Matrices Exercise 3.3 NCERT Solutions Class 12 Math Chapter 3 free PDF Download

Chapter 3 of the Class 12 NCERT Mathematics textbook, titled “Matrices,” delves into the fundamental concepts of matrices, including their types, operations, and applications. Exercise 3.3 focuses on practical problems involving matrix operations, such as addition, subtraction, and multiplication of matrices. This exercise helps students apply their understanding of matrices to solve various mathematical problems.

NCERT Solutions for Mathematics – Chapter 3 Matrices – Exercise 3.3

This section provides detailed solutions for Exercise 3.3 from Chapter 3 of the Class 12 NCERT Mathematics textbook. The exercise includes a variety of problems related to matrix operations, offering step-by-step explanations to ensure students can effectively solve these problems and grasp the underlying concepts of matrix algebra.

Question 1. Find the transpose of each of the following matrices:

(i) [Tex]\begin{bmatrix}5 \\\frac{1}{2} \\-1 \end{bmatrix}   [/Tex]

(ii) [Tex]\begin{bmatrix}1 & -1 \\2 & 3 \\\end{bmatrix}[/Tex]

(iii) [Tex]\begin{bmatrix}-1 & 5 & 6\\ \sqrt{3} & 5 & 6\\2 & 3 & -1\end{bmatrix}[/Tex]

Solution:

(i) Let A =[Tex]\begin{bmatrix}5 \\\frac{1}{2} \\-1 \end{bmatrix}[/Tex]

∴Transpose of A = A’ = A = [Tex]\begin{bmatrix}5&\frac{1}{2} &-1\\\end{bmatrix}[/Tex]

(ii) Let A =[Tex]\begin{bmatrix}1 & -1 \\2 & 3 \\\end{bmatrix}[/Tex]

∴Transpose of A = A’ = AT  =[Tex]\begin{bmatrix}1 & 2 \\-1 & 3 \\\end{bmatrix}[/Tex]

(iii) Let A =[Tex]\begin{bmatrix}-1 & 5 & 6\\ \sqrt{3} & 5 & 6\\2 & 3 & -1\end{bmatrix}[/Tex]

∴Transpose of A = A’ = AT  =[Tex]\begin{bmatrix}-1 & \sqrt{3} & 2\\ 5 & 5 & 3\\6 & 6 & -1\end{bmatrix}[/Tex]

Question 2. If A =[Tex]\begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}    [/Tex] and B = [Tex]\begin{bmatrix}-4 & 2 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix} [/Tex] then verify that:

(i) (A+B)’ = A’+B’

(ii) (A-B)’ = A’- B’

Solution:

(i) A+B =[Tex]\begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix}=\begin{bmatrix}-1-4 & 2=1 & 3-5\\5+1 & 7+2 & 9+0\\-2+1 & 1+3 & 1+1\end{bmatrix}=\begin{bmatrix}-5 & 3 & -2\\6 & 9 & 9\\-1 & 4 & 2\end{bmatrix}[/Tex]

L.H.S. = (A+B)’ = [Tex]\begin{bmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{bmatrix}[/Tex]

R.H.S. = A’+B’ = [Tex]\begin{bmatrix}-1 & 5 & -2\\2 & 7 & 1\\-2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{bmatrix}=\begin{bmatrix}-1-4 & 5+1 & -2+1\\2+1 & 7+2 & 1+3\\-2-5 & 1+0 & 1+1\end{bmatrix}=\begin{bmatrix}-5 & 6 & -1\\3 & 9 & 4\\-2 & 9 & 2\end{bmatrix}[/Tex]

∴L.H.S = R.H.S.

Hence, proved.

