Anand Classes brings you the most accurate and easy-to-understand NCERT Solutions for Matrices Exercise 3.2 Class 12 Chapter 3, designed to help students strengthen their conceptual understanding of matrix operations, properties, and solved examples. These solutions follow the latest NCERT guidelines and are crafted to simplify concepts for board exam preparation and competitive entrance test aspirants. With step-by-step answers, clear explanations, and exam-focused notes, this PDF ensures scoring confidence in the Matrices chapter. Click the print button to download study material and notes.
Access NCERT Solutions for Matrices Exercise 3.2 Class 12 Math Chapter-3
NCERT Question.8 : Find $X$ if
$$Y=\begin{bmatrix}3 & 2\\ 1 & 4\end{bmatrix}\; \text{and}\;\; 2X+Y=\begin{bmatrix}1 & 0\\ -3 & 2\end{bmatrix}$$
Solution:
Given
$$2X + Y = \begin{bmatrix}1 & 0\\ -3 & 2\end{bmatrix}$$
and
$$Y = \begin{bmatrix}3 & 2\\ 1 & 4\end{bmatrix}$$
Substituting the value of $Y$:
$$2X + \begin{bmatrix}3 & 2\\ 1 & 4\end{bmatrix} = \begin{bmatrix}1 & 0\\ -3 & 2\end{bmatrix}$$
Subtracting the matrix (Y) from both sides:
$$2X = \begin{bmatrix}1 & 0\\ -3 & 2\end{bmatrix} – \begin{bmatrix}3 & 2\\ 1 & 4\end{bmatrix}$$
Performing subtraction:
$$2X = \begin{bmatrix}1-3 & 0-2\\ -3-1 & 2-4\end{bmatrix}$$
$$2X = \begin{bmatrix}-2 & -2\\ -4 & -2\end{bmatrix}$$
Dividing both sides by 2:
$$X = \frac{1}{2}\begin{bmatrix}-2 & -2\\ -4 & -2\end{bmatrix}$$
$$X = \begin{bmatrix}-1 & -1\\ -2 & -1\end{bmatrix}$$
Matrices questions like this are important for CBSE and competitive exams. Keep practicing more chapter-wise solutions from Anand Classes for scoring high in board exams and strengthening matrix algebra concepts.
NCERT Question.9 : Find $x$ and $y$, if
$$2\begin{bmatrix}1 & 3\\ 0 & x\end{bmatrix} + \begin{bmatrix}y & 0\\ 1 & 2\end{bmatrix} = \begin{bmatrix}5 & 6\\ 1 & 8\end{bmatrix}$$
Solution
Given:
$$2\begin{bmatrix}1 & 3\\ 0 & x\end{bmatrix} = \begin{bmatrix}2 & 6\\ 0 & 2x\end{bmatrix}$$
Now add the matrices:
$$\begin{bmatrix}2 & 6\\ 0 & 2x\end{bmatrix} + \begin{bmatrix}y & 0\\ 1 & 2\end{bmatrix} = \begin{bmatrix}2+y & 6\\ 1 & 2x+2\end{bmatrix}$$
According to given equation :
$$\begin{bmatrix}2+y & 6\\ 1 & 2x+2\end{bmatrix}=\begin{bmatrix}5 & 6\\ 1 & 8\end{bmatrix}$$
Equating corresponding elements:
$$
2 + y = 5 \Rightarrow y = 3
$$
$$
2x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3
$$
So,
$$
\boxed{x = 3, y = 3}
$$
Matrices problems like these are very helpful for strengthening algebraic skills for board exams and competitive tests. Continue practicing more solutions from Anand Classes for higher accuracy and confidence.
