# Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.1 | Set 1

**Question 1: If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?**

**Solution: **

Part 1:Let the matrix be of mxn dimension.

Hence, mxn = 8.

Then, all we have to do is, find the divisors of 8, which are : 1, 2, 4, 8.

Thus, the possible orders are: 1×8, 8×1, 2×4 and 4×2.

Part 2:Following a similar approach:

Since, mxn = 5.

The divisors of 5 are : 1,5.

Thus, the possible orders are: 1×5 and 5×1.

**Question 2(i): If A = [a**_{ij}] = ** and B = [b**_{jj}] = ** then find a**_{22}+b_{21}.

_{ij}] =

_{jj}] =

_{22}+b

_{21}.

**Solution:**

We know that every element in a matrix

Aof dimensions mxn can be addressed as a_{ij}, where 1â‰¤ i â‰¤ m and 1 â‰¤ j â‰¤ n.Thus, a

_{22 }is the 2^{nd}element in the 2^{nd}row ofA, and b_{21}is the 1^{rst}element in the 2^{nd}row ofB.That implies, a

_{22}+ b_{21}= 4 + (-3) = 1.

**Question 2(ii): If A = [a**_{ij}]= ** and B = [b**_{ij}] = ** then find a**_{11}b_{11} + a_{22}b_{22}.

_{ij}]=

_{ij}] =

_{11}b

_{11}+ a

_{22}b

_{22}.

**Solution:**

As seen in the previous question, every element in a matrix

Aof dimensions mxn can be addressed as a_{ij}, where 1â‰¤ i â‰¤ m and 1 â‰¤ j â‰¤ n.Thus,

a

_{11}= 1^{rst}element in the 1^{rst }row ofA= 2.a

_{22}= 2^{nd}element in the 2^{nd}row ofA= 4.b

_{11}= 1^{rst}element in the 1^{rst}row ofB= 2.b

_{22}= 2^{nd}element in the 2^{nd}row ofB= 4.That implies, a

_{11}b_{11 }+ a_{22}b_{22}= (2×2) + (4×4) = 4 + 16 = 20.

**Question 3: Let A be a matrix of order 3×4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.**

**Solution:**

Given,

Ais a matrix of order 3×4.We know that a matrix having an order of mxn has m rows and n columns.

Thus,

Acontains, 3 rows, and each row contains 4 elements.Now, if R1 is a row, it has 4 elements, and thus its order can be written as 1×4,

And similarly, if C2 is a column, it will have 3 rows, each with 1 element, and thus its order is 3×1,

**Question 4(i): Construct a 2×3 matrix A = [a**_{ij}] whose elements a_{ij} are given by : a_{ij} = i x j.

_{ij}] whose elements a

_{ij}are given by : a

_{ij}= i x j.

**Solution:**

We know that

Ais a matrix of order 2×3.Thus

Acan be depicted as:Since each element is a product of its row number and column number (i x j):

a

_{11}= 1 a_{12 }= 2 a_{13}= 3a

_{21 }= 2 a_{22 }= 4 a_{23}= 6Thus,

Acan be depicted as :

**Question 4(ii): Construct a 2×3 matrix A = [a**_{ij}] whose elements a_{ij} are given by : a_{ij} = 2i – j.

_{ij}] whose elements a

_{ij}are given by : a

_{ij}= 2i – j.

**Solution:**

We know that

Ais a matrix of order 2×3.Thus

Acan be depicted as:Since each element can be defined as (2 x (row number)) – column number:

a

_{11}= 1 a_{12}= 0 a_{13}= -1a

_{21}= 3 a_{22}= 2 a_{23}= 1Thus,

Acan be depicted as :

**Question 4(iii): Construct a 2×3 matrix A = [a**_{ij}] whose elements a_{ij} are given by : a_{ij} = i + j.

_{ij}] whose elements a

_{ij}are given by : a

_{ij}= i + j.

**Solution:**

We know that

Ais a matrix of order 2×3.Thus

Acan be depicted as:Since each element can be defined as the sum of its row number and column number:

a

_{11}= 2 a_{12}= 3 a_{13}= 4a

_{21}= 3 a_{22}= 4 a_{23}= 5Thus,

Acan be depicted as :

**Question 4(iv): Construct a 2×3 matrix A = [a**_{ij}] whose elements a_{ij }are given by : a_{ij} = **.**

_{ij}] whose elements a

_{ij }are given by : a

_{ij}=

**Solution:**

We know that

Ais a matrix of order 2×3.Thus

Acan be depicted as:Since each element can be defined as : ,

a

_{11}= 2 a_{12}= 4.5 a_{13}= 8a

_{21}= 4.5 a_{22}= 8 a_{23}= 12.5Thus,

Acan be depicted as :

**Question 5(i): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : .

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as :Since each element can be defined as :,

a

_{11}= 2 a_{12}= 4.5a

_{21}= 4.5 a_{22}= 8Thus,

Acan be depicted as :

