Anand Classes brings brings you the most accurate and student-friendly NCERT Solutions for Matrices Exercise 3.2 Class 12 Math Chapter-3, prepared strictly according to the latest CBSE curriculum. These detailed step-by-step solutions help students understand matrix operations, properties, and problem-solving methods with clarity. Whether you are preparing for Class 12 board exams or revising important matrix concepts for competitive exams, these notes ensure strong conceptual understanding and scoring efficiency. 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.1 : Let
$$
A=\begin{bmatrix}2&4\\ 3&2\end{bmatrix}, B=\begin{bmatrix}1&3\\ -2&5\end{bmatrix}, C=\begin{bmatrix}-2&5\\ 3&4\end{bmatrix}
$$
Find each of the following:
(i) $A+B$
(ii) $A-B$
(iii) $3A-C$
(iv) $AB$
(v) $BA$
Solution:
(i) $A+B$
To add two matrices, we add their corresponding elements.
$$
A+B=\begin{bmatrix}2&4\\ 3&2\end{bmatrix}+\begin{bmatrix}1&3\\ -2&5\end{bmatrix}
$$
$$
A+B=\begin{bmatrix}2+1&4+3\\ 3+(-2)&2+5\end{bmatrix} = \begin{bmatrix}3&7\\ 1&7\end{bmatrix}
$$
(ii) $A-B$
To subtract two matrices, we subtract their corresponding elements.
$$
A-B=\begin{bmatrix}2&4\\ 3&2\end{bmatrix}-\begin{bmatrix}1&3\\ -2&5\end{bmatrix}
$$
$$
A-B=\begin{bmatrix}2-1&4-3\\ 3-(-2)&2-5\end{bmatrix} = \begin{bmatrix}1&1\\ 5&-3\end{bmatrix}
$$
(iii) $3A-C$
First, multiply matrix $A$ by scalar 3 and then subtract $C$.
$$
3A-C = 3\begin{bmatrix}2&4\\ 3&2\end{bmatrix} – \begin{bmatrix}-2&5\\ 3&4\end{bmatrix}
$$
$$
3A-C = \begin{bmatrix}6&12\\ 9&6\end{bmatrix} – \begin{bmatrix}-2&5\\ 3&4\end{bmatrix}
$$
$$
3A-C = \begin{bmatrix}6-(-2)&12-5\\ 9-3&6-4\end{bmatrix} = \begin{bmatrix}8&7\\ 6&2\end{bmatrix}
$$
(iv) $AB$
To multiply matrices $A$ and $B$, multiply the rows of $A$ by the columns of $B$.
$$
AB = \begin{bmatrix}2&4\\ 3&2\end{bmatrix} \begin{bmatrix}1&3\\ -2&5\end{bmatrix}
$$
$$
AB = \begin{bmatrix}(2\times 1)+(4\times -2) & (2\times 3)+(4\times 5)\\ (3\times 1)+(2\times -2) & (3\times 3)+(2\times 5)\end{bmatrix}
$$
$$
AB = \begin{bmatrix}2-8 & 6+20\\ 3-4 & 9+10\end{bmatrix} = \begin{bmatrix}-6&26\\ -1&19\end{bmatrix}
$$
(v) $BA$
Multiply the rows of $B$ by the columns of $A$. Note that matrix multiplication is generally not commutative ($AB \ne BA$).
$$
BA = \begin{bmatrix}1&3\\ -2&5\end{bmatrix} \begin{bmatrix}2&4\\ 3&2\end{bmatrix}
$$
$$
BA = \begin{bmatrix}(1\times 2)+(3\times 3) & (1\times 4)+(3\times 2)\\ (-2\times 2)+(5\times 3) & (-2\times 4)+(5\times 2)\end{bmatrix}
$$
$$
BA = \begin{bmatrix}2+9 & 4+6\\ -4+15 & -8+10\end{bmatrix} = \begin{bmatrix}11&10\\ 11&2\end{bmatrix}
$$
Final Result:
$$
\begin{aligned}
A+B &= \begin{bmatrix}3&7\\ 1&7\end{bmatrix},\quad
A-B = \begin{bmatrix}1&1\\ 5&-3\end{bmatrix},\\[1em]
3A-C &= \begin{bmatrix}8&7\\ 6&2\end{bmatrix},\quad
AB = \begin{bmatrix}-6&26\\ -1&19\end{bmatrix},\quad
BA = \begin{bmatrix}11&10\\ 11&2\end{bmatrix}
\end{aligned}
$$
This solution provides a clear step-by-step explanation of matrix operations including addition, subtraction, scalar multiplication, and matrix multiplication, making it perfect for students preparing for JEE, CBSE, or other competitive exams. Download notes by Anand Classes for top-quality study material and detailed examples on matrices.
