PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 3Matrices Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1

Question 1.
In the matrix A = \(=\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write:
(i) The order of the matrix
(ii) The number of elements,
(iii) The elements a13, a21, a33, a24, a23.
Solution.
(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4.
(ii) Since, the order of the matrix is 3 × 4, so there are 3 × 4 = 12 elements in it.
(iii) a13 = 19, a21 = 35, a33= – 5, a24 = 12, a23 = \(\frac{5}{2}\).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution.
We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numhers whose product is 24.
The ordered pairs are: (1, 24), (24,1), (2,12), (12, 2), (3, 8), (8, 3), (4, 6) and (6, 4).
Hence, the possible orders of a matrix having 24 elements are 1 × 24,24 × 1, 2 × 12,12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 (1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 x 13 and 13×1.

Question 3.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution.
We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6 and 6 × 3.
(1, 5)and (5, 1)are ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 4.
Construct 2 × 2 matrix, A = [aij] whose elements are given by
(i) aij = \(\frac{(i+j)^{2}}{2}\)

(ii) aij = \(\frac{i}{j}\)

(iii) aij = \(\frac{(i+2 j)^{2}}{2}\)
Solution.
(i) The order of the given matrix is 2 × 2, so A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\)
where aij = \(\frac{(i+2 j)^{2}}{2}\).
To find a11, put i = 1 and j = 1
∴ a11 = \(\frac{(1+1)^{2}}{2}\) = 2 Similarly

a12 = \(\frac{(1+2)^{2}}{2}=\frac{9}{2}\)

a21 = \(\frac{(2+1)^{2}}{2}=\frac{9}{2}\) and

a22 = \(\frac{(2+2)^{2}}{2}\) = 8

(ii) Here, A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\), where aij = \(\frac{i}{j}\).
∴ a11 = \(\frac{1}{1}\) = 1,

a12 = \(\frac{1}{2}\),

a21 = \(\frac{2}{1}\) = 2 and

a22 = \(\frac{2}{1}\) = 1

Hence, the required matrix is A = \(\left[\begin{array}{cc}
1 & 1 / 2 \\
2 & 1
\end{array}\right]_{2 \times 2}\)

(iii) Here, A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\), where aij = \(\frac{(i+2 j)^{2}}{2}\)
∴ a11 = \(\frac{(1+2)^{2}}{2}=\frac{9}{2}\)

a12 = \(\frac{(1+4)^{2}}{2}=\frac{25}{2}\)

a21 = \(\frac{(2+2)^{2}}{2}\) = 8

a22 = \(\frac{(2+4)^{2}}{2}\) = 18.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Hence, the required matrix is A = \(\left[\begin{array}{cc}
9 / 2 & 25 / 2 \\
8 & 18
\end{array}\right]_{2 \times 2}\).

Question 5.
Construct a 3 × 4 matrix, whose elements are given by:
(i) aij = \(\frac{4}{4}\) |- 3i + j|
(ii) aij = 2i – j
Solution.
In general, a 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)

(i) (i) aij = \(\frac{1}{2}\) |- 3i + j|, i = 1, 2, 3, 4 and j =1, 2, 3, 4
a11 = \(\frac{1}{2}\) |- 3 × 1 + 1|
= \(\frac{1}{2}\) |- 3 + 1|
= \(\frac{1}{2}\) |- 2|
= \(\frac{21}{2}\) = 1

a21 = \(\frac{1}{2}\) |- 3 × 2 + 1|
= \(\frac{1}{2}\) |- 6 + 1|
= \(\frac{1}{2}\) |- 5|
= \(\frac{5}{2}\)

a31 = \(\frac{1}{2}\) |- 3 × 3 + 1|
= \(\frac{1}{2}\) |- 9 + 1|
= \(\frac{1}{2}\) |- 8|
= \(\frac{8}{2}\) = 4

a12 = \(\frac{1}{2}\) |- 3 × 1 + 2|
= \(\frac{1}{2}\) |- 3 + 2|
= \(\frac{1}{2}\) |- 1|
= \(\frac{1}{2}\)

a22 = \(\frac{1}{2}\) |- 3 × 2 + 2|
= \(\frac{1}{2}\) |- 6 + 2|
= \(\frac{1}{2}\) |- 4|
= \(\frac{4}{2}\) = 2

a32 = \(\frac{1}{2}\) |- 3 × 3 + 2|
= \(\frac{1}{2}\) |- 9 + 2|
= \(\frac{1}{2}\) |- 7|
= \(\frac{7}{2}\)

a13 = \(\frac{1}{2}\) |- 3 × 1 + 3|
= \(\frac{1}{2}\) |- 3 + 3| = 0

a23 = \(\frac{1}{2}\) |- 3 × 2 + 3|
= \(\frac{1}{2}\) |- 6 + 3|
= \(\frac{1}{2}\) |- 3|
= \(\frac{3}{2}\)

a33 = \(\frac{1}{2}\) |- 3 × 3 + 3|
= \(\frac{1}{2}\) |- 9 + 3|
= \(\frac{1}{2}\) |- 6|
= \(\frac{6}{2}\) = 3