**Question 5(ii): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as :Since each element can be defined as : ,

a

_{11}= 0 a_{12}= 0.5a

_{21}= 0.5 a_{22}= 0Thus,

Acan be depicted as :

**Question 5(iii): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as :Since each element can be defined as : ,

a

_{11}= 0.5 a_{12}= 4.5a

_{21}= 0 a_{22}= 2Thus,

Acan be depicted as :

**Question 5(iv): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution: **

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as :Since each element can be defined as : ,

a

_{11}= 4.5 a_{12}= 8a

_{21}= 12.5 a_{22}= 18Thus,

Acan be depicted as :

**Question 5(v): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as : ,Since each element can be defined as : ,

a

_{11 }= 0.5 a_{12 }= 2a

_{21}= 0.5 a_{22}= 1Thus,

Acan be depicted as :

**Question 5(vi): Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as : ,Since each element can be defined as : ,

a

_{11}= 1 a_{12}= 0.5a

_{21}= 2.5 a_{22}= 2Thus,

Acan be depicted as :

**Question 5(vii) Construct a 2×2 matrix A = [a**_{ij}] whose a_{ij} are given by : ** .**

_{ij}] whose a

_{ij}are given by :

**Solution:**

We know that

Ais a matrix of order 2×2.Thus

Acan be depicted as : ,Since each element can be defined as : ,

a

_{11}= e^{2x}sinx a_{12}= e^{2x}sin2xa

_{21}= e^{4x}sinx a_{22}= e^{4x}sin2xThus,

Acan be depicted as :

**Question 6(i): Construct a 3×4 matrix A = [a**_{ij}] whose a_{ij }are given by : a_{ij} = i + j .

_{ij}] whose a

_{ij }are given by : a

_{ij}= i + j .

**Solution:**

Ais a matrix of order 3×4.Thus,

Acan be depicted as : ,Since each element can be defined as : ( row number + column number ),

a

_{11}= 2 a_{12}= 3 a_{13}= 4 a_{14}= 5a

_{21}= 3 a_{22}= 4 a_{23}= 5 a_{24}= 6a

_{31}= 4 a_{32}= 5 a_{33}= 6 a_{34}= 7Thus,

Acan be depicted as :

**Question 6(ii): Construct a 3×4 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = i – j.

_{ij}] whose a

_{ij}are given by : a

_{ij}= i – j.

**Solution:**

Ais a matrix of order 3×4.Thus,

Acan be depicted as : ,Since each element can be defined as : (row number – column number),

a

_{11}= 0 a_{12}= -1 a_{13}= -2 a_{14}= -3a

_{21}= 1 a_{22}= 0 a_{23}= -1 a_{24}= -2a

_{31}= 2 a_{32}= 1 a_{33 }= 0 a_{34}= -1Thus,

Acan be depicted as :

**Question 6(iii): Construct a 3×4 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = 2i .

_{ij}] whose a

_{ij}are given by : a

_{ij}= 2i .

**Solution:**

Ais a matrix of order 3×4.Thus,

Acan be depicted as : ,Since each element can be defined as : (2 x row number ),

a

_{11}= 2 a_{12}= 2 a_{13}= 2 a_{14}= 2a

_{21}= 4 a_{22}= 4 a_{23}= 4 a_{24}= 4a

_{31}= 6 a_{32}= 6 a_{33}= 6 a_{34}= 6Thus,

Acan be depicted as :

**Question 6(iv): Construct a 3×4 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = j.

_{ij}] whose a

_{ij}are given by : a

_{ij}= j.

**Solution:**

Ais a matrix of order 3×4.Thus,

Acan be depicted as : ,Since each element can be defined as : (column number ),

a

_{11}= 1 a_{12}= 2 a_{13}= 3 a_{14}= 4a

_{21}= 1 a_{22}= 2 a_{23}= 3 a_{24}= 4a

_{31}= 1 a_{32}= 2 a_{33}= 3 a_{34}= 4Thus,

Acan be depicted as :

**Question 6(v): Construct a 3×4 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = ** .**

_{ij}] whose a

_{ij}are given by : a

_{ij}=

**Solution:**

Ais a matrix of order 3×4.Thus,

Acan be depicted as : ,Since each element can be defined as : ,

a

_{11}= -1 a_{12}= -1/2 a_{13}= 0 a_{14}= 1/2a

_{21}= -5/2 a_{22}= -2 a_{23}= -3/2 a_{24}= -1a

_{31}= -4 a_{32}= -7/2 a_{33}= -3 a_{34 }= -5/2Thus,

Acan be depicted as :

**Question 7(i): Construct a 4×3 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = **.**

_{ij}] whose a

_{ij}are given by : a

_{ij}=

**Solution:**

Ais a matrix of order 4×3.Thus,

Acan be depicted as :Since each element can be defined as : ,

a

_{11}= 3 a_{12}= 5/2 a_{13}= 7/3a

_{21}= 6 a_{22}= 5 a_{23}= 14/3a

_{31}= 9 a_{32}= 15/2 a_{33}= 7a

_{41}= 12 a_{42}= 10 a_{43 }= 28/3Thus,

Acan be depicted as :

