Anand Classes provides reliable, exam-focused NCERT Solutions for Matrices Exercise 3.1 of Class 12 Math Chapter-3, helping students strengthen their fundamentals through clear explanations and step-by-step solutions. These well-structured notes are designed to support CBSE board exam preparation and enhance conceptual understanding with accuracy and simplicity. Ideal for students aiming for high marks, this Matrices Exercise 3.1 PDF ensures complete clarity for all important topics and formulas. Click the print button to download study material and notes.
NCERT Question.6 (i) : Find the values of $x$, $y$, and $z$ from the equation of matrices:
$$
\begin{bmatrix}
4 & 3\\
x & 5
\end{bmatrix}=
\begin{bmatrix}
y & z\\
1 & 5
\end{bmatrix}
$$
Solution :
When two matrices are equal, their corresponding entries are equal.
So equate each entry of the left matrix with the corresponding entry of the right matrix:
Compare entries in the first row, first column:
$$4 = y \quad\Rightarrow\quad y = 4.$$
First row, second column:
$$3 = z \quad\Rightarrow\quad z = 3.$$
Second row, first column:
$$x = 1 \quad\Rightarrow\quad x = 1.$$
Second row, second column:
$$5 = 5\quad\text{(holds true, no new information).}$$
All entries match consistently, so the solution is:
Final Result
$$\boxed{x = 1, y = 4, z = 3}$$
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NCERT Question.6 (ii) : Find the values of $x$, $y$, and $z$ from the equation of matrices:
$$
\begin{bmatrix}
x + y & 2 \\
5 + z & xy
\end{bmatrix}=\begin{bmatrix}
6 & 2 \\
5 & 8
\end{bmatrix}
$$
Solution :
Since the matrices are equal, their corresponding elements must also be equal:
Comparing first row, first column:
$$x + y = 6 \quad \text{(1)}$$
First row, second column:
$$2 = 2 \quad \text{(True, gives no new equation)}$$
Second row, first column:
$$5 + z = 5 \quad \Rightarrow\quad z = 0 \quad \text{(2)}$$
Second row, second column:
$$xy = 8 \quad \text{(3)}$$
From equation (1):
$$x = 6 – y \quad \text{(4)}$$
Substitute (4) into (3):
$$
(6 – y)y = 8
$$
$$
6y – y^2 = 8
$$
$$
y^2 – 6y + 8 = 0 \quad \text{(5)}
$$
Factorizing:
$$
(y – 4)(y – 2) = 0
$$
So,
$$y = 4 \quad \text{or} \quad y = 2$$
Now substitute these in equation (4):
- If $y = 4$, then:
$$
x = 6 – 4 = 2
$$ - If $y = 2$, then:
$$
x = 6 – 2 = 4
$$
Thus, there are two possible sets of solutions:
$$
(x, y, z) = (2, 4, 0) \quad \text{or} \quad (4, 2, 0)
$$
Final Result
$$
\boxed{(x, y, z) = (2, 4, 0)\ \text{or}\ (4, 2, 0)}
$$
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NCERT Question.6 (iii) : Find the values of $x$, $y$, and $z$ from the equation of matrices:
$$
\begin{bmatrix}
x + y + z\\
x + z\\
y + z
\end{bmatrix}=\begin{bmatrix}
9\\
5\\
7
\end{bmatrix}
$$
Solution :
Since both matrices are equal, we equate corresponding elements:
From the first row:
$$x + y + z = 9 \quad \text{(1)}$$
From the second row:
$$x + z = 5 \quad \text{(2)}$$
From the third row:
$$y + z = 7 \quad \text{(3)}$$
Subtract equation (2) from equation (1):
$$(x + y + z) – (x + z) = 9 – 5$$
$$y = 4$$
Now substitute $y = 4$ in equation (3):
$$4 + z = 7$$
$$z = 3$$
Now substitute $z = 3$ in equation (2):
$$x + 3 = 5$$
$$x = 2$$
Final Result
$$\boxed{x = 2, y = 4, z = 3}$$
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NCERT Question.7 : Find the values of $a$, $b$, $c$, and $d$ from the following matrices of operation :
$$
\begin{bmatrix}
a – b & 2a + c \\
2a – b & 3c + d
\end{bmatrix}=\begin{bmatrix}
-1 & 5 \\
0 & 13
\end{bmatrix}
$$
Solution :
Since both matrices are equal, we compare the corresponding elements:
$$a – b = -1 \quad (1)$$
$$2a – b = 0 \quad (2)$$
$$2a + c = 5 \quad (3)$$
$$3c + d = 13 \quad (4)$$
