Anand Classes provides comprehensive and exam-oriented NCERT Solutions for Matrices Exercise 3.3 Class 12 Math Chapter-3, designed to help students master the concept of matrix operations, inverse matrices, and related problem-solving techniques with clarity. These solutions follow the latest CBSE guidelines and present step-by-step explanations that make even the most challenging questions easy to understand for Class 12 students preparing for board exams. Click the print button to download study material and notes.

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NCERT Question 1: Find the transpose of each of the following matrices:
(i) \begin{bmatrix}5\\ \frac{1}{2}\\ -1\end{bmatrix}
(ii) \begin{bmatrix}1 & -1\\ 2 & 3\end{bmatrix}
(iii) \begin{bmatrix}-1 & 5 & 6\\ \sqrt{3} & 5 & 6\\ 2 & 3 & -1\end{bmatrix}

Solution:
(i) Let
$$A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$$

Transpose is obtained by converting column into row:

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

(ii) Let
$$A = \begin{bmatrix}1 & -1\\ 2 & 3\end{bmatrix}$$

Interchanging rows and columns:

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

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

Interchanging rows and columns:

Transpose:

$$A^T = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$$

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NCERT Question 2: If
$$A=\begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$$
then verify that:
(i) $(A+B)’ = A’ + B’$
(ii) $(A-B)’ = A’ – B’$

Solution

(i) Verification of $(A+B)’ = A’ + B’$

First compute $(A + B)$:

$$A + B = \begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}+\begin{bmatrix}-4 & 1 & -5 \\1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$$

$$A + B= \begin{bmatrix}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$$

Now take transpose:

$$ (A+B)’ = \begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix} $$

Now find $(A’)$ and $(B’)$, then add:

$$A’ = \begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}, \quad
B’ = \begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}$$

$$A’ + B’ = \begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}+ \begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix} $$

$$A’ + B’=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$$

Thus,
$$ (A+B)’ = A’ + B’ $$
Hence proved.

(ii) Verification of $(A-B)’ = A’ – B’$

Compute $A – B$:

$$A – B = \begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}-\begin{bmatrix}-4 & 1 & -5 \\1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$$

$$A – B=\begin{bmatrix}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$$

Now transpose:

$$ (A-B)’ = \begin{bmatrix}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix} $$

Now find $(A’)$ and $(B’)$, then subtract:

$$A’ = \begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}, \quad
B’ = \begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}$$

$$A’ – B’ = \begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}- \begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix} $$

$$A’ – B’ = \begin{bmatrix}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$$

Thus,
$$ (A-B)’ = A’ – B’ $$
Hence proved.

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NCERT Question.3 : Given
$$A’=\begin{bmatrix}3 & 4\\-1 & 2\\0 & 1\end{bmatrix}\qquad\text{and}\qquad
B=\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\end{bmatrix},$$
verify that
(i) $(A+B)’=A’+B’$ and
(ii) $(A-B)’=A’-B’$

Solution
First recover $A$ from $(A’=(A)^{T})$. Taking transpose of $(A’)$ gives
$$A=(A’)^{T}=\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\end{bmatrix}.$$

Both $A$ and $B$ are $2\times3$ matrices, so $A\pm B$ are defined.

(i) Check $(A+B)’=A’+B’$

Compute $(A+B)$:
$$
A+B=\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\end{bmatrix}
+\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\end{bmatrix}
=\begin{bmatrix}2 & 1 & 1\\5 & 4 & 4\end{bmatrix}.
$$

Now transpose:
$$
(A+B)’=(A+B)^{T}=\begin{bmatrix}2 & 5\\1 & 4\\1 & 4\end{bmatrix}.
$$

Compute $(A’+B’)$
First $$B’=\begin{bmatrix}-1 & 1\\2 & 2\\1 & 3\end{bmatrix}$$
Then
$$
A’+B’=
\begin{bmatrix}3 & 4\\-1 & 2\\0 & 1\end{bmatrix}
+\begin{bmatrix}-1 & 1\\2 & 2\\1 & 3\end{bmatrix}
=\begin{bmatrix}2 & 5\\1 & 4\\1 & 4\end{bmatrix}.
$$

Thus $$(A+B)’=A’+B’$$

(ii) Check $(A-B)’=A’-B’$

Compute $(A-B)$:
$$
A-B=\begin{bmatrix}3 & -1 & 0\\4 & 2 & 1\end{bmatrix}
-\begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\end{bmatrix}
=\begin{bmatrix}4 & -3 & -1\\3 & 0 & -2\end{bmatrix}.
$$

Now transpose:
$$
(A-B)’=(A-B)^{T}=\begin{bmatrix}4 & 3\\-3 & 0\\-1 & -2\end{bmatrix}.
$$

Compute $(A’-B’)$:
$$
A’-B’=
\begin{bmatrix}3 & 4\\-1 & 2\\0 & 1\end{bmatrix}
-\begin{bmatrix}-1 & 1\\2 & 2\\1 & 3\end{bmatrix}
=\begin{bmatrix}4 & 3\\-3 & 0\\-1 & -2\end{bmatrix}.
$$

Thus $$(A-B)’=A’-B’$$

Final Result

$$\boxed{(A+B)’=A’+B’\quad\text{and}\quad (A-B)’=A’-B’}$$

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NCERT Question 4: If
$$A’=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix} \;\;\text{and} \quad B=\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}$$
find $(A+2B)’$.