(ii) A-B = [Tex]\begin{bmatrix}-1 & 2 & 3\\5 & 7 & 9\\-2 & 1 & 1\end{bmatrix}-\begin{bmatrix}-4 & 2 & -5\\1 & 2 & 0\\1 & 3 & 1\end{bmatrix}=\begin{bmatrix}3 & 1 & 8\\4 & 5 & 9\\-3 & -2 & 0\end{bmatrix}[/Tex]

L.H.S. = (A-B)’[Tex]=\begin{bmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{bmatrix}[/Tex]

R.H.S. = A’-B’ =[Tex]\begin{bmatrix}-1 & 5 & -2\\2 & 7 & 1\\-2 & 1 & 1\end{bmatrix}-\begin{bmatrix}-4 & 1 & 1\\1 & 2 & 3\\-5 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 & 4 & -3\\1 & 5 & -2\\8 & 9 & 0\end{bmatrix}[/Tex]

∴ L.H.S. = R.H.S.

Hence, proved.

Question 3.  If A’ =[Tex]\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}    [/Tex] and B = [Tex]\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix} [/Tex], then verify that:

(i) (A+B)’=A’+B’

(ii) (A-B)’=A’-B’

Solution:

Given A’=[Tex]\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}    [/Tex]and B=[Tex]\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}[/Tex]

then, (A’)’ = A =[Tex]\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}[/Tex]

(i) A+B =[Tex]\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}+\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}=\begin{bmatrix}2 &1 & 1\\5& 4 & 4\\\end{bmatrix}[/Tex]

∴ L.H.S. =  (A+B)’=[Tex]\begin{bmatrix}2 & 5\\1 & 4 \\1 & 4 \end{bmatrix}[/Tex]

R.H.S.= A’+B’ = [Tex]\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}+\begin{bmatrix}-1 & 1  \\2 & 2 \\1 & 3 \end{bmatrix}=\begin{bmatrix}2 & 5\\1 & 4 \\1 & 4 \end{bmatrix}[/Tex]

∴ L.H.S. = R.H.S.

Hence, proved.

(ii) A-B = [Tex]\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\\\end{bmatrix}-\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\\\end{bmatrix}=\begin{bmatrix}4 &-3 & -1\\3 & 0 & -2\\\end{bmatrix}[/Tex]

∴ L.H.S. =  (A-B)’=[Tex]\begin{bmatrix}4 & 3 \\-3 & 0 \\-1 & -2 \end{bmatrix}[/Tex]

R.H.S.= A’-B’ = [Tex]\begin{bmatrix}3 & 4 \\-1 & 2 \\0 & 1 \end{bmatrix}-\begin{bmatrix}-1 & 1  \\2 & 2 \\1 & 3 \end{bmatrix}=\begin{bmatrix}4 & 3\\-3 & 0 \\-1 & -2 \end{bmatrix}[/Tex]

∴ L.H.S. = R.H.S.

Hence, proved.

Question 4. If A’ = [Tex]\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}     [/Tex]and B = [Tex]\begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}  [/Tex]then find (A+2B)’.

Solution:

Given: A’ =[Tex]\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}     [/Tex]and B =[Tex]\begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}[/Tex]

then (A’)’ =A=[Tex]\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix} [/Tex]

Now, A+2B = [Tex]\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}+2\begin{bmatrix}-1 & 0 \\1 & 2 \\\end{bmatrix}=\begin{bmatrix}-2 & 3 \\1 & 2 \\\end{bmatrix}+\begin{bmatrix}-2 & 0 \\2 & 4 \\\end{bmatrix}=\begin{bmatrix}-2-2 & 1+0 \\3+2 & 2+4 \\\end{bmatrix}=\begin{bmatrix}-4 & 1 \\5 & 6 \\\end{bmatrix}[/Tex]

∴(A+2B)’ = [Tex]\begin{bmatrix}-4 & 5 \\1 & 6 \\\end{bmatrix}[/Tex]

Question 5. For the matrices A and B, verify that (AB)′ = B′A′, where

(i) A =[Tex]\begin{bmatrix}1 \\-4 \\3 \end{bmatrix}[/Tex] and B = [Tex]\begin{bmatrix}-1 & 2 & 1\\\end{bmatrix}[/Tex]

