Anand Classes brings you the most accurate and exam-oriented NCERT Solutions for Matrices Exercise 3.2 Class 12 Chapter-3, designed to strengthen your conceptual understanding and improve problem-solving skills for board exams. These expertly prepared solutions follow the latest NCERT guidelines and help students clarify every type of question asked in Exercise 3.2, ensuring strong preparation and higher performance in Class 12 Mathematics. Click the print button to download study material and notes.
NCERT Question.14 : Show that:
(i)
$$\begin{bmatrix}5 & -1\\ 6 & 7\end{bmatrix} \begin{bmatrix}2 & 1\\ 3 & 4\end{bmatrix} \ne
\begin{bmatrix}2 & 1\\ 3 & 4\end{bmatrix} \begin{bmatrix}5 & -1\\ 6 & 7\end{bmatrix}$$
(ii)
$$\begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0\end{bmatrix}
\begin{bmatrix}-1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4\end{bmatrix} \ne
\begin{bmatrix}-1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4\end{bmatrix}
\begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0\end{bmatrix}$$
Solution
(i) Let
$$A=\begin{bmatrix}5 & -1\\ 6 & 7\end{bmatrix}, \quad B=\begin{bmatrix}2 & 1\\ 3 & 4\end{bmatrix}$$
Compute
$$AB=\begin{bmatrix}5 & -1\\ 6 & 7\end{bmatrix}\begin{bmatrix}2 & 1\\ 3 & 4\end{bmatrix}$$
$$AB=\begin{bmatrix}(5\cdot2)+(-1\cdot3) & (5\cdot1)+(-1\cdot4)\\(6\cdot2)+(7\cdot3) & (6\cdot1)+(7\cdot4)\end{bmatrix}$$
$$AB=\begin{bmatrix}10-3 & 5-4\\ 12+21 & 6+28\end{bmatrix}
=\begin{bmatrix}7 & 1\\ 33 & 34\end{bmatrix}$$
Now compute
$$BA=\begin{bmatrix}2 & 1\\ 3 & 4\end{bmatrix} \begin{bmatrix}5 & -1\\ 6 & 7\end{bmatrix}$$
$$BA=\begin{bmatrix}(2\cdot5)+(1\cdot6) & (2\cdot-1)+(1\cdot7)\\ (3\cdot5)+(4\cdot6) & (3\cdot-1)+(4\cdot7)\end{bmatrix}$$
$$BA=\begin{bmatrix}10+6 & -2+7\\ 15+24 & -3+28\end{bmatrix}
=\begin{bmatrix}16 & 5\\ 39 & 25\end{bmatrix}$$
Since
$$\begin{bmatrix}7 & 1\\ 33 & 34\end{bmatrix} \ne \begin{bmatrix}16 & 5\\ 39 & 25\end{bmatrix}$$
Thus
$$AB \ne BA$$
(ii) Let
$$C=\begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0\end{bmatrix}, \quad
D=\begin{bmatrix}-1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4\end{bmatrix}$$
Now compute
$$CD=\begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0\end{bmatrix}
\begin{bmatrix}-1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4\end{bmatrix} $$
Calculate
$$CD=\begin{bmatrix}1\cdot(-1)+2\cdot0+3\cdot2 & 1\cdot1+2\cdot(-1)+3\cdot3 & 1\cdot0+2\cdot1+3\cdot4\\
0\cdot(-1)+1\cdot0+0\cdot2 & 0\cdot1+1\cdot(-1)+0\cdot3 & 0\cdot0+1\cdot1+0\cdot4\\
1\cdot(-1)+1\cdot0+0\cdot2 & 1\cdot1+1\cdot(-1)+0\cdot3 & 1\cdot0+1\cdot1+0\cdot4\end{bmatrix}$$
$$CD=\begin{bmatrix}-1+0+6 & 1-2+9 & 0+2+12\\ 0+0+0 & 0-1+0 & 0+1+0\\ -1+0+0 & 1-1+0 & 0+1+0\end{bmatrix}$$
$$CD=\begin{bmatrix}5 & 8 & 14\\ 0 & -1 & 1\\ -1 & 0 & 1\end{bmatrix}$$
Now compute
$$DC=\begin{bmatrix}-1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4\end{bmatrix}
\begin{bmatrix}1 & 2 & 3\\ 0 & 1 & 0\\ 1 & 1 & 0\end{bmatrix}$$
$$DC=\begin{bmatrix}-1\cdot1+1\cdot0+0\cdot1 & -1\cdot2+1\cdot1+0\cdot1 & -1\cdot3+1\cdot0+0\cdot0\\