Subtract (1) from (2):
$$(2a – b) – (a – b) = 0 – (-1)$$
$$a = 1$$
Substitute $a = 1$ in (3):
$$2(1) + c = 5$$
$$c = 3$$
Substitute $a = 1$ in (1):
$$1 – b = -1$$
$$b = 2$$
Substitute $c = 3$ in (4):
$$3(3) + d = 13$$
$$9 + d = 13 \Rightarrow d = 4$$
Final Result
$$\boxed{a = 1, b = 2, c = 3, d = 4}$$
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NCERT Question.8 : A = $[a_{ij}]_{m \times n}$ is a square matrix, if:
(A) (m < n)
(B) (m > n)
(C) (m = n)
(D) None of these
Solution :
Since a matrix is square only when the number of rows equals the number of columns:
$$m = n$$
Correct Option: C
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Question 9. Which of the given values of $x$ and $y$ make the following pair of matrices equal
$$
\begin{bmatrix}
3x + 7 & 5 \\
y + 1 & 2 – 3x
\end{bmatrix}, \;
\begin{bmatrix}
0 & 8 \\
y – 2 & 4
\end{bmatrix}
$$
(A) $x = -\frac{1}{3}, y = 7$
(B) Not possible to find
(C) $y = 7, x = -\frac{2}{3}$
(D) $x = -\frac{1}{3}, y = -\frac{2}{3}$
Solution:
Given matrices:
$$
\begin{bmatrix}3x+7 & 5 \\ y+1 & 2-3x\end{bmatrix}=\begin{bmatrix}0 & y-2 \\ 8 & 4\end{bmatrix}$$
Since both matrices are equal, their corresponding elements must be equal:
$$3x + 7 = 0 \quad \text{…(1)}$$
$$5 = y – 2 \quad \text{…(2)}$$
$$y + 1 = 8 \quad \text{…(3)}$$
$$2 – 3x = 4 \quad \text{…(4)}$$
From (2):
$$5 = y – 2 \Rightarrow y = 7$$
From (3):
$$y + 1 = 8 \Rightarrow y = 7$$
So $y = 7$ is consistent.
Now check for $x$:
From (1):
$$3x + 7 = 0 \Rightarrow x = -\frac{7}{3}$$
From (4):
$$2 – 3x = 4 \Rightarrow -3x = 2 \Rightarrow x = -\frac{2}{3}$$
Values of $x$ are not same, so no common pair $(x, y)$ can satisfy all conditions simultaneously.
Final Answer:
$$\boxed{\text{Not possible to find values of } x \text{ and } y}$$
Correct Option: B
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NCERT Question.10 : The number of all possible matrices of order $3 \times 3$ with each entry either 0 or 1 is:
(A) 27 (B) 18 (C) 81 (D) 512
Solution:
A matrix of order $3 \times 3$ has 9 elements:
$$3 \times 3 = 9$$
Each element can be either 0 or 1, so there are 2 choices for each element.
By the multiplication principle, the total number of matrices is:
$$\text{Total matrices} = 2^9$$
Calculating:
$$2^9 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512$$
Final Answer:
$$\boxed{512}$$
Correct Option: D
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✅ FAQ Section
1. What is Exercise 3.1 of Matrices in Class 12 NCERT?
Exercise 3.1 introduces the basics of matrices, including definition, types of matrices, and fundamental matrix concepts essential for more advanced operations in later exercises.
2. Are these Matrices Exercise 3.1 solutions suitable for Class 12 board exam preparation?
Yes. The solutions follow NCERT guidelines strictly and are ideal for CBSE board preparation, competitive exams, and revision before tests.
3. Is the Matrices NCERT Solutions PDF free to download?
Yes. Anand Classes provides the Matrices Exercise 3.1 Class 12 NCERT Solutions PDF completely free for students.
4. Does the PDF include step-by-step explanations?
Absolutely. All solutions are written step-by-step to help students understand the concepts clearly and learn the correct method of solving matrix-based questions.
5. Can I print the Matrices Exercise 3.1 solutions?
Yes. A print-friendly version of the PDF is available. Just click the print button to download or print the study material and notes.
6. Who can use these NCERT Solutions for Matrices?
These solutions are ideal for CBSE students, competitive exam aspirants, tutors, and anyone looking to strengthen their conceptual understanding of Class 12 matrices.