NCERT Question.2 : Compute the following
$\\[1em](i) \begin{bmatrix}a&b\\ -b&a\end{bmatrix}+\begin{bmatrix}a&b\\ b&a\end{bmatrix}$
$\\[1em]
(ii)\begin{bmatrix}a^2+b^2&b^2+c^2\\ a^2+c^2&a^2+b^2\end{bmatrix}
+
\begin{bmatrix}2ab&2bc\\ -2ac&-2ab\end{bmatrix}$
$\\[1em]
(iii)\begin{bmatrix}-1&4&-6\\ 8&5&16\\ 2&8&5\end{bmatrix}
+
\begin{bmatrix}12&7&6\\ 8&0&5\\ 3&2&4\end{bmatrix}$
$\\[1em]
(iv)\begin{bmatrix}\cos^2 x&\sin^2 x\\sin^2 x&\cos^2 x\end{bmatrix}
+
\begin{bmatrix}\sin^2 x&\cos^2 x\\cos^2 x&\sin^2 x\end{bmatrix}$
Solution:
(i)
$$
\begin{bmatrix}a&b\\ -b&a\end{bmatrix}+\begin{bmatrix}a&b\\ b&a\end{bmatrix}
$$
$$
\begin{bmatrix}a&b\\ -b&a\end{bmatrix}+\begin{bmatrix}a&b\\ b&a\end{bmatrix}$$
\begin{bmatrix}a+a&b+b\\ -b+b&a+a\end{bmatrix}
\begin{bmatrix}2a&2b\\ 0&2a\end{bmatrix}
(ii)
$$
\begin{bmatrix}a^2+b^2&b^2+c^2\\ a^2+c^2&a^2+b^2\end{bmatrix}
+
\begin{bmatrix}2ab&2bc\\ -2ac&-2ab\end{bmatrix}
$$
$$
\begin{bmatrix}a^2+b^2+2ab & b^2+c^2+2bc\\ a^2+c^2-2ac & a^2+b^2-2ab\end{bmatrix}
$$
$$\begin{bmatrix}(a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2\end{bmatrix}
$$
Use the identities in above equation :
$(x+y)^2=x^2+y^2+2xy$ and $(x-y)^2=x^2+y^2-2xy$
(iii)
$$\begin{bmatrix}-1&4&-6\\ 8&5&16\\ 2&8&5\end{bmatrix}
+
\begin{bmatrix}12&7&6\\ 8&0&5\\ 3&2&4\end{bmatrix}$$
\begin{bmatrix}-1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4\end{bmatrix}
\begin{bmatrix}11&11&0\\ 16&5&21\\ 5&10&9\end{bmatrix}
(iv)
$$
\begin{bmatrix}\cos^2 x&\sin^2 x\\sin^2 x&\cos^2 x\end{bmatrix}
+
\begin{bmatrix}\sin^2 x&\cos^2 x\\cos^2 x&\sin^2 x\end{bmatrix}
$$
\begin{bmatrix}\cos^2 x+\sin^2 x & \sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2 x+\sin^2 x\end{bmatrix}
\begin{bmatrix}1&1\\ 1&1\end{bmatrix}
Use $\sin^2 \theta+\cos^2 \theta=1$ in above equation
Final Result:
$$
\boxed{
\begin{aligned}
(i)&\begin{bmatrix}2a&2b\\ 0&2a\end{bmatrix}\\[1em]
(ii)&\begin{bmatrix}(a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2\end{bmatrix}\\[1em]
(iii)&\begin{bmatrix}11&11&0\\ 16&5&21\\ 5&10&9\end{bmatrix}\\[1em]
(iv)&\begin{bmatrix}1&1\\ 1&1\end{bmatrix}
\end{aligned}
}
$$
Master matrix addition and algebraic identities easily with these step-by-step solutions. For more such high-quality CBSE and JEE study notes and solved questions, follow Anand Classes for the best learning resources.
