Matrices

The given table can be expressed as a matrix where each column denotes the week and weekend rates for Adult, Children and Senior Citizen.

The dimension of above matrix is 3 × 2.

Also, the same information can also be expressed as a matrix where each row denotes the week and weekend rates for Adult, Children and Senior Citizen.

The dimension of above matrix is 2 × 3.

Representing the given information in a 3 × 1 matrix, we get,

Also, representing the given information in a 1 × 3 matrix, we get,

⇒ (6 8 13)

(i) Order is 2 × 3

(ii) Order is 3 × 1

(iii) Order is 3 × 3

(iv) Order is 1 × 3

(v) Order is 4 × 2

Since there are 8 elements, we can make the multiples of 8, which are:- 1, 8, 2, 4

Therefore, the possible orders of a matrix are 1×8; 8×1; 2×4 and 4×2.

Since there are 30 elements, we can make the multiples of 30, which are:- 1×30; 30×1; 2×15; 15×2; 3×10; 10×3; 5×6 and 6×5

Therefore, the possible orders of a matrix are 1×30; 30×1; 2×15; 15×2; 3×10; 10×3; 5×6 and 6×5

(i) Since a_ij = i × j, and the general of matrix is:

On substituting the values, we get,

(ii) Since the general of matrix is:

On substituting the values, we get,

(iii) Since the general of matrix is:

On substituting the values, we get,

(i) Since the general of matrix is:

On substituting the values, we get,

(ii) Since the general of matrix is:

On substituting the values, we get,

(iii) Since the general of matrix is:

On substituting the values, we get,

(i) Order of matrix is 3 × 4.

(ii) Since the general of matrix is:

⇒ a₂₄ = 4 and a₃₂ = 0

(iii) Element 7 occurs in 2^nd row 3^rd column

For the transpose of a matrix, we know that,

A^T_ij = A_ji, therefore the transpose of matrix A is:-

For the transpose of a matrix, we know that,

A^T_ij = A_ji, therefore the transpose of matrix A is:-

Now, applying transpose on A^T , we get,

∴ We see that (A^T)^T = A

Since given is the matrix equation we would equate the right hand side elements with left hand side elements

⇒ 5x + 2 = 12 , y – 4 = – 8 and 4z + 6 = 2

⇒ 5x = 10 , y = – 4 and 4z = – 4

⇒ x = 2 , y = – 4 and z = – 1

Since given is the matrix equation we would equate the right hand side elements with left hand side elements

2x + y = 5 …1

and x – 3y = 13 ….2

Multiplying equation 2 by 2

we get (x – 3y = 13 ) 2

2x – 6y = 26 ….3

Subtracting equation 1 and 3

Or

Y = – 3

Substituting value of y in 1

2x – 3 = 5 ⇒ 2x = 8 ⇒ x = 4

Hence the solution is x = 4 and y = – 3

Let us first solve for the value of matrix A =

⇒ A =

The additive inverse of A = negative of the matrix

– A = additive inverse of A =

Given A = and B =

We have to find matrix C where C = 2A + B

Now 2A = 2

We would multiply each term of A with 2

2A =

2A + B = =

C =

Given A =

B =

6A =

3B = 3

6A – 3B =

6A – 3B =

⇒

⇒

Equating both the sides of the matrix equation

We get

2a –b = 10 …..1

And 3a + b = 5 …..2

Adding equation 1 and 2

⇒ a = 3

Putting value of a in 1

2a –b = 10

⇒ 6 –b = 10 ⇒ b = – 4

Thus the required solutions are a = 3 and b = – 4

Given 2x + 3y = ….1

and 3x + 2y = ...2

Adding 1 and 3 equations we get,

5x + 5y =

Dividing both the sides by 5

x + y = …3

Now subtracting 1 from 2

X – y = ……4

Adding 3 and 4

2x =

X =

Subtracting 4 from 3

2y =

Y =

Hence x = = and y =

given

⇒

⇒

Equating each element of the matrix equation with corresponding element

x² + 6x = – 9

⇒ x² + 6x + 9 = 0

⇒ x( x + 3) + 3(x + 3) = 0

⇒ (x + 3) (x + 3) = 0

x = – 3 , – 3

y² – 3y = 4

⇒ y² – 3y – 4 = 0

⇒ y( y – 4) + 1( y – 4) = 0

⇒ (y + 1) (y – 4) = 0

⇒ y = – 1 or 4

Hence the values of x = – 3, – 3 and y = – 1 , 4

9) given

A =

(i) To verify A + B = B + A

LHS:

A + B =

RHS = B + A

=

Here, LHS = RHS hence proved

(ii) A + ( – A) = 0 = ( – A) + A

– A =

LHS:

= A + (– A)

=

RHS ( – A) + A =

Hence LHS = RHS = 0 proved

10) given

A = , B = and C

LHS = A + (B + C)

⇒

⇒

⇒

RHS = (A + B) + C

⇒

⇒

⇒

LHS = RHS

Hence proved

Here we consider the types of items sold along the column and the store in which they are stored along the rows

Thus week 1 and week 2 matrices can be written as

W₁ = and W₂ =

The sum of the items sold in two weeks is the sum of the above two matrices, which is sum of each corresponding elements of the two matrices

W₁ + W₂ =

⇒

TV DVD video CD

⇒ The sum of items sold in two weeks =

Here we consider the type of member along the columns and the timings in the rows. Thus members and non – members matrices can be written as:

M = and N = respectively

The additional cost for non – members as compared to the members is the difference of the above two matrices, which is the difference of each element of the matrices to its corresponding element in the other matrix

N –M =

⇒

The additional cost of non – members as compared to the members is

(i) The multiplication of 2 matrices is possible if number of columns in first matrix is equal to number of rows in second.

