exercise chapter 3 Class 12 consist of questions from all the different topics of this chapter. If you have done with all the exercises problems, you should be able to these miscellaneous exercise chapter 3 Class 12 problems. Miscellaneous exercises are considered to be difficult as compared to other exercises but students who have solved NCERT exercises on their own can solve these exercises easily.

You may find difficulties while these problems, you can take help from NCERT Solutions for Class 12 Maths Chapter 3 Miscellaneous Exercise. Miscellaneous exercises are not very important for the board exams as over 90% of the questions in the board exams are not asked from Class 12 Maths chapter 3 miscellaneous exercise solutions. Miscellaneous exercise class 12 chapter 3 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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## Matrices Miscellaneous Exercise:

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove : $(aI + bA)^n = a^n I + na^{n-1} bA$

For n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for n=1.

Let result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

$(aI + bA)^{k+1} = (aI + bA)^k (aI + bA)$

$= (a^k I + ka^{k-1} bA)$$(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

Hence, the result is true for n=k+1.

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$ where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$,$n \in N$.

Given :

$A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$

To prove:

$A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$

For n=1, we have

$A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix}$$=\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix}$$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$$\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$ where $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$.

Given :

$A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$

To prove:

$A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$

For n=1, we have

$A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}$$= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A$

Thus, result is true for n=1.

Now, take result is true for n=k,

$A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$$\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

$=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}$

$=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}$

$=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}$

$=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$.

If A, B are symmetric matrices then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew symmetric.

Let be a A is symmetric matrix , then $A'=A$

Consider, $(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by $A$

$=B'(AB)$

i.e. $(B'AB)'$ $=B'(AB)$

Thus, if A is a symmetric matrix than $B'(AB)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A'=-A$.

$(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by -$A$,

$=B'(-AB)$

$= - B'AB$

i.e. $(B'AB)'$ $= - B'AB$.

Thus, if A is a skew-symmetric matrix then $- B'AB$ is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

$A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$

$A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$

$A'A = I$

$\begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$$\begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} 2x^{2} & 0& 0\\ 0& 6y^{2} & 0\\ 0 & 0 &3z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

Thus equating the terms element wise

$2x^{2} = 1$ $6y^{2} = 1$ $3z^{2} = 1$

$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2}=\frac{1}{3}$

$x = \pm \frac{1}{\sqrt{2}}$ $y= \pm \frac{1}{\sqrt{6}}$ $z=\pm \frac{1}{\sqrt{3}}$

$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 0+4+4x \end{bmatrix} = O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, value of x is -1.

$A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$\begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 9-1 &3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$

$I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

To prove: $A^2 -5A + 7I= 0$

L.H.S : $A^2 -5A + 7I$

$= \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$$-5 \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$+ 7 \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 8-15+7 &5-5+0 \\ -5+5+0& 3-10+7 \end{bmatrix}$

$=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S$

Hence, we proved that

$A^2 -5A + 7I= 0$.

$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x +0-2& 0-10+0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x -2& -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix}x (x -2)-40+(2x-8) \end{bmatrix} = 0$

$\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0$

$\therefore \, \, x ^{2}-48= 0$

$x ^{2}=48$

thus the value of x is

$x =\pm 4\sqrt{3}$

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are  2.50,  1.50 and  1.00, respectively, find the total revenue in each market with the help of matrix algebra.

The unit sale prices of x, y and z are  2.50,  1.50 and  1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 10000\times 2.50+2000\times 1.50+18000\times 1.00$

$= 25000+3000+18000$

$= 46000$

The total revenue in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$= 15000+30000+8000$

$= 53000$

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are  2.00,  1.00 and 50 paise respectively. Find the gross profit.

The unit costs of the above three commodities are  2.00,  1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 10000\times 2.00+2000\times 1.00+18000\times 0.50$

$= 20000+2000+9000$

$= 31000$

Total revenue in the market I is 46000 , gross profit in the market is $= 46000-31000$$=Rs. 15000$

The total cost price in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 6000\times 2.0+20000\times 1.0+8000\times 0.50$

$= 12000+20000+4000$

$= 36000$

Total revenue in market II is 53000, gross profit in the market is$= 53000-36000= Rs. 17000$

$X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

The matrix given on R.H.S is $2\times 3$ matrix and on LH.S is $2\times 3$ matrix.Therefore, X has to be $2\times 2$ matrix.

Let X be $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$

$\begin{bmatrix} a & c\\ b & d \end{bmatrix}$$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$\begin{bmatrix} a+4c & 2a+5c &3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$a+4c=-7$ $2a+5c=-8$ $3a+6c=-9$

$b+4d=2$ $2b+5d=4$ $3b+6d=6$

Taking, $a+4c=-7$

$a=-4c-7$

$2a+5c=-8$

$-8c-14+5c=-8$

$-3c=6$

$c=-2$

$a=-4\times -2-7$

$a=8-7=1$

$b+4d=2$

$b=-4d+2$

$2b+5d=4$

$\Rightarrow$ $-8d+4+5d=4$

$\Rightarrow -3d=0$

$\Rightarrow d=0$

$b=-4d+2$

$\Rightarrow b=-4\times 0+2=2$

Hence, we have $a=1, b=2,c=-2,d=0$

Matrix X is $\begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix}$.