NCERT Question.3 : Compute the indicated products $\\[1em]$
(i) $\begin{bmatrix}a&b\\ -b&a\end{bmatrix}\begin{bmatrix}a&-b\\ b&a\end{bmatrix}$ $\\[1em]$
(ii)$ \begin{bmatrix}1\\ 2\\ 3\end{bmatrix}
\begin{bmatrix}2 & 3 & 4\end{bmatrix}$ $\\[1em]$
(iii) $
\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}$ $\\[1em]$
(iv)$
\begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}
\begin{bmatrix}1 & -3 & 5\\0&2&4\\3&0&5\end{bmatrix}$ $\\[1em]$
(v) $ \begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}
\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}$ $\\[1em]$
(vi) $\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}
\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}$ $\\[1em]$
Solution :
(i)
$$
\begin{bmatrix}a&b\\ -b&a\end{bmatrix}\begin{bmatrix}a&-b\\ b&a\end{bmatrix}$$
\begin{bmatrix}
a^2+bb & -ab+ba\\
-ba+ab & b^2+ a^2
\end{bmatrix}
\begin{bmatrix}a^{2}+b^{2}&0\\ 0&a^{2}+b^{2}\end{bmatrix}
(ii)
$$
\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}
\begin{bmatrix}2 & 3 & 4\end{bmatrix}$$
\begin{bmatrix}
1\times 2 & 1\times 3 & 1\times 4\\
2\times 2 & 2\times 3 & 2\times 4\\
3\times 2 & 3\times 3 & 3\times 4
\end{bmatrix}
\begin{bmatrix}2 & 3 & 4\\4 & 6 & 8\\6 & 9 & 12\end{bmatrix}
(iii)
$$
\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}$$
\begin{bmatrix}
(1\cdot1)+(-2\cdot2) & (1\cdot2)+(-2\cdot3) & (1\cdot3)+(-2\cdot1)\\
(2\cdot1)+(3\cdot2) & (2\cdot2)+(3\cdot3) & (2\cdot3)+(3\cdot1)
\end{bmatrix}
\begin{bmatrix}-3 & -4 & 1\\8 & 13 & 9\end{bmatrix}
(iv)
$$
\begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}
\begin{bmatrix}1 & -3 & 5\\0&2&4\\3&0&5\end{bmatrix}$$
\begin{bmatrix}
2\cdot1+3\cdot0+4\cdot3 & 2\cdot(-3)+3\cdot2+4\cdot0 & 2\cdot5+3\cdot4+4\cdot5\\
3\cdot1+4\cdot0+5\cdot3 & 3\cdot(-3)+4\cdot2+5\cdot0 & 3\cdot5+4\cdot4+5\cdot5\\
4\cdot1+5\cdot0+6\cdot3 & 4\cdot(-3)+5\cdot2+6\cdot0 & 4\cdot5+5\cdot4+6\cdot5
\end{bmatrix}
\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}
(v)
$$
\begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}
\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}$$
\begin{bmatrix}
2\cdot1+1\cdot(-1) & 2\cdot0+1\cdot2 & 2\cdot1+1\cdot1\\
3\cdot1+2\cdot(-1) & 3\cdot0+2\cdot2 & 3\cdot1+2\cdot1\\
(-1)\cdot1+1\cdot(-1) & (-1)\cdot0+1\cdot2 & (-1)\cdot1+1\cdot1
\end{bmatrix}
\begin{bmatrix}1&2&3\\1&4&5\\-2&2&0\end{bmatrix}
(vi)
$$
\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}
\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}$$
\begin{bmatrix}
3\cdot2+(-1)\cdot1+3\cdot3 & 3\cdot(-3)+(-1)\cdot0+3\cdot1\\
(-1)\cdot2+0\cdot1+2\cdot3 & (-1)\cdot(-3)+0\cdot0+2\cdot1
\end{bmatrix}
\begin{bmatrix}14&-6\\4&5\end{bmatrix}
These worked examples cover common matrix multiplication types — conjugate-like products, column-by-row outer products, rectangular multiplications, and square matrix products — useful for CBSE and competitive exam practice. Find more solved exercises and downloadable notes from Anand Classes to strengthen your matrix algebra skills.