a14 = \(\frac{1}{2}\) |- 3 × 1 + 4|
= \(\frac{1}{2}\) |- 3 + 4|
= \(\frac{1}{2}\) |1|
= \(\frac{1}{2}\)

a24 = \(\frac{1}{2}\) |- 3 × 2 + 4|
= \(\frac{1}{2}\) | – 6 + 4|
= \(\frac{1}{2}\) |- 2|
= \(\frac{2}{2}\) = 1

a34 = \(\frac{1}{2}\) |- 3 × 3 + 4|
= \(\frac{1}{2}\) |- 9 + 4|
= \(\frac{1}{2}\) |- 5|
= \(\frac{5}{2}\)

Therfore, the required matrix is A = \(\left[\begin{array}{cccc}
1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & =2 & \frac{3}{2} & 1 \\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{array}\right]_{3 \times 4}\)

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

(ii) aij = 2i – j, i = 1, 2, 3, 4 and j = 1, 2, 3, 4
a11 = 2 × 1 – 1 = 2 – 1 = 1
a12 = 2 × 2 – 1 = 4 – 1 = 3
a13 = 2 × 3 – 1 = 6 – 1 = 5

a21 = 2 × 1 – 2 = 2 – 2 = 0
a22 = 2 × 2 – 2 = 4 – 2 = 2
a23 = 2 × 3 – 2 = 6 – 2 = 4

a31 = 2 × 1 – 3 = 2 – 3 = – 1
a32 = 2 × 2 – 3 = 4 – 3 = 1
a33 = 2 × 3 – 3 = 6 – 3 = 3

a41 = 2 × 1 – 4 = 2 – 4 = – 2
a42 = 2 × 2 – 4 = 4 – 4 = 0
a43 = 2 × 3 – 4 = 6 – 4 = 0

Therfore, the required matrix is A = \(\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]_{3 \times 4}\).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 6.
Find the values of x, y, and z from the following equations:
(i) \(\left[\begin{array}{ll}
\mathbf{4} & \mathbf{3} \\
\boldsymbol{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\boldsymbol{y} & \boldsymbol{z} \\
\mathbf{1} & \mathbf{5}
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)

(iii) \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
Solution.
(i) \(\left[\begin{array}{ll}
\mathbf{4} & \mathbf{3} \\
\boldsymbol{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\boldsymbol{y} & \boldsymbol{z} \\
\mathbf{1} & \mathbf{5}
\end{array}\right]\)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding eleme nts, we get x = 1, y = 4, and z = 3

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

(ii) \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get x + y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5
⇒ z = 0
We know that,
(x – y)2 = (x + y)2 – 4xy
⇒ (x – y)2 = 36 – 32 = 4
⇒ x – y = ±2
Now, when x – y – 2 and x + y = 6, we get x = 4 and y = 2
When x – y = – 2 and x + y = 6, we get x = 2 and y = 4
∴ x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0.

(iii) \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get
x + y + z = 9 ………..(i)
x + z = 5 ………….(ii)
y + z = 7
From Eqs. (i) and (ii), we have
y = 9 – 5
⇒ y = 4
Then, from Eq. (iii), we have:
4 + z = 7
⇒ z = 3
Now, x + z = 5
⇒ x = 5 – 3 = 2
∴ x – 2, y = 4 and z = 3.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 7.
Find the value of a, b, c and d from the following equation:
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution.
We have,
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get
a – b = – 1 ……………(i)
2a – b = 0 …………….(ii)
2a + c = 5 …………..(iii)
3c + d = 13 ………..(iv)
From Eq. (ii), we have
b = 2a
Then, from Eq. (i), we have
a – 2a = – 1
⇒ a -1
⇒ b = 2
Now, from Eq. (iii), we have
2 x 1 + c = 5
⇒ c = 5 – 2 = 3
From Eq. (iv) we have
3 × 3 + d = 13
⇒ 9 + d = 13
⇒ d = 13 – 9 = 4
Hence, a = 1, b = 2, c = 3 and d = 4.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 8.
A = [aij]m × n is a square matrix, if
(A) m < n (B) m > n
(C) m = n
(D) None of these
Solution.
It is known that a given matrix is said to be a square matrix, if the number of rows is equal to the number of columns.
Therefore, A = [aij]m × n is a square matrix, if m – n.
Hence, the correct answer is (C).

Question 9.
Which of the given values of x and y make the following pair of matrices equal?
\(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
(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.
It is given that \(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
On equating the corresponding elements, we get
3x + y = 0
⇒ x = – 3
5 = y – 2
⇒ y = 7
y + 1 = 8
⇒ y = 7
2 – 3x = 4
⇒ x = \(\frac{-2}{3}\)

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.
Hence, it is not possible to find the values of x and y for which the given matrices are equal.
Hence, the correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Q. 10.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is
(A) 27
(B) 18
(C) 81
(D) 512
Solution.
The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512.
Hence, the correct answer is (D).

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