Ncert Solutions For Class 12 Maths Ex 3.1

Ncert Solutions For Class 12 Maths Chapter 3 Ex 3.1

Q-1: In a matrix given below:

A = \(\begin{bmatrix} 3 & 6 & 20 & -8 \\ 36 & -3 & \frac{ 7 }{ 2 } & 13 \\ \sqrt{ 5 } & 2 & -6 & 18 \end{bmatrix}\), write

(a) the order of the matrix.

(b) the number of element.

(c) Write the elements of a13, a21, a23, a24, a33

Solution:

(a) In the matrix given in the question, there are 3 rows and 4 columns. Hence, the order of the matrix will be 3 × 4.

(b) As the order of the given matrix is 3 × 4, there are 3 × 4 = 12 elements in the matrix.

(c) a13 = 20, a21 = 35, a23 = \(\frac{ 7 }{ 2 }\), a24 = 13, a33 = -5

 

 

Q-2: Let us consider a matrix A having 24 elements. Then, find the possible order for the matrix. What if there are 15 element of the same matrix?

Solution:

We know it well that if the order of the matrix is m × n, then the number of elements in the matrix will be mn. Then, to get all the possible orders of the matrix which having 24 elements, we need to find all of the possible pairs of the natural numbers having product 24.

Thus,

The possible ordered pairs will be:

( 1, 24 ), ( 2, 12 ) , ( 3, 8 ) , ( 4, 6 ) , ( 6, 4 ) , ( 8, 3 ), ( 12, 2 ), ( 24, 1 )

Therefore, the possible orders for the matrix with 24 elements are:

( 1 × 24 ), ( 2 × 12 ) , ( 3 × 8 ) , ( 4 × 6 ) , ( 6 × 4 ) , ( 8 × 3 ), ( 12 × 2 ), ( 24 × 1 )

If the number of elements in the matrix is 15, then the possible ordered pairs will be:

( 1, 15 ), ( 3, 5 ), ( 5, 3 ), ( 15, 1 )

Therefore, the possible orders for the matrix with 13 elements are:

( 1 × 15 ), ( 3 × 5 ) , ( 5 × 3 ) , ( 15 × 1 )

 

 

Q-3: Let us consider a matrix A having 21 elements. Then, find the possible order for the matrix. What if there are 10 element of the same matrix?

Solution:

We know it well that if the order of the matrix is m × n, then the number of elements in the matrix will be mn. Then, to get all the possible orders of the matrix which having 21 elements, we need to find all of the possible pairs of the natural numbers having product 21.

Thus,

The possible ordered pairs will be:

( 1, 21 ), ( 3, 7 ) , ( 7, 3 ), ( 21, 1 )

Therefore, the possible orders for the matrix with 24 elements are:

( 1 × 21 ), ( 3 × 7 ), ( 7 × 3 ), ( 21 × 1 )

If the number of elements in the matrix is 10, then the possible ordered pairs will be:

( 1, 10 ), ( 2, 5 ), ( 5, 2 ), ( 10, 1 )

Therefore, the possible orders for the matrix with 13 elements are:

( 1 × 10 ), ( 2 × 5 ) , ( 5 × 2 ) , ( 10 × 1 )

 

 

Q-4: Construct a 2 × 2 matrix, A = [ mij ], whose elements will be given by

(a) mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)

(b) mij = \(\frac{ i }{ j }\)

(c) mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)

 

Solution:

(a) As per the data given in the question:

mij = \(\frac{ \left ( i + j \right )^{ 2 } }{ 2 }\)

A ( 2 × 2 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ \left ( 1 + 1 \right )^{ 2 } }{ 2 } = \frac{ 4 }{ 2 } = 2\)

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

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

a22 = \(\frac{ \left ( 2 + 2 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)

Hence, the required matrix is A = \(\begin{bmatrix} 2 & \frac{ 9 }{ 2 } \\ \frac{ 9 }{ 2 } & 8 \end{bmatrix}\).