NCERT Question.4 : If the following matrices $$A=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}, \; B=\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix},\; \text{and} \\[1em] C=\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}$$ Compute $(A + B)$ and $(B – C)$.
Also verify that $A + (B – C) = (A + B) – C$.
Solution
1. Compute $(A + B)$
$$A + B=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix}$$
$$A + B = \begin{bmatrix}1+3 & 2 + (-1) & -3+2\\5+4 & 0+2 & 2+5\\1+2 & -1+0 & 1+3\end{bmatrix} = \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix}$$
2. Compute $(B – C)$
$$B-C=\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}$$
$$B – C = \begin{bmatrix}3-4 & -1-1 & 2-2\\4-0 & 2-3 & 5-2\\2-1 & 0-(-2) & 3-3\end{bmatrix} = \begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}$$
Compute $(A + B)$ and $(B – C)$.
3. Verification of $A + (B – C) = (A + B) – C$
LHS : $A + (B – C)$
$$A + (B – C) = \begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix} + \begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}$$
$$A + (B – C)= \begin{bmatrix}1+(-1) & 2+(-2) & -3+0\\5+4 & 0+(-1) & 2+3\\1+1 & -1+2 & 1+0\end{bmatrix} = \begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}$$
RHS : $(A + B) – C$
$$(A + B) – C = \begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix} – \begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}$$
$$(A + B) – C = \begin{bmatrix}4-4 & 1-1 & -1-2\\9-0 & 2-3 & 7-2\\3-1 & -1-(-2) & 4-3\end{bmatrix} = \begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}$$
Since LHS = RHS, the verification is correct.
Matrices addition and subtraction concepts like this are essential for CBSE and competitive exams. Stay tuned for more high-quality study material and detailed NCERT solutions by Anand Classes.
NCERT Question.5 : Compute $3A – 5B$ where
$$
A=\begin{bmatrix}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{bmatrix},
\quad
B=\begin{bmatrix}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{bmatrix}
$$
Solution :
Given matrix
$$
A=\begin{bmatrix}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{bmatrix}$$
$$3A=\begin{bmatrix}
3 \cdot \frac{2}{3} & 3 \cdot 1 & 3 \cdot \frac{5}{3} \\
3 \cdot \frac{1}{3} & 3 \cdot \frac{2}{3} & 3 \cdot \frac{4}{3} \\
3 \cdot \frac{7}{3} & 3 \cdot 2 & 3 \cdot \frac{2}{3}
\end{bmatrix}$$
$$3A=\begin{bmatrix}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{bmatrix}$$
Given matrix
$$B=\begin{bmatrix}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{bmatrix}
$$
$$5B=\begin{bmatrix}
5\cdot \frac{2}{5} & 5\cdot \frac{3}{5} & 5\cdot 1 \\
5\cdot \frac{1}{5} & 5\cdot \frac{2}{5} & 5\cdot \frac{4}{5} \\
5\cdot \frac{7}{5} & 5\cdot \frac{6}{5} & 5\cdot \frac{2}{5}
\end{bmatrix}$$
$$5B=\begin{bmatrix}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{bmatrix}$$
Now compute,
$$3A – 5B=\begin{bmatrix}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{bmatrix}-\begin{bmatrix}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{bmatrix}$$
$$3A – 5B=\begin{bmatrix}
2-2 & 3-3 & 5-5 \\
1-1 & 2-2 & 4-4 \\
7-7 & 6-6 & 2-2
\end{bmatrix}$$
$$3A – 5B=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
This chapter on matrices is important for strengthening your basics of linear algebra. Continue practicing with Anand Classes to excel in your CBSE and competitive exams like JEE — get high-quality study material and solutions for better understanding!