0\cdot1+(-1)\cdot0+1\cdot1 & 0\cdot2+(-1)\cdot1+1\cdot1 & 0\cdot3+(-1)\cdot0+1\cdot0\\
2\cdot1+3\cdot0+4\cdot1 & 2\cdot2+3\cdot1+4\cdot1 & 2\cdot3+3\cdot0+4\cdot0\end{bmatrix}$$
$$DC=\begin{bmatrix}-1+0+0 & -2+1+0 & -3+0+0\\ 0+0+1 & 0-1+1 & 0+0+0\\ 2+0+4 & 4+3+4 & 6+0+0\end{bmatrix}$$
$$DC=\begin{bmatrix}-1 & -1 & -3\\ 1 & 0 & 0\\ 6 & 11 & 6\end{bmatrix}$$
Since
$$\begin{bmatrix}5 & 8 & 14\\0 & -1 & 1\\ -1 & 0 & 1\end{bmatrix} \ne \begin{bmatrix}-1 & -1 & -3\\ 1 & 0 & 0\\ 6 & 11 & 6\end{bmatrix}$$
Thus
$$CD \ne DC$$
Final Result
Matrix multiplication is not commutative, i.e.
$$AB \ne BA, \quad CD \ne DC$$
Students preparing for CBSE and competitive exams should practice such matrix operations regularly, as non-commutativity is a key concept in linear algebra. For more solved NCERT exercises and revision notes, explore high-quality study material from Anand Classes.
NCERT Question.15 : Find $A^{2}-5A+6I$
if
$$A=\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}$$ where (I) is identity matrix of order 3.
Solution
Step 1 : Calculate $A^{2}$
$$A^{2}=A \times A=\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}$$
Performing multiplication:
$$
A^{2}=
\begin{bmatrix}
(2)(2)+(0)(2)+(1)(1) & (2)(0)+(0)(1)+(1)(-1) & (2)(1)+(0)(3)+(1)(0) \\
(2)(2)+(1)(2)+(3)(1) & (2)(0)+(1)(1)+(3)(-1) & (2)(1)+(1)(3)+(3)(0) \\
(1)(2)+(-1)(2)+(0)(1) & (1)(0)+(-1)(1)+(0)(-1) & (1)(1)+(-1)(3)+(0)(0)
\end{bmatrix}
$$
$$
A^{2}=
\begin{bmatrix}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{bmatrix}
$$
Step 2 : Calculate $5A$
$$5A=5\begin{bmatrix}2&0&1\\ 2&1&3\\ 1&-1&0\end{bmatrix}$$
$$5A=\begin{bmatrix}
10&0&5\\
10&5&15\\
5&-5&0
\end{bmatrix}
$$
Step 3 : Calculate $6I$
$$I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$
$$6I=
\begin{bmatrix}
6&0&0\\
0&6&0\\
0&0&6
\end{bmatrix}
$$
Step 4 : Compute $A^{2}-5A+6I$
$$ A^{2}-5A=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}
10&0&5\\
10&5&15\\
5&-5&0
\end{bmatrix}=\begin{bmatrix}-5&-1&-3\\-1&-7&-10\\-5&4&-2\end{bmatrix}$$
Now add $6I$:
$$A^{2}-5A+6I=\begin{bmatrix}-5&-1&-3\\-1&-7&-10\\-5&4&-2
\end{bmatrix}+\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$$
$$A^{2}-5A+6I=\begin{bmatrix}
1&-1&-3\\
-1&-1&-10\\
-5&4&4
\end{bmatrix}
$$
Final Result
$$\boxed{\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}}$$
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NCERT Question.16 : If
$$
A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},
$$
prove that
$$
A^{3}-6A^{2}+7A+2I=O
$$
Solution
Step 1 — Compute $A^{2}=A\cdot A$
$$
A^{2}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
$$
Multiply $A$ by $A$ and compute each entry (row × column):
$$A^{2}=\begin{bmatrix}
(1)(1)+(0)(0)+(2)(2) & (1)(0)+(0)(2)+(2)(0) & (1)(2)+(0)(1)+(2)(3)\\
(0)(1)+(2)(0)+(1)(2) & (0)(0)+(2)(2)+(1)(0) & (0)(2)+(2)(1)+(1)(3)\\
(2)(1)+(0)(0)+(3)(2) & (2)(0)+(0)(2)+(3)(0) & (2)(2)+(0)(1)+(3)(3)
\end{bmatrix}
$$
Now evaluate each entry:
- Top row: (1+0+4=5,; 0+0+0=0,; 2+0+6=8).