**Question 7(ii): Construct a 4×3 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = **.**

_{ij}] whose a

_{ij}are given by : a

_{ij}=

**Solution:**

Ais a matrix of order 4×3.Thus,

Acan be depicted as:Since each element can be defined as : ,

a

_{11}= 0 a_{12}= -1/3 a_{13}= -1/2a

_{21}= 1/3 a_{22}= 0 a_{23}= -1/5a

_{31}= 1/2 a_{32}= 1/5 a_{33}= 0a

_{41}= 3/5 a_{42}= 1/3 a_{43}= 1/7Thus,

Acan be depicted as :

**Question 7(iii): Construct a 4×3 matrix A = [a**_{ij}] whose a_{ij} are given by : a_{ij} = i.

_{ij}] whose a

_{ij}are given by : a

_{ij}= i.

**Solution:**

Ais a matrix of order 4×3.Thus,

Acan be depicted as :Since each element can be defined as : (row number)

a

_{11}= 1 a_{12}= 1 a_{13}= 1a

_{21}= 2 a_{22}= 2 a_{23}= 2a

_{31}= 3 a_{32}= 3 a_{33}= 3a

_{41}= 4 a_{42}= 4 a_{43}= 4Thus,

Acan be depicted as :

**Question 8: Find x, y, a and b if: **

**Solution:**

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a

_{11}: 3x+4y = 2 ………………(eq.1) a_{12}: 2 = 2 a_{13}: x-2y = 4 ………………(eq.2)a

_{21}: a+b = 5 ………………(eq.3) a_{22}: 2a-b = -5……………..(eq.4) a_{23}: -1 = -1Thus, (eq.1) and (eq.2) form one system of equations comprising variables x and y.

Solving (eq.1) and (eq.2): (eq.1) + 2x(eq.2)

=> (3x+2x) + (4y-2(2y)) = 2+ (2(4))

=> 5x = 10

=>

x = 2Substituting (x=2) in (eq.1) :

=> (3(2)) + 4y = 2

=> 4y = 2-6 = -4

=>

y=-1Similarly, (eq.3) and (eq.4) form a system of equations comprising variables a and b.

Solving (eq.3) and (eq.4) : (eq.1) + (eq.2)

=> (a+2a) + (b-b) = 5 – 5

=> 3a = 0

=>

a = 0Substituting (a=0) in (eq.3):

=> 0 + b = 5

=>

b = 5Thus, a=0, b=5, x=2 and y=-1.

**Question 9: Find x, y, a and b if : ****.**

**Solution:**

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on theRHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a

_{11}: 2x-3y = 1………………(eq.1) a_{12}: a-b = -2………………(eq.2) a_{13}: 3 = 3a

_{21}: 1 = 1 a_{22}: x+4y = 6……………..(eq.3) a_{23}: 3a+4b = 29………………(eq.4)Thus, (eq.1) and (eq.3) form one system of equations comprising of variables x and y.

Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)

=> (2x-2x) + (-3y-2(4y)) = 1- (2(6))

=> -11y = -11

=> y = 1Substituting (y=1) in (eq.1) :

=> 2x – 3(1) = 1

=> 2x = 3+1 = 4

=> x = 2Similarly, (eq.2) and (eq.4) form a system of equations comprising of variables a and b.

Solving (eq.2) and (eq.4) : 4x(eq.1) + (eq.2)

=> (4a+3a) + (-4(b)+4b) = 4(-2) + 29

=> 7a = 21

=> a = 3Substituting (a=3) in (eq.2) :

=> 3 – b = -2

=>b = 5Thus, a=3, b=5, x=2 and y=1.

**Question 10: Find a, b, c and d if : **

**Solution:**

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×2.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a

_{11 }: 2a+b = 4 ………….(eq.1)a

_{12 }: a-2b = -3 …………(eq.2)a

_{21 }: 5c-d = 11 …………(eq.3)a

_{22 }: 4c+3d = 24 ……..(eq.4)Thus, (eq.1) and (eq.2) form one system of equations comprising of variables a and b.

Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)

=> (2a-2a) + (b+4b) = 4 + (-2(-3))

=> 5b = 10

=>

b = 2Substituting (b=2) in (eq.1) :

=> 2a + 2 = 4

=> 2a = 4-2 = 2

=>

a=1Similarly, (eq.3) and (eq.4) form a system of equations comprising of variables c and d.

Solving (eq.3) and (eq.4) : 3x(eq.1) + (eq.2)

=> (15c+4c) + (-3d+3d) = 33 + 24

=> 19c = 57

=>

c = 3Substituting (c=3) in (eq.4) :

=> (4(3)) + 3d = 24

=>3d = 24 – 12 = 12

=>

d = 4Thus, a=1, b=2, c=3 and d=4.

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