Anand Classes presents the most accurate and exam-focused Matrices NCERT Solutions for Exercise 3.3, Class 12 Math Chapter 3, designed to help students master matrix operations, elementary transformations, and core problem-solving skills. These solutions follow the latest CBSE guidelines and provide step-by-step explanations that simplify even the most challenging questions. Ideal for last-minute revision as well as deep conceptual learning, these notes ensure maximum score improvement for Class 12 board exams. Click the print button to download study material and notes.

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NCERT Question 6: If
$$
\text{(i)}\quad A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix},
\qquad
\text{(ii)}\quad A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix},
$$
verify that in each case $A’ A = I$ (where $A’$ denotes the transpose and $I$ is the $2\times2$ identity matrix).

Solution

We will compute $A’ A$ for each case and simplify using the identities
$$\sin^2\alpha+\cos^2\alpha=1,\qquad \sin\alpha\cos\alpha=\cos\alpha\sin\alpha.$$

(i)
Given
$$A=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}.$$
Its transpose is
$$A’=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}.$$

Compute the product $A’A$:
$$
A’A=
\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}
\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}.
$$

Multiply (row by column) and simplify each entry:

• Top-left:
  $$\cos\alpha\cos\alpha + (-\sin\alpha)(-\sin\alpha)
  =\cos^2\alpha+\sin^2\alpha=1.$$
• Top-right:
  $$\cos\alpha\sin\alpha + (-\sin\alpha)\cos\alpha
  =\cos\alpha\sin\alpha-\sin\alpha\cos\alpha=0.$$
• Bottom-left:
  $$\sin\alpha\cos\alpha + \cos\alpha(-\sin\alpha)
  =\sin\alpha\cos\alpha-\sin\alpha\cos\alpha=0.$$
• Bottom-right:
  $$\sin\alpha\sin\alpha + \cos\alpha\cos\alpha
  =\sin^2\alpha+\cos^2\alpha=1.$$

Thus
$$
A’A=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.
$$

(ii)
Given
$$A=\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}.$$
Its transpose is
$$A’=\begin{bmatrix}\sin\alpha & -\cos\alpha\\\cos\alpha & \sin\alpha\end{bmatrix}.$$

Compute $A’A$:
$$
A’A=
\begin{bmatrix}\sin\alpha & -\cos\alpha\\\cos\alpha & \sin\alpha\end{bmatrix}
\begin{bmatrix}\sin\alpha & \cos\alpha\\-\cos\alpha & \sin\alpha\end{bmatrix}.
$$

Multiply and simplify:

• Top-left:
  $$\sin\alpha\sin\alpha + (-\cos\alpha)(-\cos\alpha)
  =\sin^2\alpha+\cos^2\alpha=1.$$
• Top-right:
  $$\sin\alpha\cos\alpha + (-\cos\alpha)\sin\alpha
  =\sin\alpha\cos\alpha-\cos\alpha\sin\alpha=0.$$
• Bottom-left:
  $$\cos\alpha\sin\alpha + \sin\alpha(-\cos\alpha)
  =\cos\alpha\sin\alpha-\cos\alpha\sin\alpha=0.$$
• Bottom-right:
  $$\cos\alpha\cos\alpha + \sin\alpha\sin\alpha
  =\cos^2\alpha+\sin^2\alpha=1.$$

So again
$$
A’A=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.
$$

Final Result

$$
\boxed{A’A=I\ \text{for both (i) and (ii).}}
$$

Practice these verification problems to get comfortable with matrix multiplication and trigonometry identities — great preparation for NCERT exercises and competitive exams. Download clear, step-by-step notes and solved problems from Anand Classes to sharpen your matrix algebra skills and boost exam performance.

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NCERT Question.7 :
(i) Show that $$A=\displaystyle\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}$$ is a symmetric matrix.
(ii) Show that $$A=\displaystyle\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}$$ is a skew symmetric matrix.