NCERT Question.10 : Solve the equation for $x, y, z$ and $t$, if :
$$2\begin{bmatrix}x & z\\ y & t\end{bmatrix} + 3\begin{bmatrix}1 & -1\\ 0 & 2\end{bmatrix} = 3\begin{bmatrix}3 & 5\\ 4 & 6\end{bmatrix}$$
Solution
Perform scalar multiplication:
$$2\begin{bmatrix}x & z\\ y & t\end{bmatrix} = \begin{bmatrix}2x & 2z\\ 2y & 2t\end{bmatrix}$$
$$3\begin{bmatrix}1 & -1\\ 0 & 2\end{bmatrix} = \begin{bmatrix}3 & -3\\ 0 & 6\end{bmatrix}$$
Substitute:
$$2\begin{bmatrix}x & z\\ y & t\end{bmatrix} + 3\begin{bmatrix}1 & -1\\ 0 & 2\end{bmatrix}=\begin{bmatrix}2x & 2z\\ 2y & 2t\end{bmatrix}+\begin{bmatrix}3 & -3\\ 0 & 6\end{bmatrix}=\begin{bmatrix}2x + 3 & 2z – 3\\ 2y + 0 & 2t + 6\end{bmatrix} $$
According to given equation :
$$\begin{bmatrix}2x + 3 & 2z – 3\\ 2y + 0 & 2t + 6\end{bmatrix}=3\begin{bmatrix}3 & 5\\ 4 & 6\end{bmatrix}$$
$$\begin{bmatrix}2x + 3 & 2z – 3\\ 2y + 0 & 2t + 6\end{bmatrix}= \begin{bmatrix}9 & 15\\ 12 & 18\end{bmatrix}$$
Equating corresponding elements:
$$
2x + 3 = 9,\quad 2z – 3 = 15,\quad 2y = 12,\quad 2t + 6 = 18
$$
Solving:
$$
2x = 6 \Rightarrow x = 3
$$
$$
2y = 12 \Rightarrow y = 6
$$
$$
2z = 18 \Rightarrow z = 9
$$
$$
2t = 12 \Rightarrow t = 6
$$
$$
\boxed{x = 3,\ y = 6,\ z = 9,\ t = 6}
$$
Mastering matrix equations is essential for strong fundamentals in algebra and competitive exam preparation. Keep practicing with Anand Classes for top-quality results!
NCERT Question.11 : If
$$x\begin{bmatrix}2\\ 3\end{bmatrix} + y\begin{bmatrix}-1\\ 1\end{bmatrix} = \begin{bmatrix}10\\ 5\end{bmatrix}$$
find the values of x and y.
Solution
Given matrix equation:
$$x\begin{bmatrix}2\\ 3\end{bmatrix} + y\begin{bmatrix}-1\\ 1\end{bmatrix} = \begin{bmatrix}10\\ 5\end{bmatrix}$$
Scalar multiplication:
$$\begin{bmatrix}2x\\ 3x\end{bmatrix} + \begin{bmatrix}-y\\ y\end{bmatrix} = \begin{bmatrix}10\\ 5\end{bmatrix}$$
Add corresponding elements:
$$\begin{bmatrix}2x – y\\ 3x + y\end{bmatrix} = \begin{bmatrix}10\\ 5\end{bmatrix}$$
Equate entries:
$$
2x – y = 10 \quad …(1)
$$
$$
3x + y = 5 \quad …(2)
$$
Add (1) and (2):
$$
5x = 15 \Rightarrow x = 3
$$
Substitute $x = 3$ in (2):
$$
3(3) + y = 5 \Rightarrow 9 + y = 5 \Rightarrow y = -4
$$
$$
\boxed{x = 3,\quad y = -4}
$$
NCERT Question.12 : Given
$$
3\begin{bmatrix}x & y \\z & w\end{bmatrix}=
\begin{bmatrix}x & 6 \\ -1 & 2w\end{bmatrix}
+
\begin{bmatrix}4 & x+y \\ z+w & 3\end{bmatrix}
$$
find the values of $x, y, z$ and $w$.
Solution
Given
$$
3\begin{bmatrix}x & y \\z & w\end{bmatrix}=
\begin{bmatrix}x & 6 \\ -1 & 2w\end{bmatrix}
+
\begin{bmatrix}4 & x+y \\ z+w & 3\end{bmatrix}
$$
Multiply the left-hand side by 3:
$$
\begin{bmatrix}3x & 3y \\ 3z & 3w\end{bmatrix}=\begin{bmatrix}x+4 & 6+(x+y) \\ -1+(z+w) & 2w+3\end{bmatrix}
$$
$$
\begin{bmatrix}3x & 3y \\ 3z & 3w\end{bmatrix}= \begin{bmatrix}x+4 & x+y+6 \\ z+w-1 & 2w+3\end{bmatrix}
$$
Equate corresponding elements:
$$
3x = x + 4 \quad …(1)
$$
$$
3y = x + y + 6 \quad …(2)
$$
$$
3z = z + w – 1 \quad …(3)
$$
$$
3w = 2w + 3 \quad …(4)
$$
Solve them one by one:
From (1):
$$
3x – x = 4 \Rightarrow 2x = 4 \Rightarrow x = 2
$$
From (4):
$$
3w – 2w = 3 \Rightarrow w = 3
$$
Substitute $x = 2$ into (2):
$$
3y = 2 + y + 6 \Rightarrow 3y – y = 8 \Rightarrow 2y = 8 \Rightarrow y = 4
$$
Substitute $w = 3$ into (3):
$$
3z = z + 3 – 1 \Rightarrow 3z – z = 2 \Rightarrow 2z = 2 \Rightarrow z = 1
$$
Final Answer
$$
\boxed{x = 2,\quad y = 4,\quad z = 1,\quad w = 3}
$$
Boost your matrix algebra skills with expertly curated notes, solved NCERT problems, and exam-oriented practice material from Anand Classes — perfect for CBSE and JEE aspirants aiming for top scores.
