Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.2 | Set 2

Chapter 5 of RD Sharma's Class 12 Mathematics textbook continues to explore the Algebra of Matrices. Exercise 5.2 Set 2 typically delves deeper into matrix operations, focusing on more complex problems involving matrix addition, subtraction, multiplication, and related concepts. This set often includes questions that require a combination of matrix operations and properties, helping students develop a more comprehensive understanding of matrix algebra.

Question 11: Find matrix A, if+ A =.

Solution:

Given,+ A =.

=> A =

=> A =

=> A =

Question 12: If A =, B =, find C such that 5A + 3B + 2C is a null matrix.

Solution:

Given 5A + 3B + 2C =O, where O is the null matrix.

=> 2C = O - 5A - 3B.

=> 5A =

=> 3B =

=> 2C =

=> 2C =

=> 2C =

=> C =

=> C =

Question 13: If A =, B =, find matrix X such that 2A + 3X = 5B.

Solution:

Given 2A + 3X = 5B.

=> 3X = 5B - 2A.

=> 5B =

=> 2A =

=> 3X =

=> 3X =

=> 3X =

=> X =

=> X =

Question 14: If A =and B =, find the matrix C such that A + B + C is a zero matrix.

Solution:

Given that A + B + C = O, where O is a null matrix.

=> C = O - A - B.

=> C =

=> C =

=> C =

Question 15(i): Find x, y satisfying the matrix equation

Solution:

Given that,

We can arrive at 2 equations from the above matrix equation.

=> x - y + 3 = 6

=> x - y = 3 ......(eq.1)

=> x + 0 = 2x + y

=> -x = y ..........(eq.2)

Solving (eq.1) and (eq.2) for x and y.

=> 2x = 3

=> x = 3/2

Substitute x in (eq.2)

=> y = -3/2

Question 15(ii): Find x, y and z satisfying the matrix equation

Solution:

Given that,

=>

=>

We can arrive at 3 equations from the above matrix equation.

=> x + y = 4 ......(eq.1)

=> y + 6 = 9 ......(eq.2)

=> z + 2 = 12 ....(eq.3)

From (eq.2),

=> y = 9 - 6

=> y = 3

From (eq.3),

=> z = 12 - 2

=> z = 10

Substitute the value of y in (eq.1),

=> x + 3 = 4

=> x = 4 - 3

=> x = 1

Question 15(iii): Find x and y satisfying the matrix equation O.

Solution:

Given that,

We can arrive at 2 equations from the above matrix equation.

=> 2x + 3y - 8 = 0

=> 2x + 3y = 8 .......(eq.1)

=> x + 5y -11 = 0

=> x + 5y = 11 .......(eq.2)

Solving for x and y , (eq.1) - 2.(eq.2),

=> 2x -2x + 3y - 10y = 8 - 22

=> -7y = -14

=> y = 2

Substitute y in (eq.2),

=> x + 5(2) = 11

=> x = 11 - 10

=> x = 1

Question 16: If, find x and y.

Solution:

Given that,

We can arrive at 2 equations from the above matrix equation.

=> 2x + 1 = 5.......(eq.1)

=> 8 + y = 0........(eq.2)

Solving for x,

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

Solving for y,

=> y = -8

Question 17: Find the value of, a non-zero scalar, if

Solution:

Given that,

=> We can arrive at several equations to solve forhowever lets take one.

=>

=>

If we substitute, in the matrix we see that the equation remains consistent.

Hence,.

Question 18(i): Find a matrix X such that 2A + B + X = O, where A =, B =.

Solution:

Given that, 2A + B + X = O.

=> 2A =

=> X = O - 2A - B

=> X =

=> X =

=> X =

Question 18(ii): If A =and B =, then find the matrix X of order 3x2 such that 2A + 3X = 5B.

Solution:

Given that 2A + 3X = 5B.

=> 3X = 5B - 2A.

=> 5B =

=> 2A =

=> 3X =

=> 3X =

=> 3X =

=> X =

=> X =

Question 19(i): Find x, y, z and t, if .

Solution:

Given that,

We can arrive at 4 different equations from the above matrix equation,

=> 3x = x + 4 ............(eq.1)

=> 3y = 6 + x + y ....(eq.2)

=> 3z = -1 + z + t ...(eq.3)

=> 3t = 2t + 3 ...........(eq.4)

From (eq.1),

=> 2x = 4

=> x = 2

Substitute x=2 in (eq.2),

=> 3y = 6 + 2 + y

=> 2y = 8

=> y = 4

From (eq.4),

=> t = 3

Substitute t=3 in (eq.3),

=> 3z = -1 + z + 3

=> 2z = 2

=> z = 1

Question 19(ii): Find x, y, z and t, if .

Solution:

Given that,

We can arrive at 2 equations from the above matrix equation,

=> 2x + 3 = 7 ....................(eq.1)

=> 2 (y - 3) + 2 = 14 ....(eq.2)

From (eq.1),

=> 2x = 7 - 3

=> 2x = 4

=> x = 2

From (eq.2),

=> 2y - 6 + 2 = 14

=> 2y = 14 + 4

=> 2y = 18

=> y = 9

Question 20: If X and Y are 2x2 matrices, then solve the following matrix equations for X and Y, 2X + 3Y =, 3X + 2Y =.

Solution:

Let 2X + 3Y =be (eq.1) and let 3X + 2Y =, be (eq.2) .

=> 2(2X + 3Y) - 3(3X + 2Y) = 4X + 6Y - 9X - 6Y = -5X.

=> -5X =

=> -5X =

=> -5X =

=> -5X =

=> 5X =

=> X =

=> X =

Substitute the matrix X in (eq.1),

=> 3Y =

=> 3Y =

=> 3Y =

=> 3Y =

=> Y =

=> Y =

Question 21: In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind of in all the colleges.

Solution:

Let the different posts in each college be represented as :

Now the total posts will be computed as follows:

Question 22: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves 15000 per month, find their monthly incomes using the matrix method.

Solution:

The problem can be solved by considering two matrices, one for expenditure and one for income.

=> The income matrix is: where x is a constant.

=> The expenditure matrix is:where y is a constant.

=>

We arrive at 2 equations from the above matrix equation.

=> 3x - 5y = 15000........(eq.1)

=> 4x - 7y = 15000........(eq.2)

Solving for y by 4(eq.1) - 3(eq.2),

=> 12x - 20y - 12x + 21y = 4(15000) - 3(15000)

=> y = 15000

Substitute the value of y in (eq.1),

=> 3x = 15000 + 5(15000)

=> 3x = 15000 + 75000

=> 3x = 90000

=> x = 30000

=> Their incomes and expenditures are,

=> 3x = 3(30000) = 90000 and 5y = 5(15000) = 75000

=> 4x = 4(30000) = 120000 and 7y = 7(15000) = 105000