- Middle row: (0+0+2=2,; 0+4+0=4,; 0+2+3=5).
- Bottom row: (2+0+6=8,; 0+0+0=0,; 4+0+9=13).
So
$$
A^{2}=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
$$
Step 2 — Compute $A^{3}=A^{2}\cdot A$
Multiply the matrix found for $A^{2}$ by $A$:
$$A^{3}=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
$$
Compute entries (row × column):
- Top row:
- $(1,1): (5\cdot1+0\cdot0+8\cdot2=5+0+16=21).$
- $(1,2): (5\cdot0+0\cdot2+8\cdot0=0).$
- $(1,3): (5\cdot2+0\cdot1+8\cdot3=10+0+24=34).$
- Middle row:
- $(2,1): (2\cdot1+4\cdot0+5\cdot2=2+0+10=12).$
- $(2,2): (2\cdot0+4\cdot2+5\cdot0=0+8+0=8).$
- $(2,3): (2\cdot2+4\cdot1+5\cdot3=4+4+15=23).$
- Bottom row:
- $(3,1): (8\cdot1+0\cdot0+13\cdot2=8+0+26=34).$
- $(3,2): (8\cdot0+0\cdot2+13\cdot0=0).$
- $(3,3): (8\cdot2+0\cdot1+13\cdot3=16+0+39=55).$
So
$$
A^{3}=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}
$$
Step 3 — Form the expression $A^{3}-6A^{2}+7A+2I$.
Compute each scalar multiple:
$$
6A^{2}=6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}
=\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix},
$$
$$ 7A=7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}
=\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix},
$$
$$
2I=2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}.
$$
Now compute
$$ A^{3}-6A^{2}+7A+2I=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\\[1em]+
\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}.$$
Do the arithmetic entrywise (showing one representative row and then the result):
- Top row:
- $(1,1): (21-30+7+2 = 0).$
- $(1,2): (0-0+0+0 = 0).$
- $(1,3): (34-48+14+0 = 0).$
- Middle row:
- $(2,1): (12-12+0+0 = 0).$
- $(2,2): (8-24+14+2 = 0).$
- $(2,3): (23-30+7+0 = 0).$
- Bottom row:
- $(3,1): (34-48+14+0 = 0).$
- $(3,2): (0-0+0+0 = 0).$
- $(3,3): (55-78+21+2 = 0).$
Thus every entry is zero, so
$$
A^{3}-6A^{2}+7A+2I=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=O
$$
For more worked proofs like this (with detailed entrywise verification and explanations of each multiplication step), check Anand Classes’ comprehensive NCERT solutions and practice sets — great for mastering matrix identities and exam preparation.
NCERT Question.17 : If
$$
A=\begin{bmatrix}3&-2\\ 4&-2\end{bmatrix},\quad
I=\begin{bmatrix}1&0\\ 0&1\end{bmatrix}
$$
find $k$ so that $A^{2}=kA-2I.$
Solution
First compute $A^{2}=A\cdot A$:
$$A^{2}=\begin{bmatrix}3&-2\\ 4&-2\end{bmatrix}
\begin{bmatrix}3&-2\\ 4&-2\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}
3\cdot3+(-2)\cdot4 & 3\cdot(-2)+(-2)\cdot(-2)\\
4\cdot3+(-2)\cdot4 & 4\cdot(-2)+(-2)\cdot(-2)
\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}$$
$$A^{2}=\begin{bmatrix}1 & -2\\ 4 & -4\end{bmatrix}$$
Now compute:
$$kA=\begin{bmatrix}3k & -2k\\4k & -2k\end{bmatrix},
\quad
2I=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$$
Substitute into $$A^{2}=kA-2I$$
$$
\begin{bmatrix}1 & -2\\ 4 & -4\end{bmatrix}=\begin{bmatrix}3k-2 & -2k\\ 4k & -2k-2\end{bmatrix}$$
Equating corresponding entries:
$$
1=3k-2,\quad -2=-2k,\quad 4=4k,\quad -4=-2k-2
$$
Each equation gives:
$$
\boxed{k=1}
$$
Final Result
$$
\boxed{k=1}
$$
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NCERT Question.18 : Let
$$
A=\begin{bmatrix}0 & -\tan\frac{\alpha}{2}\\ \tan\frac{\alpha}{2} & 0\end{bmatrix},\qquad I=\begin{bmatrix}1&0\\0&1\end{bmatrix}
$$
Show that
$$
I+A=(I-A)\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}
$$
Solution :
Set $(t=\tan\frac{\alpha}{2})$. Then
$$
A=\begin{bmatrix}0 & -t\\t & 0\end{bmatrix},\quad
I+A=\begin{bmatrix}1 & -t\\t & 1\end{bmatrix},\quad
I-A=\begin{bmatrix}1 & t\\-t & 1\end{bmatrix}.