NCERT Question.6 : Simplify
$$\cos\theta \begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
+
\sin\theta
\begin{bmatrix}
\sin\theta & \cos\theta \\
-\cos\theta & \sin\theta
\end{bmatrix}
$$
Solution :
$$\cos\theta \begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
+
\sin\theta
\begin{bmatrix}
\sin\theta & \cos\theta \\
-\cos\theta & \sin\theta
\end{bmatrix}
$$
Multiply scalars inside both matrices:
$$\begin{bmatrix}
\cos\theta \cos\theta & -\cos\theta \sin\theta \\
\cos\theta \sin\theta & \cos^2\theta
\end{bmatrix}
+
\begin{bmatrix}
\sin\theta \sin\theta & \sin\theta \cos\theta \\
-\sin\theta \cos\theta & \sin^2\theta
\end{bmatrix}
$$
Now simplify:
$$
\begin{bmatrix}
\cos^2\theta + \sin^2\theta & -\cos\theta \sin\theta + \sin\theta \cos\theta \\
\cos\theta \sin\theta – \sin\theta \cos\theta & \cos^2\theta + \sin^2\theta
\end{bmatrix}
$$
$$\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
$$
Use the identity $\cos^2\theta + \sin^2\theta = 1$
This matrix problem is part of strengthening your fundamentals in trigonometric matrices. Continue learning with Anand Classes for high-quality CBSE and JEE preparation notes and solutions!
NCERT Question.7 : Find $X$ and $Y$ if $\\[1em]$
(i) $ X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix} \;$ and $\;X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}$ $\\[1em]$
(ii) $2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\;$ and $\; 3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}$
Solution
(i) $ X+Y=\begin{bmatrix}7&0\\2&5\end{bmatrix} \;$ and $\;X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}$ $\\[1em]$
Add the two given equations:
$$(X+Y)+(X-Y)=2X=\begin{bmatrix}7&0\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}10&0\\2&8\end{bmatrix}
$$
Thus
$$
X=\frac{1}{2}\begin{bmatrix}10&0\\2&8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}
$$
Now $Y=(X+Y)-X$:
$$
Y=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\1&4\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}
$$
(ii) We have
$$
2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix} …..(1)$$
$$3X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}…..(2) $$
Multiply the first equation (1) by 2 and the second (2) by 3 and subtract to eliminate $(Y)$:
$$
(4X+6Y)-(9X+6Y)= \begin{bmatrix}4&6\\8&0\end{bmatrix}-\begin{bmatrix}6&-6\\-3&15\end{bmatrix}$$
$$-5X=\begin{bmatrix}-2&12\\11&-15\end{bmatrix}$$
So
$$
X=-\frac{1}{5}\begin{bmatrix}-2&12\\11&-15\end{bmatrix}
=\begin{bmatrix}\tfrac{2}{5}&-\tfrac{12}{5}\\-\tfrac{11}{5}&3\end{bmatrix}$$
Now solve for $Y$ from equation $2X+3Y$:
$$ 3Y=(2X+3Y)-2X $$
$$3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}-2 \begin{bmatrix}\tfrac{2}{5}&-\tfrac{12}{5}\\-\tfrac{11}{5}&3\end{bmatrix} $$
$$3Y=
$$Y=\frac{1}{3}\begin{bmatrix}\tfrac{6}{5}&\tfrac{39}{5}\\\tfrac{42}{5}&-6\end{bmatrix}.
$$
$$Y=\begin{bmatrix}\tfrac{2}{5}&\tfrac{13}{5}\\\tfrac{14}{5}&-2\end{bmatrix}.
$$
Strengthen your matrix equation skills with more solved examples — download complete NCERT solutions and exam-focused notes from Anand Classes for clear methods and plenty of practice.
✅ FAQ Section
Q1. Where can I download Class 12 Matrices Exercise 3.2 NCERT Solutions PDF?
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Q2. Are these Matrices NCERT Solutions suitable for the Class 12 board exam?
Yes. Anand Classes prepares the solutions strictly according to the latest CBSE Class 12 syllabus and exam pattern. They are ideal for last-minute revision and scoring high marks.
Q3. Does this PDF include step-by-step solutions?
Absolutely. Each question from NCERT Exercise 3.2 is solved step-by-step to help students understand matrix operations clearly.
Q4. Is this Matrices Class 12 PDF free to download?
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Q5. Do I need additional books apart from these NCERT Solutions?
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