(ii) A =[Tex]\begin{bmatrix}0 \\1 \\2 \end{bmatrix}[/Tex] and B =[Tex]\begin{bmatrix}1 & 5 & 7\\\end{bmatrix}[/Tex]

Solution:

(i) AB = =[Tex]\begin{bmatrix}1 \\-4 \\3 \end{bmatrix}\begin{bmatrix}-1 & 2 & 1\\\end{bmatrix}=\begin{bmatrix}-1 & 2 & 1\\4 & -8 & -4\\-3 & 6 & 3\end{bmatrix}[/Tex]

∴  L.H.S. = (AB)′ =[Tex]\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{bmatrix}[/Tex]

R.H.S.= B′A’ = [Tex]\begin{bmatrix}-1 \\2 \\1 \end{bmatrix}\begin{bmatrix}1 & -4 & 3\\\end{bmatrix}=\begin{bmatrix}-1 & 4 & -3\\2 & -8 & 6\\1 & -4 & 3\end{bmatrix}[/Tex]

∴ L.H.S. = R.H.S.

Hence, proved.

(ii) AB =[Tex]\begin{bmatrix}0 \\1 \\2 \end{bmatrix}\begin{bmatrix}1 & 5 & 7\\\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\1 & 5 & 7\\2 & 10 & 14\end{bmatrix}[/Tex]

∴  L.H.S. = (AB)′ =[Tex]\begin{bmatrix}0 & 1 & 2\\0 & 5 & 10\\0 & 7 & 14\end{bmatrix}[/Tex]

Now, R.H.S.=B’A’ = [Tex]\begin{bmatrix}1 \\5 \\7 \end{bmatrix}\begin{bmatrix}0 & 1 & 2\\\end{bmatrix}=\begin{bmatrix}0 & 1 & 2\\0 & 5 & 7\\0 & 7 & 14\end{bmatrix}[/Tex]

∴ L.H.S. = R.H.S.

Hence, proved.

Question 6. If (i) A =[Tex]\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}    [/Tex]  , then verify that A′ A = I.

(ii) A =[Tex]\begin{bmatrix}sinα & cosα \\-cosα & sinα \\\end{bmatrix}    [/Tex]  ,then verify that A′ A = I.

Solution:

(i) [Tex]\begin{bmatrix}cosα &-sinα \\sinα & cosα \\\end{bmatrix}\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}=\begin{bmatrix}cos ^{2}α+sin^{2}α & cosαsinα-sinαcosα\\sinαcosα-cosαsinα & sin ^{2}α+cos^{2}α \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}[/Tex]

= I = R.H.S.

∴ L.H.S. = R.H.S.

(ii) [Tex]\begin{bmatrix}sinα &-cosα \\cosα & sinα \\\end{bmatrix}\begin{bmatrix}sinα & cosα \\-cosα & sinα \\\end{bmatrix}=\begin{bmatrix}sin ^{2}α+sin^{2}α & sinαcosα-cosαsinα\\cosαsinα-sinαcosα & cos ^{2}α+sin^{2}α \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}[/Tex]

= I = R.H.S.

∴ L.H.S. = R.H.S.

Question 7. (i) Show that the matrix A[Tex]\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}    [/Tex] = is a symmetric matrix.

(ii) Show that the matrix A[Tex]\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}    [/Tex] = is a symmetric matrix.

(i) Given: A =[Tex]\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}    [/Tex]     

Now, A’=[Tex]\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}’=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}    [/Tex] 

∵ A = A’

∴ A is a symmetric matrix.

(ii) Given: A = [Tex]\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}[/Tex]

Now, A’=[Tex]\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}’=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}[/Tex]

∵ A = A’

∴ A is a symmetric matrix.