$$
Compute the right hand side $(I-A)R$ where $$R=\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}$$
$$(I-A)R=\begin{bmatrix}1 & t\\-t & 1\end{bmatrix}
\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix} $$
$$(I-A)R=\begin{bmatrix}
\cos\alpha + t\sin\alpha & -\sin\alpha + t\cos\alpha\\
-t\cos\alpha + \sin\alpha & t\sin\alpha + \cos\alpha
\end{bmatrix}$$
So we must verify the four scalar identities
$$
\cos\alpha + t\sin\alpha = 1,\qquad -\sin\alpha + t\cos\alpha = -t,
$$
(the bottom row then follows by symmetry).
Use the half-angle formulas expressed in $(t=\tan\frac{\alpha}{2})$:
$$
\sin\alpha=\frac{2t}{1+t^{2}},\qquad
\cos\alpha=\frac{1-t^{2}}{1+t^{2}}.
$$
Now compute the first entry:
$$
\cos\alpha + t\sin\alpha
=\frac{1-t^{2}}{1+t^{2}}+t\cdot\frac{2t}{1+t^{2}}
=\frac{1-t^{2}+2t^{2}}{1+t^{2}}
=\frac{1+t^{2}}{1+t^{2}}=1.
$$
Compute the second entry:
$$
-\sin\alpha + t\cos\alpha
=-\frac{2t}{1+t^{2}}+t\cdot\frac{1-t^{2}}{1+t^{2}}$$
$$-\sin\alpha + t\cos\alpha=\frac{-2t+t-t^{3}}{1+t^{2}}
=\frac{-t(1+t^{2})}{1+t^{2}}=-t$$
Thus
$$
(I-A)R=\begin{bmatrix}1 & -t\\t & 1\end{bmatrix}=I+A.
$$
Therefore
$$
\boxed{I+A=(I-A)\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}}
$$
holds for all $(\alpha)$ (where the half-angle tangent is defined).
For more neat derivations of trigonometry–matrix identities and many solved NCERT examples, check Anand Classes for clear, exam-oriented notes and practice problems.
NCERT Question.19 : A trust fund has ₹ 30,000 to invest in two types of bonds paying 5% and 7% annual interest. Using matrix multiplication, determine how to allocate the ₹ 30,000 to achieve a total annual interest of
(a) ₹ 1,800 and
(b) ₹ 2,000
Step 1: Define Variables and Matrices
Let the investment in the first bond be $x$.
The investment in the second bond is then $30,000 – x$.