NCERT Question.13 : Show that if
$$F(x)=\begin{bmatrix}\cos x & -\sin x & 0\\\sin x & \cos x & 0\\0 & 0 & 1\end{bmatrix}$$
then $F(x)F(y)=F(x+y)$.
Solution
Compute the product $F(x)F(y)$ where
$$F(x)=\begin{bmatrix}\cos x & -\sin x & 0\\\sin x & \cos x & 0\\0 & 0 & 1\end{bmatrix}$$
Multiply the matrices (showing each entry):
$$ F(x)F(y)=\begin{bmatrix}
\cos x & -\sin x & 0\\
\sin x & \cos x & 0\\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
\cos y & -\sin y & 0\\
\sin y & \cos y & 0\\
0 & 0 & 1
\end{bmatrix} $$
$$ =\begin{bmatrix}
(\cos x)(\cos y)+(-\sin x)(\sin y) & (\cos x)(-\sin y)+(-\sin x)(\cos y) & 0\\
(\sin x)(\cos y)+(\cos x)(\sin y) & (\sin x)(-\sin y)+(\cos x)(\cos y) & 0\\
0 & 0 & 1
\end{bmatrix}$$
Now simplify each entry using trigonometric addition formulas:
- Top-left entry:
$$\cos x\cos y-\sin x\sin y=\cos(x+y).$$ - Top-middle entry:
$$-\cos x\sin y-\sin x\cos y=-(\sin x\cos y+\cos x\sin y)=-\sin(x+y).$$ - Middle-left entry:
$$\sin x\cos y+\cos x\sin y=\sin(x+y).$$ - Middle-middle entry:
$$-\sin x\sin y+\cos x\cos y=\cos x\cos y-\sin x\sin y=\cos(x+y).$$ - Other (rightmost column and bottom row) entries remain $0$ or $1$ as in the product above.
Thus
$$
F(x)F(y)=
\begin{bmatrix}
\cos(x+y) & -\sin(x+y) & 0\\
\sin(x+y) & \cos(x+y) & 0\\
0 & 0 & 1
\end{bmatrix}.
$$
But this is exactly $F(x+y)$ by definition. Therefore the equality holds.
Final Result
$$
\boxed{F(x)F(y)=F(x+y)=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}}
$$
This identity shows that the family ${F(x)}$ is closed under multiplication and corresponds to rotations about the $z$-axis in 3D — excellent practice for linear algebra and geometry problems. For more clear geometric interpretations, worked examples, and downloadable NCERT-style notes, see Anand Classes’ detailed resources tailored for CBSE and competitive exam preparation.
FAQ Section
Q1. What is covered in Matrices Exercise 3.2 of Class 12 NCERT?
Exercise 3.2 focuses on matrix multiplication, properties of multiplication, and solving numerical problems based on matrix order and rules of multiplication.
Q2. Are these NCERT Solutions useful for CBSE Class 12 board exams?
Yes, the solutions are fully aligned with the CBSE exam pattern and help students score high by strengthening conceptual clarity.
Q3. Can I download the Matrices Exercise 3.2 PDF for free?
Yes, the PDF is available for free download and includes step-wise solutions prepared by Anand Classes.
Q4. Are these solutions useful for competitive exams like JEE?
Yes, understanding matrices is essential for JEE Main, CUET, NDA, and other entrance exams, and these solutions help build strong basics.
Q5. Do I need to refer to any additional book besides NCERT?
For most students, NCERT is sufficient. However, for advanced practice, you may refer to RD Sharma or previous years’ papers.