Question 8.  For the matrix A =[Tex]\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}    [/Tex], verify that:

(i) (A + A′) is a symmetric matrix

(ii) (A – A′) is a skew symmetric matrix

Solution:

(i) Given: A =[Tex]\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}[/Tex]

Let B = (A+A’) = [Tex]\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}+\begin{bmatrix}1 & 6 \\5 & 7 \\\end{bmatrix}=\begin{bmatrix}1+1 & 5+6 \\6+5 & 7+7 \\\end{bmatrix}=\begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}[/Tex]

Now, B’ = (A+A’)’ = [Tex]\begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}’=\begin{bmatrix}2 & 11 \\11 & 14 \\\end{bmatrix}[/Tex]

∵ B = B’

∴ B=(A+A’) is a symmetric matrix.

(ii) Given: A =[Tex]\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}[/Tex]

Let B = (A-A’) =[Tex]\begin{bmatrix}1 & 5 \\6 & 7 \\\end{bmatrix}-\begin{bmatrix}1 & 6 \\5 & 7 \\\end{bmatrix}=\begin{bmatrix}1-1 & 5-6 \\6-5 & 7-7 \\\end{bmatrix}=\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}[/Tex]

Now, B’ = (A-A’)’ =[Tex]\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}’=\begin{bmatrix}0 & 1 \\-1 & 0 \\\end{bmatrix}=-\begin{bmatrix}0 & -1 \\1 & 0 \\\end{bmatrix}[/Tex]

∵ -B = B’

∴ B=(A-A’) is a skew symmetric matrix.

Question 9. Find 1/2(A+A’) and 1/2(A-A’) ,when A =[Tex]\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}    [/Tex].

Solution:

Given: A = [Tex]\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}[/Tex]

∴  A’ = [Tex]\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}’=\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}[/Tex]

Now,  A+A’ = +[Tex]\frac{1}{2}\{\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}+\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 +0& a-a & b-b\\-a+a & 0+0 & c-c\\-b+b & -c+c & 0+0\end{bmatrix}=\frac{1}{2}\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}[/Tex]

Now, A-A’ =[Tex]\frac{1}{2}\{\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}-\begin{bmatrix}0 & -a & -b\\a & 0 & -c\\b & c & 0\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 -0& a+a & b+b\\-a-a & 0-0 & c+c\\-b-b & -c-c & 0-0\end{bmatrix}=\frac{1}{2}\begin{bmatrix}0 & 2a & 2b\\-2a & 0 & 2c\\-2b & -2c & 0\end{bmatrix}=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{bmatrix}[/Tex]

Question 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) [Tex]\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}[/Tex]

(ii) [Tex]\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}[/Tex]

(iii) [Tex]\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}[/Tex]

(iv) [Tex]\begin{bmatrix}1 & 5 \\-1& 2 \\\end{bmatrix}[/Tex]

Solution:

(i) Given : A =[Tex]\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}[/Tex]

⇒ A’=[Tex]\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}[/Tex]

Let P = [Tex]\frac{1}{2}(A+A’)[/Tex]

and Q = [Tex]\frac{1}{2}(A-A’)[/Tex]

Now, P =[Tex]\frac{1}{2} \{\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}+\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}6 & 6 \\6 & -2 \\\end{bmatrix}=\begin{bmatrix}3 & 3 \\3 & -1 \\\end{bmatrix}    [/Tex]…..(1)

& P’ = [Tex]\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}’=\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}[/Tex]

∵ P=P’

∴ P is a symmetric matrix.

Now, Q =[Tex]\frac{1}{2}(A-A’)=\frac{1}{2} \{\begin{bmatrix}3 & 5 \\1 & -1 \\\end{bmatrix}-\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}0 & 4 \\-4 & 0 \\\end{bmatrix}=\begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}    [/Tex]…..(2)

& Q’ = [Tex]\begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}’=-\begin{bmatrix}0 & -2 \\2 & 0 \\\end{bmatrix}[/Tex]

∵ -Q=Q’ 

∴ Q is a skew symmetric matrix.