A and B are square matrices of the same order such that $AB = BA$,

To prove : $AB^n = B^n A$, $n \in N$

For n=1, we have $AB^1 = B^1 A$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $AB^k = B^k A$

Now, taking n=k+1 , we have $AB^{k+1} = AB^k .B$

$AB^{k+1} = (B^kA) .B$

$AB^{k+1} = (B^k) .AB$

$AB^{k+1} = (B^k) .BA$

$AB^{k+1} = (B^k.B) .A$

$AB^{k+1} = (B^{k+1}) .A$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$, $n \in N$.

To prove: $(AB)^n = A^n B^n$

For n=1, we have $(AB)^1 = A^1 B^1$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $(AB)^k = A^k B^k$

Now, taking n=k+1 , we have $(AB)^{k+1} = (A B)^k.(AB)$

$(AB)^{k+1} = A^k B^k.(AB)$

$(AB)^{k+1} = A^{K}( B^kA)B$

$(AB)^{k+1} = A^{K}( AB^k)B$

$(AB)^{k+1} = (A^{K}A)(B^kB)$

$(AB)^{k+1} = (A^{k+1})(B^{k+1})$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ and $(AB)^n = A^n B^n$for all $n \in N$.

If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$

(A) $1 + \alpha^2 + \beta \gamma = 0$

(B) $1 - \alpha^2 + \beta \gamma = 0$

(C) $1 - \alpha^2 - \beta \gamma = 0$

(D) $1 + \alpha^2 - \beta \gamma = 0$

$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$

$A^2 = I$

$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

Thus we obtained that

$\alpha^{2} +\beta \gamma=1$

$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$

Option C is correct.

(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these

If the matrix A is both symmetric and skew-symmetric, then

$A'=A$ and $A'=-A$

$A'=A'$

$\Rightarrow \, \, \, \, \, \, \, A=-A$

$\Rightarrow \, \, \, \, \, \, \, A+A=0$

$\Rightarrow \, \, \, \, \, \, \, 2A=0$

$\Rightarrow \, \, \, \, \, \, \, A=0$

Hence, A is a zero matrix.

Option B is correct.

(A) A
(B) I – A
(C) I
(D) 3A

A is a square matrix such that $A^{2}=A$

$(I + A)^3 - 7 A$

$=I^{3}+A^{3}+3I^{2}A+3IA^{2}-7A$

$=I+A^{2}.A+3A+3A^{2}-7A$

$=I+A.A+3A+3A-7A$ (Replace $A^{2}$ by $A$)

$=I+A^{2}+6A-7A$

$=I+A-A$

$=I$

Hence, we have $(I + A)^3 - 7 A=I$

Option C is correct.

More about NCERT Solutions for Class 12 Maths Chapter 3 Miscellaneous Exercise:-

There are 3 miscellaneous solved examples are given in the NCERT textbook before themiscellaneous exercise which could be solved if you have conceptual clarity. You can solve the 12 long answer type questions and 3 multiple choice types questions given in the miscellaneous exercise of NCERT syllabus chapter 3 Class 12. Solving these problems will check your understanding of this chapter.

Also Read| Matrices Class 12 Maths Chapter Notes

Benefits of NCERT for Class 12 Maths Chapter 3 Miscellaneous Exercise:-

• Class 12 Maths chapter 3 miscellaneous exercise solutions are available at one place which are helpful for the students who are facing problems while solving them.
• First, try to solve previous exercises then you will be able to solve the miscellaneous exercise.
• Class 12 Maths chapter 3 miscellaneous exercise solutions are also helpful for the students who are preparing for the engineering entrance exams as some questions in these exams are directly asked from miscellaneous exercises.
• You can use Class 12 Maths chapter 3 miscellaneous solutions for quick revision before the exam.
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## Key Features Of NCERT Solutions For Class 12 Chapter 3 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 3, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 3 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 3 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 3 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Also see-

## NCERT solutions of class 12 subject wise

Subject wise NCERT Exampler solutions

Happy learning!!!

1. If matrix B is the inverse of matrix A, then what is inverse of B ?

If matrix B is the inverse of matrix A, then matrix A is also the inverse of matrix B.

2. Which is the Best Book for NCERT Class 12 Maths ?

NCERT textbook is the most important book for the students who are preparing for board exams. CBSE usually use syllabus similar to the NCERT syllabus. For more questions from the chapter Matrices NCERT exemplar for Class 12 Maths can be used.

3. Where can I get the NCERT exemplar for Class 12 Maths chapter 3?
4. Number of chapters in the NCERT class 12 maths ?

There are 13 chapters in the NCERT Class 12 maths.

5. what is the weightage of calculus in CBSE class 12 maths ?

Calculus carries 35 marks weighatge in the CBSE final board exam.

6. what is the weightage of Algebra in CBSE class 12 maths ?

Algebra carries 10 marks weighatge in the CBSE final board exam.

7. What are miscellaneous exercises ?

As the name suggests miscellaneous exercises consist of a mixture of questions from all the exercises of the chapter.

8. Do miscellaneous exercises are important for competitive exams ?

Yes,  miscellaneous exercises are very important for competitive exams like JEE, SRMJEE, etc.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

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.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.

Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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