Solution
A matrix $A$ is symmetric iff $A^{T}=A$. We compute the transpose in each case.
(i) Let
$$
A=\begin{bmatrix}1 & -1 & 5\\-1 & 2 & 1\\5 & 1 & 3\end{bmatrix}.
$$
The transpose $A^{T}$ is obtained by interchanging rows and columns:
$$
A^{T}=
\begin{bmatrix}
1 & -1 & 5\\
-1 & 2 & 1\\
5 & 1 & 3
\end{bmatrix}.
$$
We see $A^{T}=A$. Hence $A$ is symmetric.

(ii) Let
$$
A=\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}.
$$
Compute the transpose:
$$
A^{T}=
\begin{bmatrix}
0 & -1 & 1\\
1 & 0 & -1\\
-1 & 1 & 0
\end{bmatrix}.
$$
Reordering entries (row/column interchange) gives not the same matrix as the original:
$$
A^{T}=-\begin{bmatrix}0 & 1 & -1\\-1 & 0 & 1\\1 & -1 & 0\end{bmatrix}=-A.
$$
Therefore $A$ is skew symmetric.

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NCERT Question 8: For the matrix
$$A=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}$$
verify that
(i) $(A + A’)$ is a symmetric matrix, and
(ii) $(A – A’)$ is a skew-symmetric matrix.

Solution
Recall: for any matrix $M$, the transpose is denoted $M’$.

A matrix $(S)$ is symmetric if $(S’ = S)$.

A matrix $(K)$ is skew-symmetric if $(K’ = -K)$.

Given
$$
A=\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}
\quad\text{so}\quad
A’=\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}.
$$

(i) Check $(A + A’)$
Compute
$$
A + A’= \begin{bmatrix}1 & 5\\6 & 7\end{bmatrix} + \begin{bmatrix}1 & 6\\5 & 7\end{bmatrix} $$

$$
A + A’=\begin{bmatrix}1+1 & 5+6\\6+5 & 7+7\end{bmatrix} $$

$$
A + A’=\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}$$

Take transpose:
$$
(A + A’)’ =
\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}’$$

$$
(A + A’)’ =\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}$$

$$(A + A’)’ = A + A’$$

Since $(A + A’)’ = A + A’$, the matrix $(A + A’)$ is symmetric.

(ii) Check $(A – A’)$

Compute
$$
A – A’=
\begin{bmatrix}1 & 5\\6 & 7\end{bmatrix}-\begin{bmatrix}1 & 6\\5 & 7\end{bmatrix}$$

$$A – A’=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}.$$

Take transpose:
$$ (A – A’)’ =\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}’$$

$$ (A – A’)’ =\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}$$

$$ (A – A’)’ = -\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$$

$$ (A – A’)’ = -(A – A’).$$

Since $(A – A’)’ = -(A – A’)$, the matrix $(A – A’)$ is skew-symmetric.

Final Answer

$$
\boxed{A + A’=\begin{bmatrix}2 & 11\\11 & 14\end{bmatrix}\ \text{(symmetric)},\quad
A – A’=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}\ \text{(skew-symmetric)}}
$$

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NCERT Question.9 : Find $\frac{1}{2}(A+A’)$ and $\frac{1}{2}(A-A’)$ when
$$ A=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0\end{bmatrix}$$

Solution
Given

$$ A=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0\end{bmatrix}$$

First compute the transpose $A’$. Interchanging rows and columns gives
$$
A’=\begin{bmatrix}
0 & -a & -b\\
a & 0 & -c\\
b & c & 0
\end{bmatrix}.
$$

Now compute the sum and difference.

Compute $(A+A’)$ :

$$A+A’=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0\end{bmatrix}+\begin{bmatrix}
0 & -a & -b\\
a & 0 & -c\\
b & c & 0
\end{bmatrix}$$

$$A+A’=\begin{bmatrix}
0+0 & a+(-a) & b+(-b)\\-a + a & 0+0 & c+(-c)\\-b + b & -c + c & 0+0
\end{bmatrix}$$

$$A+A’=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}.
$$

$$\frac{1}{2}(A+A’)=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

Therefore
$$
\boxed{\frac{1}{2}(A+A’)=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}.
$$