By adding (1) and (2), we get,

[Tex]\begin{bmatrix}3 & 1 \\5 & -1 \\\end{bmatrix}+\begin{bmatrix}0 & 2 \\-2 & 0 \\\end{bmatrix}=\begin{bmatrix}3 & 5\\1 & -1 \\\end{bmatrix}[/Tex]

Therefore, A =P + Q

(ii) Given : [Tex]\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}[/Tex]

⇒ A’=[Tex]\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}[/Tex]

P = [Tex]\frac{1}{2}(A+A’)=\frac{1}{2}(\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}+\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix})=\frac{1}{2}\begin{bmatrix}12 & -4 & 4\\-4 & 6 & -2\\4 & -2 & 6\end{bmatrix}=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}[/Tex]

…..(1)

Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2}(\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}-\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix})=\frac{1}{2}\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}[/Tex]

……(2)

By adding (1) and (2), we get,

\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}[Tex]\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}+\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}[/Tex]

Therefore, A =P + Q

(iii) Given: A =[Tex]\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}[/Tex]

⇒ A’=[Tex]\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix}[/Tex]

P = }[Tex]\frac{1}{2}(A+A’)=\frac{1}{2}(\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}+\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix})=\begin{bmatrix}6 & 1 & -5\\1 & -4 & -4\\-5 & -4 & 4\end{bmatrix}=\begin{bmatrix}3 & 1/2 & -5/2\\1/2 & -2 & -2\\-5/2 & -2 & 2\end{bmatrix}    [/Tex]…..(1)

Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2}(\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}-\begin{bmatrix}3 & -2 & -4\\3 & -2 & -5\\-1 & 1 & 2\end{bmatrix})=\frac{1}{2}\begin{bmatrix}0 & 5 & 3\\-5 & 0 & 6\\-3 & -6 & 0\end{bmatrix}=\begin{bmatrix}0 & 5/2 & 3/2\\-5/2 & 0 & 3\\-3/2 & -3 & 0\end{bmatrix}    [/Tex]……(2)

By adding (1) and (2), we get

}[Tex]\begin{bmatrix}3 & 1/2 & -5/2\\1/2 & -2 & -2\\-5/2 & -2 & 2\end{bmatrix}+\begin{bmatrix}0 & 5/2 & 3/2\\-5/2 & 0 & 3\\-3/2 & -3 & 0\end{bmatrix}=\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}[/Tex]

Therefore, A =P + Q

(iv) Given: A = [Tex]\begin{bmatrix}1 & 5 \\-1& 2 \\\end{bmatrix}[/Tex]

⇒ A’= [Tex]\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}[/Tex]

P =[Tex]\frac{1}{2} \{\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}+\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}2 & 4 \\4 & 4 \\\end{bmatrix}=\begin{bmatrix}1 & 2 \\2 & 2 \\\end{bmatrix}[/Tex]

…..(1)

Q = [Tex]\frac{1}{2}(A-A’)=\frac{1}{2} \{\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}-\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}\}=\frac{1}{2}\begin{bmatrix}1 & -1 \\5 & 2 \\\end{bmatrix}=\begin{bmatrix}0 & 3 \\-3 & 0 \\\end{bmatrix}[/Tex]

…..(2)

By adding (1) and (2), we get

[Tex]\begin{bmatrix}1 & 2 \\2 & 2 \\\end{bmatrix}+\begin{bmatrix}0 & 3 \\-3 & 0 \\\end{bmatrix}=\begin{bmatrix}1 & 5 \\-1 & 2 \\\end{bmatrix}[/Tex]

Therefore, A =P + Q

Question 11. If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix (B) Symmetric matrix

(C) Zero matrix (D) Identity matrix 

Solution:

Given: A and B are symmetric matrices.

⇒ A=A’

⇒ B=B’

Now, ( AB – BA)’ =(AB)’-(BA)’              [∵ (X-Y)’=X’-Y’]

                          =B’A’-A’B’                [∵ (XY)’=Y’X’]

                         =BA-AB                   [∵ Given]

                        = -(AB-BA)

∴(AB-BA) is a skew symmetric matrix.

∴ The option (A) is correct.

