# RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices

Here we provide RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices

### Question 1(i): Compute the following sum:.

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2×2.

=>

=>

### Question 1(ii): Compute the following sum:.

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3×3.

=>

=>

### (i): 2A – 3B

Solution:

Both the matrices A and B are of the same order which is 2×2, hence the operation can be performed.

=> 2A =

=> 3B =

=> 2A – 3B =

=> 2A – 3B =

### (ii): B – 4C

Solution:

Both the matrices B and C are of the same order which is 2×2, hence the operation can be performed.

=> B =

=> 4C =

=> B – 4C =

=> B – 4C =

### (iii): 3A – C

Solution:

Both the matrices A and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =

=> C =

=> 3A – C =

=> 3A – C =

### (iv): 3A -2B + 3C

Solution:

The matrices A, B and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =

=> 2B =

=> 3C =

=> 3A – 2B + 3C =

=> 3A – 2B + 3C =

### (i): A + B and B + C

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

B+C can be computed and is solved as follows:

=> B + C =

=> B + C =

### (ii): 2B + 3A and 3C – 4B

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3, and thus 2B + 3A can not be calculated.

3C – 4B can be computed and is solved as follows:

=> 3C =

=> 4B =

=> 3C – 4B =

=> 3C – 4B =

### Question 4: Let A =, B =and C =. Compute 2A – 3B + 4C.

Solution:

The result can be computed since A, B and C are of the same order which is 2×3.

=> 2A =

=> 3B =

=> 4C =

=> 2A – 3B + 4C =

=> 2A – 3B + 4C =

### (i): A – 2B

Solution:

In the given question A and B are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A – 2B = diag(2-2, -5-2, 9+8)

=> A – 2B = diag(0, -7, 17)

### (ii): B + C – 2A

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C – 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C – 2A = diag(-9, 14, -18)

### (iii): 2A + 3B – 5C

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B – 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B – 5C = diag(37, -22, -14)

### Question 6: Given the matrices A =, B =and C =. Verify that (A + B) + C = A + (B + C).

Solution:

Given L.H.S :

=> (A + B) =

=> (A + B) =

=> (A + B) =

=> (A + B) + C =

=> (A + B) + C =

=> (A + B) + C =

Given R.H.S :

=> (B + C) =

=> (B + C) =

=> (B + C) =

=> A + (B + C) =

=> A + (B + C) =

=> A + (B + C) =

Hence R.H.S = L.H.S has been verified.

### Question 7: Find the matrices X and Y, if X + Y =and X – Y =.

Solution:

We know that (X + Y) + (X – Y) = 2X.

=> (X + Y) + (X – Y) =

=> 2X =

=> 2X =

=> X =

=> X =

Now Y = (X + Y) – X

=> Y =

=> Y =

=> Y =

Solution:

Given 2X + Y =

=> 2X +=

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =

### Question 9: Find matrices X and Y, if 2X – Y =and X + 2Y =.

Solution:

We know that 2 (2X – Y) + (X + 2Y) = 4X – 2Y + X + 2Y = 5X .

=> 2 (2X – Y) =

=> 2 (2X -Y) =

=> 2 (2X – Y) + (X + 2Y) =

=> 5X =

=> 5X =

=> X =

=> X =

As (X + 2Y) =

=>

=> 2Y =

=> 2Y =

=> Y =

=> Y =

### Question 10: If X – Y =and X + Y =, find X and Y.

Solution:

We know that (X + Y) + (X – Y) = 2X.

=> 2X =

=> 2X =

=> 2X =

=> X =

=> X =

Also (X + Y) – (X -Y) = 2Y.

=> 2Y =

=> 2Y =

=> 2Y =

=> Y =

=> Y =

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 3×2 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 2×2 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

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