(b) As per the data given in the question:

mij = \(\frac{ i }{ j }\)

A ( 2 × 2 ) matrix is usually given by:

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ 1 }{ 1 } = 1\)

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

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

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

Hence, the required matrix is A = \(\begin{bmatrix} 1 & \frac{ 1 }{ 2 } \\ 2 & 1 \end{bmatrix}\)

 

(c) As per the data given in the question:

mij = \(\frac{ \left ( i + 2j \right )^{ 2 } }{ 2 }\)

A ( 2 × 2 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } \\ a_{ 21 } & a_{ 22 } \end{bmatrix}\)

Then,

a11 = \(\frac{ \left ( 1 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 9 }{ 2 }\)

a12 = \(\frac{ \left ( 1 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 25 }{ 2 }\)

a21 = \(\frac{ \left ( 2 + 2 \times 1 \right )^{ 2 } }{ 2 } = \frac{ 16 }{ 2 } = 8\)

a22 = \(\frac{ \left ( 2 + 2 \times 2 \right )^{ 2 } }{ 2 } = \frac{ 36 }{ 2 } = 18\)

Hence, the required matrix is A = \(\begin{bmatrix} \frac{ 9 }{ 2 } & \frac{ 25 }{ 2 } \\ 8 & 18 \end{bmatrix}\)

 

Q-5: Construct a 3 × 4 matrix, A = [ mij ], whose elements will be given by

(a) mij = \(\frac{ 1 }{ 2 }| -3i + j |\)

(b) mij = \(2i – j\)

Solution:

(a) As per the data given in the question:

mij = \(\frac{ 1 }{ 2 }| -3i + j |\)

A ( 3 × 4 ) matrix is usually given by:

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,

a11 = \(\frac{ 1 }{ 2 }| -3 \times 1 + 1 | = 1\)

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

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

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

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

a22 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 2 |= 2\)

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

a24 = \( \frac{ 1 }{ 2 }| -3 \times 2 + 4 | = 1\)

a31 = \(\frac{ 1 }{ 2 }| -3 \times 3 + 1 | = 4\)

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

a33 = \( \frac{ 1 }{ 2 }| -3 \times 3 + 3 |= 3\)

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

Hence, the required matrix is A = \(\begin{bmatrix} 1 & \left ( \frac{ 1 }{ 2 } \right ) & 0 & \left ( \frac{ 1 }{ 2 } \right ) \\ \\ \left ( \frac{ 5 }{ 2 } \right ) & 2 & \left ( \frac{3 }{ 2 } \right ) & 1 \\ \\ 4 & \left ( \frac{ 7 }{ 2 } \right ) & 3 & \left ( \frac{ 5 }{ 2 } \right ) \end{bmatrix}\)

 

(b) As per the data given in the question:

mij = \(2i + j\)

A ( 3 × 4 ) matrix is usually given by,

A = \(\begin{bmatrix} a_{ 11 } & a_{ 12 } & a_{ 13 } & a_{ 14 } \\ a_{ 21 } & a_{ 22 } & a_{ 23 } & a_{ 24 } \\ a_{ 31 } & a_{ 32 } & a_{ 33 } & a_{ 34 } \end{bmatrix}\)Then,

a11 = \(2 \times 1 – 1 = 1\)

a12 = \(2 \times 1 – 2 = 0\)

a13 = \( 2 \times 1 – 3 = -1\)

a14 = \( 2 \times 1 – 4 = -2\)

a21 = \( 2 \times 2 – 1 = 3\)

a22 = \( 2 \times 2 – 2 = 2\)

a23 = \( 2 \times 2 – 3 = 1\)

a24 = \( 2 \times 2 – 4 = 0\)

a31 = \( 2 \times 3 – 1 = 5\)

a32 = \( 2 \times 3 – 2 = 4\)

a33 = \( 2 \times 3 – 3 = 3\)

a34 = \( 2 \times 3 – 4 = 2\)

Hence, the required matrix is A = \(\begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}\)

Q-6: What will be the value of a, b and c in the following equation:

(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)

(b) \(\begin{bmatrix} a + b & 2 \\ 5 + c & ab \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)

(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)

 

Solution:

(a) \(\begin{bmatrix} 6 & 3 \\ a & 7 \end{bmatrix} = \begin{bmatrix} b & c \\ 1 & 7 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get:

a = 1, b = 6 and c = 3

 

(b) \(\begin{bmatrix} a + b & 2 \\ 5 + c & ab \end{bmatrix} = \begin{bmatrix} 9 & 2 \\ 6 & 8\end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get

a + b = 9 ………………… (i)

5 + c = 6 ………………… (ii)

And,

ab = 8 ………………… (iii)

From (ii), we will get:

c = 6 – 5 = 1

From (i), we have

a = 9 – b ………………… (iv)

Substituting (iv) in equation (iii), we will get:

( 9 – b )b = 8

⟹ 9b – b2 = 8

⟹ b2 – 9b + 8 = 0

⟹ b2 – 8b – b + 8 = 0

⟹ b( b – 8 ) -1( b – 8 ) = 0

⟹ ( b – 8 )( b – 1 ) = 0

b = 1, 8

For b = 1, a = 9 – 1 = 8

For b = 8, a = 9 – 8 = 1

Hence, a = 1, b = 8 and c = 1 or, a = 8, b = 1 and c = 1

 

(c) \(\begin{bmatrix} a + b + c \\ a + c \\ b + c \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get

a + b + c = 9 ………………(i)

a + c = 5 .………………(ii)

b + c = 7.………………(iii)

From equation (i), we have:

a + ( b + c ) = 9 ………………..(iv)

From equation (iii) and (iv), we will get:

a + 7 = 9

a = 2

Putting the value of an in equation (ii), we will get:

2 + c = 5

c = 3

Putting the value of c in equation (iii), we will get:

b + 3 = 7

b = 4

Hence, a = 2, b = 4 and c= 3.

 

 

Q-7: What will be the value of p, q, r and s in the following equation:

\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)

Solution:

\(\begin{bmatrix} p – q & 2p – r \\ 2p – q & 3r + s \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ 0 & 14 \end{bmatrix}\)

Since, the given matrices are equal, then their corresponding elements will also be equal.

Now,

By comparing the corresponding elements, we will get:

p – q = -2 …………(i)

2p – r = 0 …………(ii)

2p + r = 6 …………(iii)

3r + s = 14 …………(iv)

From equation (ii), we have:

r = 2p

Putting it in equation (iii), we will get:

2p + 2p = 6

4p = 6

⟹ p = \(\frac{ 6 }{ 4 }\)

⟹ p = \(\frac{ 3 }{ 2 }\)

Putting the value of p in equation (ii), we will get:

r = 2 × \(\frac{ 3 }{ 2 }\) = 3

Putting the value of r in equation (iv), we will get:

3 × 3 + s = 14

s = 5

Putting the value of p in equation (i), we will get:

\(\frac{ 3 }{ 2 }\) – q = -2

⟹ q = \(\frac{ 3 }{ 2 }\) + 2

⟹ q = \(\frac{ 3 + 2 \times 2 }{ 2 } = \frac{ 7 }{ 2 }\)

Hence, p = \(\frac{ 3 }{ 2 }\), q = \(\frac{ 7 }{ 2 }\), r = 3 and s = 5.

 

 

Q-8: Let us consider a matrix A = [aij]m × n. The matrix will be a square matrix, if

(a) m < n (b) m > n (c) m = n (d) None of these

Solution:

We know that a matrix can be a square matrix if and only if the number of rows and number of the columns of the matrix.

i.e., m = n

Hence, A = [aij]m × n will be a square matrix when m = n

Therefore, (c) is the correct answer.

Q-9. What will be the value of a and b in the following equation:

\(\begin{bmatrix} 3a + 7 & 5 \\ b + 1 & 2 – 3a \end{bmatrix} = \begin{bmatrix} 0 & b – 2 \\ 8 & 4 \end{bmatrix}\)

(a) a = \(\frac{ -1 }{ 3 }\), b = 7

(b) Not possible to find

(c) y = 7, x \(\frac{ -2 }{ 3 }\)

(d) x = \(\frac{ -1 }{ 3 }, y = \frac{ -2 }{ 3 }\)

 

Solution:

From the matrix, we have:

3a + 7 = 0

a = \(\frac{ -7 }{ 3 }\)

b + 1 = 8

b = 7

Also,

5 = b – 2

⟹ b = 3

2 – 3a = 4

⟹ 3a = -2

⟹ a = \(\frac{ -2 }{ 3 }\)

Hence, the correct answer is (b).

 

 

Q-10. The number of all the possible matrix of order 3 × 3 having each entry as 0 and 1 is-

(a) 512 (b) 81 (c) 18 (d) 27

 

Solution:

As the given matrix has order 3 × 3, i.e., the matrix have 9 elements and each of the element of this matrix will be either 0 or 1.

As we have only two digits and there are 9 elements in the matrix, so, the matrix will be filled in two possible ways only.

Hence, by the principle of multiplication, required number for the possible matrices will be 29 = 512

Therefore, the correct answer is (a).

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