Question 12. If A =[Tex]\begin{bmatrix}cosα & -sinα \\sinα & cosα \\\end{bmatrix}   [/Tex], and A + A′ = I, then the value of α is

(A)π/6    (B) π/3

(C) π    (D)3π/2

Solution:

[Tex]\begin{bmatrix}cosα & -sinα \\sinα & cosα \\\end{bmatrix}+\begin{bmatrix}cosα & sinα \\-sinα & cosα \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}[/Tex]

[Tex]\begin{bmatrix}2cosα & 0 \\0 & 2cosα \\\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1 \\\end{bmatrix}[/Tex]

On comparing both sides, we get

           2cosα = 1

⇒      cosα = [Tex]\frac{1}{2}[/Tex]

⇒      cosα = cos[Tex]\frac{π}{3}[/Tex]

⇒      α = [Tex]\frac{π}{3}[/Tex]

∴ The option (B) is correct.

Summary

Chapter 3 of the Class 12 NCERT Mathematics textbook, “Matrices,” explores essential concepts such as matrix operations and their applications. Exercise 3.3 focuses on practical problems involving matrix addition, subtraction, and multiplication. This exercise provides step-by-step solutions to help students understand and apply matrix operations effectively. Key topics include matrix addition and subtraction, matrix multiplication, determinants, and finding matrix inverses.

FAQs on Matrices

What are matrices, and why are they important in mathematics?

Matrices are rectangular arrays of numbers arranged in rows and columns. They are important in mathematics because they provide a systematic way to handle and solve systems of linear equations, perform linear transformations, and represent data in various applications.

Can matrices be multiplied if their dimensions do not match?

Matrices can only be multiplied if the number of columns in the first matrix is equal to the number of rows in the second matrix. If this condition is not met, matrix multiplication is not possible.

How do you find the inverse of a matrix, and when does a matrix have an inverse?

To find the inverse of a matrix A, the matrix must be square (same number of rows and columns) and have a non-zero determinant. The inverse of A can be found using various methods, such as the adjoint method or Gaussian elimination. The matrix A has an inverse if and only if its determinant is non-zero.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

CBSE Class 12 Maths Syllabus 2025-26 with Marks Distribution

The table below shows the marks weightage along with the number of periods required for teaching. The Maths theory paper is of 80 marks, and the internal assessment is of 20 marks which totally comes out to be 100 marks.

CBSE Class 12 Maths Syllabus And Marks Distribution 2023-24

Max Marks: 80

No.UnitsMarks
I.Relations and Functions08
II.Algebra10
III.Calculus35
IV.Vectors and Three – Dimensional Geometry14
V.Linear Programming05
VI.Probability08
Total Theory80
Internal Assessment20
Grand Total100

Unit-I: Relations and Functions

1. Relations and Functions

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operations on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability

Continuity and differentiability, derivative of composite functions, chain rule, derivative of inverse trigonometric functions like sin-1 x, cos-1 x and tan-1 x, derivative of implicit functions. Concept of exponential and logarithmic functions.
Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives

Applications of derivatives: rate of change of quantities, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).

3. Integrals 

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

dy/dx + py = q, where p and q are functions of x or constants.

dx/dy + px = q, where p and q are functions of y or constants.

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

Students can go through the CBSE Class 12 Syllabus to get the detailed syllabus of all subjects. Get access to interactive lessons and videos related to Maths and Science with ANAND CLASSES’S App/ Tablet.

Frequently Asked Questions on CBSE Class 12 Maths Syllabus 2025-26

Q1

Is Calculus an important chapter in the CBSE Class 12 Maths Syllabus?

Yes, Calculus is an important chapter in the CBSE Class 12 Maths Syllabus. It is for 35 marks which means that if a student is thorough with this chapter will be able to pass the final exam.

Q2

How many units are discussed in the CBSE Class 12 Maths Syllabus?

In the CBSE Class 12 Maths Syllabus, about 6 units are discussed, which contains a total of 13 chapters.

Q3

How many marks are allotted for internals in the CBSE Class 12 Maths syllabus?

About 20 marks are allotted for internals in the CBSE Class 12 Maths Syllabus. Students can score it with ease through constant practice.