Compute $(A-A’)$ :

$$A-A’=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0\end{bmatrix}-\begin{bmatrix}
0 & -a & -b\\
a & 0 & -c\\
b & c & 0
\end{bmatrix}$$

$$A-A’=\begin{bmatrix}
0-0 & a-(-a) & b-(-b)\\-a – a & 0-0 & c-(-c)\\-b – b & -c -c & 0-0
\end{bmatrix}$$

$$A-A’=\begin{bmatrix}
0 & 2a & 2b\\
-2a & 0 & 2c\\
-2b & -2c & 0
\end{bmatrix}.
$$

$$\frac{1}{2}(A-A’)=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0
\end{bmatrix}$$

Hence
$$
\boxed{\frac{1}{2}(A-A’)=\begin{bmatrix}0 & a & b\\-a & 0 & c\\-b & – c & 0
\end{bmatrix}=A}$$

So $\tfrac{1}{2}(A+A’)$ is the zero matrix and $\tfrac{1}{2}(A-A’)$ returns the original skew-symmetric matrix $A$.

Continue practicing these decompositions — download complete NCERT solution notes and solved examples by Anand Classes, perfect for CBSE revision and JEE practice to build strong matrix algebra fundamentals.

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NCERT Question 10 : Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
$$
\text{(i)}\quad
A=\begin{bmatrix}3 & 5\\1 & -1\end{bmatrix}\qquad
\text{(ii)}\quad
A=\begin{bmatrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{bmatrix}
$$
$$
\text{(iii)}\quad
A=\begin{bmatrix}3 & 3 & -1\\-2 & -2 & 1\\-4 & -5 & 2\end{bmatrix}\qquad
\text{(iv)}\quad
A=\begin{bmatrix}1 & 5\\-1 & 2\end{bmatrix}
$$

Solution : Method (used for every part).
Any square matrix $A$ can be written uniquely as
$$ A = P + Q $$

where
$$P=\tfrac{1}{2}(A+A’)\ \text{(symmetric)}$$

$$Q=\tfrac{1}{2}(A-A’)\ \text{(skew-symmetric)}$$

We apply this formula to each matrix.

(i)

$$A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$$

$$A’=\begin{bmatrix}3&1\\5&-1\end{bmatrix}$$

$$P=\frac{1}{2}(A+A’)$$

$$P=\frac{1}{2}\begin{bmatrix}6&6\\6&-2\end{bmatrix}$$

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

$$Q=\frac{1}{2}(A-A’)$$

$$Q=\frac{1}{2}\begin{bmatrix}0&4\\-4&0\end{bmatrix}$$

$$Q=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$$

$$
\boxed{A=\begin{bmatrix}3&3\\3&-1\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}}
$$

(ii)

$$A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}$$

Here $A’=A$ (matrix is already symmetric).

$$
P=\frac{1}{2}(A+A’)=A,\qquad Q=\frac{1}{2}(A-A’)=0,
$$

$$
\boxed{A=\begin{bmatrix}6&-2&2\\-2&3&-1\\2&-1&3\end{bmatrix}+\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}}
$$

(iii)

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

Compute $$A’=\begin{bmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{bmatrix}$$

Now
$$P=\frac{1}{2}(A+A’)$$

$$P=\begin{bmatrix}
3 & \tfrac{1}{2} & -\tfrac{5}{2}\\\tfrac{1}{2} & -2 & -2\\
-\tfrac{5}{2} & -2 & 2
\end{bmatrix}$$

$$Q=\frac{1}{2}(A-A’)$$

$$Q=\begin{bmatrix}
0 & \tfrac{5}{2} & \tfrac{3}{2}\\-\tfrac{5}{2} & 0 & 3\\-\tfrac{3}{2} & -3 & 0
\end{bmatrix}.
$$

(You can check $P’=P$ and $Q’=-Q$)

$$
\boxed{A=
\begin{bmatrix}
3 & \tfrac{1}{2} & -\tfrac{5}{2}\\
\tfrac{1}{2} & -2 & -2\\
-\tfrac{5}{2} & -2 & 2
\end{bmatrix}
+
\begin{bmatrix}
0 & \tfrac{5}{2} & \tfrac{3}{2}\\
-\tfrac{5}{2} & 0 & 3\\
-\tfrac{3}{2} & -3 & 0
\end{bmatrix}}
$$