⇒ Here A[a_ij]_{4 x 3} and B = [b_ij]_3x2

⇒ Number of columns in A = 3

⇒ Number of rows in B = 3

Thus the product is defined and the order if product is

Number of rows in A × Number of columns in B

∴ AB = 4 × 3

(ii) The multiplication of 2 matrices is possible if number of columns in first matrix is equal to number of rows in second.

⇒ Here P[p_ij]_{4 x 3} and Q = [q_ij]_4x3

⇒ Number of columns in P = 3

⇒ Number of rows in Q = 4

Thus the product is not defined.

(iii) The multiplication of 2 matrices is possible if number of columns in first matrix is equal to number of rows in second.

⇒ Here M[m_ij]_{3 x 1} and N = [n_ij]_1x5

⇒ Number of columns in M = 1

⇒ Number of rows in N = 1

Thus the product is defined and the order if product is

Number of rows in M × Number of columns in N

∴ MN = 3 × 5

(iv) The multiplication of 2 matrices is possible if number of columns in first matrix is equal to number of rows in second.

⇒ Here R[r_ij]_{2 x 2} and S = [s_ij]_2x2

⇒ Number of columns in R = 2

⇒ Number of rows in S = 2

Thus the product is defined and the order if product is

Number of rows in R × Number of columns in S

∴ RS = 2 × 2

(i) ⇒ let A : [2 -1] ∴ A[a_ij]_{1 × 2}

⇒ let B : ∴ B[b_ij]_{2 × 1}

Number of columns in A = 2

Number of rows in B = 2

Thus the product is defined and the order if product is

Number of rows in A × Number of columns in B

∴ AB = 1 × 1

⇒ [2 × 5 + (-1) × 4]

⇒ [10-4]

⇒ [6]

(ii) ⇒ let A : ∴ A[a_ij]_{2 × 2}

⇒ let B : ∴ B[b_ij]_{2 × 2}

Number of columns in A = 2

Number of rows in B = 2

Thus the product is defined and the order if product is

Number of rows in A × Number of columns in B

∴ AB = 2 × 2

⇒

⇒

(iii) ⇒ let A : ∴ A[a_ij]_{2 × 3}

⇒ let B : ∴ B[b_ij]_{3 × 2}

Number of columns in A = 3

Number of rows in B = 3

Thus the product is defined and the order if product is

Number of rows in A × Number of columns in B

∴ AB = 2 × 2

⇒

⇒

(iv) let A : ∴ A[a_ij]_{2 × 1}

let B : [2 7] ∴ B[b_ij]_{1 × 2}

Number of columns in A = 1

Number of rows in B = 1

Thus the product is defined and the order if product is

Number of rows in A × Number of columns in B

∴ AB = 2 × 2

× [2 -7]

⇒

⇒

⇒ Let the Sales matrix be A =

⇒ Selling price matrix B =

AB =

=

=

These are the amounts earned on each day.

Total amount earned = 1750 + 1600 + 1650 = 5000 Rs

=

=

=

Comparing with

3x = 9

∴ x = 3

And Y = 0

x = 2, y = - 5

⇒ AX =

=

Comparing with

5x + 3y = -5 ---1

7x + 5y = -11 ---2

Multiply 1 by 5 and multiply 2 by 3 and subtract,

x = 2 and y = -5

⇒ A² = AA =

=

=

⇒ 4A =

=

⇒ 5I =

⇒ A² - 4A + 5I₂ =

=

=

⇒ AB =

=

=

⇒ BA =

=

=

AB not equal to BA

⇒ AB =

=

=

⇒ (AB)C =

=

=

⇒ BC =

=

=

⇒ A(BC) =

=

=

⇒ (AB) C = A (BC)

Thus verified.

⇒ AB =

=

=

⇒ (AB)^T =

⇒ B^T =

⇒ A^T =

⇒ B^T A^T =

=

=

⇒ (AB)^T = B^T A^T

This proved.

⇒ AB =

=

=

⇒ BA =

=

=

Thus A and B are inverse to each other

Let A = [x 1] B = C =

⇒ BC =

=

=

⇒ A(BC) = [x 1]

=

= = 0

∴ x = -3, x = 5

⇒ A = , B = and C =

⇒ A + B =

=

⇒ A + B =

⇒ (A + B)C =

=

=

⇒ AC =

=

=

⇒ BC =

=

=

⇒ AC + BC =

=

⇒ Thus (A + B)C = AC + BC is true

In the above question we see the following terms,

Scalar Matrix, Square Matrix, Diagonal Matrix

To answer this question, we have to know the definition of the above terms.