The investments can be represented by the row matrix
$$P = [x \quad 30,000 – x]$$
The annual interest rates (5% and 7%) can be represented by the column matrix
$$R = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$$
Step 2: Set up the Matrix Equation
The total annual interest is given by the matrix product $P \times R$:
$$\text{Total Interest} = [x \quad 30,000 – x] \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$$
$$\text{Total Interest} = x \cdot 0.05 + (30,000 – x) \cdot 0.07$$
(a) Target Interest: ₹ 1,800
Setting the total interest to ₹ 1,800:
$$1800 = 0.05x + 0.07(30,000 – x)$$
$$1800 = 0.05x + 2100 – 0.07x$$
$$1800 = 2100 – 0.02x$$
$$0.02x = 300$$
$$x = 15,000$$
Investment in the first bond = ₹ 15,000
Investment in the second bond = ₹ 15,000
(b) Target Interest: ₹ 2,000
Setting the total interest to ₹ 2,000:
$$2000 = 0.05x + 0.07(30,000 – x)$$
$$2000 = 0.05x + 2100 – 0.07x$$
$$2000 = 2100 – 0.02x$$
$$0.02x = 100$$
$$x = 5,000$$
Investment in the first bond = ₹ 5,000
Investment in the second bond = ₹ 25,000
NCERT Question 20 : A school bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books. The selling prices are ₹80, ₹60, and ₹40 per book, respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution
1️⃣ Quantities Matrix (A)
10 dozen chemistry books = 120 books
8 dozen physics books = 96 books
10 dozen economics books = 120 books
So,
$$A = [120 \quad 96 \quad 120]$$
2️⃣ Prices Matrix (B)
$$B = \begin{bmatrix}80 \\ 60 \\ 40 \end{bmatrix}$$
3️⃣ Total Amount Calculation
Multiply the matrices:
$$AB = [120 \quad 96 \quad 120] \begin{bmatrix}80 \\ 60 \\ 40\end{bmatrix}$$
$$AB = (120 \cdot 80) + (96 \cdot 60) + (120 \cdot 40)$$
$$AB = 9600 + 5760 + 4800$$
$$AB = 20160$$
✔ Final Answer
The total amount the bookshop will receive is:
$$\boxed{₹ 20,160}$$
NCERT Question.21 : Assume $X, Y, Z, W$, and $P$ are matrices of order $2 × n$, $3 × k$, $2 × p$, $n × 3$, and $p × k$, respectively. Choose the correct answer. The restriction on $n,k,p$ so that $PY+WY$ will be defined are:
(A) $k=3,p=n$
(B) $k$ arbitrary, $p=2$
(C) $p$ arbitrary, $k=3$
(D) $k=2,p=3$
Solution
For matrix operations to be defined, specific conditions must be met:
Matrix Addition: The sum of two matrices, say C and D (say C + D), is defined only if both matrices have the same order. The resulting matrix has the same order as C and D.
Matrix Multiplication: The product of two matrices, say A and B (A x B), is defined only if the number of columns in A equals the number of rows in B. The resulting matrix has the order of (rows of A)
x (columns of B).
Matrix $P$ is of order
$$p\times k$$
Matrix $Y$ is of order
$$3\times k$$
For $PY$ to be defined:
$$k=3$$
Then
$$PY \Rightarrow (p\times 3)$$
Matrix $W$ is of order
$$n\times 3$$
Matrix $Y$ is
$$3\times k = 3\times 3$$
So $WY$ is defined and is of order:
$$n\times 3$$
For $PY + WY$ to be defined, their orders must be the same:
$$p\times 3 = n\times 3 \Rightarrow p=n$$
$$
\boxed{\text{Correct Option: (A) } k=3, p=n}
$$
NCERT Question.22 : Assume $X, Y, Z, W$, and $P$ are matrices of order $2 × n$, $3 × k$, $2 × p$, $n × 3$, and $p × k$, respectively. Choose the correct answer.
If $n=p$, then the order of the matrix $7X-5Z$ is:
(A) $p\times 2$
(B) $2\times n$
(C) $n\times 3$
(D) $p\times n$
Solution
Matrix $X$ is of order:
$$2\times n$$
So matrix $7X$ is:
$$2\times n$$
Matrix $Z$ is of order:
$$2\times p = 2\times n \;\;(\text{since } p=n)$$
So matrix $5Z$ is:
$$2\times n$$
Therefore:
$$7X – 5Z$$
is defined and is of order:
$$
\boxed{2\times n}
$$
Thus, the correct option is:
$$
\boxed{\text{(B) } 2\times n}
$$
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✅ FAQ Section
1. What is covered in NCERT Class 12 Matrices Exercise 3.2?
Exercise 3.2 focuses on matrix operations such as addition, subtraction, and scalar multiplication, helping students understand the foundational algebra of matrices.
2. Are these solutions helpful for CBSE Class 12 board exams?
Yes. The solutions strictly follow NCERT guidelines and are fully aligned with the CBSE exam pattern, making them highly useful for board exam preparation.
3. Can I download Matrices Exercise 3.2 Solutions in PDF format?
Absolutely. Anand Classes provides a free, high-quality PDF download of the complete NCERT Solutions for Matrices Exercise 3.2 (Set-3).
4. Who prepares these NCERT solutions for Matrices?
All solutions are prepared by experienced faculty at Anand Classes with years of teaching and board-exam training experience.
5. Are multiple sets available for Exercise 3.2?
Yes. Anand Classes provides Set-1, Set-2, and Set-3, each containing comprehensive solutions prepared for deeper practice and greater exam readiness.