(iv)

$$A=\begin{bmatrix}1&5\\-1&2\end{bmatrix}$$

$$A’=\begin{bmatrix}1&-1\\5&2\end{bmatrix}$$

$$
P=\frac{1}{2}(A+A’)=\begin{bmatrix}1&2\\2&2\end{bmatrix}$$

$$Q=\frac{1}{2}(A-A’)=\begin{bmatrix}0&3\\-3&0\end{bmatrix}$$

$$
\boxed{A=\begin{bmatrix}1&2\\2&2\end{bmatrix}+\begin{bmatrix}0&3\\-3&0\end{bmatrix}}
$$

Quick recap: for each given matrix $A$ we have produced $P=\tfrac12(A+A’)$ (symmetric) and $Q=\tfrac12(A-A’)$ (skew-symmetric) so that $A=P+Q$.

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NCERT Question 11: If $A, B$ are symmetric matrices of the same order, then $(AB-BA)$ is a
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Solution
Since $A$ and $B$ are symmetric matrices,

$$A’ = A\qquad\text{and}\qquad B’ = B$$

Consider the transpose of $(AB-BA)$.

Using $(X-Y)’=X’-Y’$ and $(XY)’=Y’X’$

$$
(AB-BA)’ = (AB)’- (BA)’ $$

$$ (AB-BA)’ = B’A’ – A’B’ $$

since $A’=A, B’=B$

$$(AB-BA)’= B A – A B $$

$$(AB-BA)’= -(AB-BA)$$

We have shown
$$ (AB-BA)’ = -(AB-BA) $$
which is precisely the condition for a matrix to be skew-symmetric.

Answer: $(\boxed{\text{(A) Skew symmetric matrix}})$

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NCERT Question 12 : If
$$A=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}$$
and $A+A’ = I,$ find the value of $\alpha$ is
(A)π/6    (B) π/3    (C) π    (D)3π/2

Solution
Given
$$A=\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}$$

Write the transpose $A’$:
$$A’=\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}.$$

Compute $A+A’$:
$$ A+A’=
\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}
+
\begin{bmatrix}\cos\alpha & \sin\alpha\\-\sin\alpha & \cos\alpha\end{bmatrix}$$

$$ A+A’=\begin{bmatrix}2\cos\alpha & 0\\0 & 2\cos\alpha\end{bmatrix}.
$$

Given $A+A’=I$,

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

$$ \begin{bmatrix}2\cos\alpha & 0\\0 & 2\cos\alpha\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}.
$$

Compare entries:

$$2\cos\alpha=1\quad\Rightarrow\quad \cos\alpha=\tfrac{1}{2}.$$

Therefore
$$\alpha=\frac{\pi}{3}$$
In the principal values; any solution of $\cos\alpha=\tfrac12$ is $\alpha=2\pi m\pm\frac{\pi}{3}$ for integer $m$, but the standard answer is $\alpha=\dfrac{\pi}{3}$.

Final Answer : Correct Option (B)

$$
\boxed{\alpha=\dfrac{\pi}{3}}
$$

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⭐ FAQ Section

Q1. What topics are covered in Matrices Exercise 3.3 of Class 12?

Exercise 3.3 focuses on matrix multiplication, transpose rules, properties of matrix operations, and applying identities to simplify expressions.

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Q2. Are the NCERT Solutions for Matrices Exercise 3.3 helpful for board exams?

Yes. The solutions follow the NCERT method and cover all important concepts frequently asked in CBSE Class 12 board exams.

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Q3. Can I download the Matrices Exercise 3.3 Class 12 PDF for free?

Absolutely. Anand Classes provides free downloadable PDFs of NCERT Solutions, study notes, and practice material.

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Q4. Who prepares the NCERT Solutions provided here?

All solutions are created by experienced faculty at Anand Classes with years of expertise in CBSE Mathematics.

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Q5. Do these solutions follow the latest NCERT and CBSE updates?

Yes, the solutions strictly follow the latest NCERT textbook and the updated CBSE examination pattern.

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Q6. Are step-by-step matrix operations included in the solutions?

Yes. Each question includes clear, step-wise calculations to help students understand every concept thoroughly.