Solution

From the given $A’$ we get $A$ by transposing $A’$:
$$A=(A’)’=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}.$$

Compute $(2B)$:
$$2B=2\begin{bmatrix}-1 & 0\\1 & 2\end{bmatrix}=\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}.$$

Add $A$ and $2B$:
$$
A+2B=\begin{bmatrix}-2 & 1\\3 & 2\end{bmatrix}+\begin{bmatrix}-2 & 0\\2 & 4\end{bmatrix}$$

$$A+2B=\begin{bmatrix}-4 & 1\\5 & 6\end{bmatrix}.$$

Finally take transpose to get $(A+2B)’$:
$$
(A+2B)’=\begin{bmatrix}-4 & 1\\5 & 6\end{bmatrix}’=
\begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix}.
$$

Answer:
$$\boxed{(A+2B)’=\begin{bmatrix}-4 & 5\\1 & 6\end{bmatrix}}$$

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NCERT Question 5. For the matrices $A$ and $B$,
verify that $(AB)’=B’A’$, where $\\$
(i) $A=\begin{bmatrix}1\\-4\\3\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 2 & 1\end{bmatrix}$ $\\$
(ii) $A=\begin{bmatrix}0\\1\\2\end{bmatrix}$ and $B=\begin{bmatrix}1 & 5 & 7\end{bmatrix}$

Solution :
We will compute $AB$, then $(AB)’$, and compare with $(B’A’)$ for each part.

(i) Compute $AB$ (outer product of a $(3\times1)$ and a $(1\times3)$ matrix):
$$
AB=
\begin{bmatrix}1\\-4\\3\end{bmatrix}
\begin{bmatrix}-1 & 2 & 1\end{bmatrix}$$

$$
AB=\begin{bmatrix}
1\cdot(-1) & 1\cdot2 & 1\cdot1\\
(-4)\cdot(-1) & (-4)\cdot2 & (-4)\cdot1\\
3\cdot(-1) & 3\cdot2 & 3\cdot1
\end{bmatrix}$$

$$
AB=\begin{bmatrix}
-1 & 2 & 1\\
4 & -8 & -4\\
-3 & 6 & 3
\end{bmatrix}.
$$

Transpose:
$$
(AB)’=(AB)^T=
\begin{bmatrix}
-1 & 4 & -3\\
2 & -8 & 6\\
1 & -4 & 3
\end{bmatrix}.
$$

Now compute $B’$ and $A’$ and multiply $B’A’$:
$$
B’=\begin{bmatrix}-1\\2\\1\end{bmatrix},\qquad
A’=\begin{bmatrix}1 & -4 & 3\end{bmatrix}.
$$

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

$$
B’A’=\begin{bmatrix}
-1\cdot1 & -1\cdot(-4) & -1\cdot3\\
2\cdot1 & 2\cdot(-4) & 2\cdot3\\
1\cdot1 & 1\cdot(-4) & 1\cdot3
\end{bmatrix}$$

$$
B’A’=\begin{bmatrix}
-1 & 4 & -3\\
2 & -8 & 6\\
1 & -4 & 3
\end{bmatrix}.
$$

Hence $$(AB)’=B’A’$$

(ii) Compute $(AB)$:
$$
AB=
\begin{bmatrix}0\\1\\2\end{bmatrix}
\begin{bmatrix}1 & 5 & 7\end{bmatrix}$$

$$
AB=\begin{bmatrix}
0\cdot1 & 0\cdot5 & 0\cdot7\\
1\cdot1 & 1\cdot5 & 1\cdot7\\
2\cdot1 & 2\cdot5 & 2\cdot7
\end{bmatrix}$$

$$
AB=\begin{bmatrix}
0 & 0 & 0\\
1 & 5 & 7\\
2 & 10 & 14
\end{bmatrix}.
$$

Transpose:
$$
(AB)’=
\begin{bmatrix}
0 & 1 & 2\\
0 & 5 & 10\\
0 & 7 & 14
\end{bmatrix}.
$$

Compute $(B’A’)$:
$$
B’=\begin{bmatrix}1\\5\\7\end{bmatrix},\qquad
A’=\begin{bmatrix}0 & 1 & 2\end{bmatrix},
$$

$$
B’A’=
\begin{bmatrix}1\\5\\7\end{bmatrix}
\begin{bmatrix}0 & 1 & 2\end{bmatrix}$$

$$
B’A’=\begin{bmatrix}
1\cdot0 & 1\cdot1 & 1\cdot2\\
5\cdot0 & 5\cdot1 & 5\cdot2\\
7\cdot0 & 7\cdot1 & 7\cdot2
\end{bmatrix}$$

$$
B’A’=\begin{bmatrix}
0 & 1 & 2\\
0 & 5 & 10\\
0 & 7 & 14
\end{bmatrix}$$

Thus $$(AB)’=B’A’$$

Final Verification: In both cases $\boxed{(AB)’=B’A’}$ holds.

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✅ FAQ Section

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

Exercise 3.3 focuses on the computation of inverse matrices using elementary transformations and related matrix properties essential for solving board-level questions.

Q2. Are the NCERT Solutions for Exercise 3.3 sufficient for board exam preparation?

Yes. The solutions provided by Anand Classes are step-by-step, accurate, and aligned with CBSE guidelines, making them ideal for scoring high in the Class 12 board exam.

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

Absolutely. Anand Classes offers the complete Exercise 3.3 NCERT Solutions PDF for free download to help students revise anytime.

Q4. Who should refer to these Class 12 Matrices solutions?

These solutions are perfect for Class 12 CBSE students, competitive exam aspirants, and anyone needing a clear understanding of matrix operations.

Q5. Are these solutions helpful for quick revision?

Yes. The neatly explained steps and simplified methods help students revise Matrices quickly and efficiently before exams.