Square Matrix:

If the rows and columns of the matrices are equal then it will constitute square like structure. So, it is called as square matrix.

E.g.

Diagonal Matrix:

In a square matrix all the elements are zero except the diagonal elements of the matrix. Then that matrix is said to be diagonal matrix.

E.g.

Scalar Matrix:

In a diagonal matrix, if all the diagonal elements are same then that matrix is said to be a scalar matrix.

E.g.

Option(A):

A scalar matrix should be a square matrix so, it is true

Option(B):

Diagonal matrix forms only with the square matrix so, it is true

Option(C):

A scalar matrix consists of zero except the diagonal elements so it is true.

Option(D):

A scalar matrix should comprise of same diagonal elements but in diagonal matrix it may (or) may not contains same diagonal elements. So it is False.

Option(D) is not True in the given.

Given that A is a matrix with m and n as their rows and columns respectively.

We know that for a square matrix both m and n should be equal

So, m = n is the correct answer.

Given,

=

If the one matrix is equivalent to other matrix, then their elements should be equal.

⇒ 3x + 7 = 1 & 5 = y-2

3x = -6 ; y = 7

x = -2 ; y = 7

∴ Option (A) is the answer.

For matrix addition operation their no. of rows and no. of columns should be equal. Otherwise addition is not possible.

In the given matrix A has 1 row and 3 columns

But matrix B has 3 rows and 1 column.

Since rows and columns are not equal it is not possible to add.

So the answer is (d)

Given that a matrix with order 2x3

We know that the no. of elements in the matrix is equal to the product of no. of rows and no. of columns

i.e., No. of elements = No. of rows × No. of columns

∴ No. of elements = 2 x 3

= 6

∴ There will be 6 elements in the given matrix

So, option (B) is the correct answer.

Given,

Since both the matrix is equal to each other, the elements in it also be equal.

By equalizing we will get x = 4

So, option (D) is correct answer.

Given matrix orders

[A] with 3 x 4 and [B] with 4 x 3

We have to find the order of [B].[A]

For matrix multiplication the columns of 1^st matrix and the rows of 2^nd matrix should be equal.

E.g. [A]_3x2.[B]_2x2

The order of new matrix formed after multiplication of matrices will be 1^st matrix row will be taken as new matrix rows. The no. of columns in 2^nd matrix will be taken as new matrix columns.

E.g. [A]_3x2.[B]_2x2 = [AB]_3x2

In the same way [B]_4x3.[A]_3x4 = [AB]_4x4

∴ option (B) is correct answer

Given,

A x

Let us consider as matrix B, it has an order of 2x2

Let us consider the resultant matrix as C, it has an order of 1x2

For matrix multiplication the columns of 1^st matrix and the rows of 2^nd matrix should be equal.

So, matrix A has got same Columns as Rows of matrix B

The order of new matrix formed after multiplication of matrices will be 1^st matrix row will be taken as new matrix rows. The no. of columns in 2^nd matrix will be taken as new matrix columns

So, matrix A has got same Rows as Rows of matrix C

So, Matrix A will consists of 1x2 order

∴ option (C) is correct answer

It is based on a property about multiplication inverse. Only product of a matrix and its own inverse matrix then the resultant matrix will be identity matrix.

AB = I and BA = I

So, B is the multiplicative inverse matrix of A

∴ option (C) is correct answer.

Given, =

=

By equating all the elements in the matrix

We will get 2 equations

x + 2y = 2 ⇒ 1

2x + y = 4 ⇒ 2

By solving the both equations we will get

x = 2; y = 0

∴ option (A) is correct answer

Given that

A + B = O ⇒ B = -A

∴ B =

B =

∴ option (B) is correct answer

Given A =

A² = A×A = x

=

=

=

∴ option (D) is correct answer

Given matrix of A is m x n ⇒ [A]_mxn

Matrix of B is p x q ⇒ [B]_pxq

We know that addition is possible is possible only the order is same for both the matrices

So, m = p, n = q

So, option (D) is true

=

By multiplying the matrices, we will get the value of ‘a’

=

By equating the elements in the matrices

2a-3 = 5

2a = 8

a = 4

So, option (B) is correct answer

Given A =

A² = I

I is the identity matrix

=

By comparing the elements in the matrix

⇒

∴ Option (C) is correct answer.

Given A = [a_ij]_2x2 and a_ij = i + j

It means matrix A with order 2x2 and the elements in is given by a_ij = i + j

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

A = =

∴ Option (B) is correct answer.

Given

By multiplying the matrices, we will get the values of a, b, c & d

By comparing the matrices on both sides

-a = 1; -b = 0; c = 0; d = -1

∴ Option (A) is correct answer.

Given

A + B = ; A =

B =

B =

=

=

∴ Option (C) is correct answer.

Given

By equating the elements

13 – x = 20

x = -7

∴ Option (B) is correct answer.

This is Reversal law for transpose of matrices

∴ (AB)^T = B^TA^T

∴ Option (D) is correct answer.