Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices - Exercise 5.1 | Set 2

Chapter 5 of RD Sharma's Class 12 Mathematics textbook continues exploring the Algebra of Matrices, with Exercise 5.1 Set 2 building upon the foundational concepts introduced in Set 1. This set likely delves deeper into matrix operations, properties, and applications. Students will encounter more complex problems involving matrix algebra, further developing their skills in manipulating and analyzing matrices, which are crucial for various fields of mathematics and its applications.

Question 11: Find x, y and z so that A = B, where A= .

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : x-2 = y .............(eq.1)

a₁₂ : z = 3    .............(eq.2)

a₁₃ : 2z = 6  .............(eq.3)

a₂₁ : 18z = 6y ..........(eq.4)

a₂₂ : y+2 = x ...........(eq.5)

a₂₃ : 6z = 2y ............(eq.6)

From (eq.2) and (eq.3),

=> z = 3

From (eq.4) and (eq.6),

=> y = 3z

=> y = 3(3)

=> y = 9

Substitute ( y=9 ) in (eq.1),

=> x-2 = 9

=>x = 11

Thus x=11, y=9 and z=3.

Question 12: If , find x, y, z and w.

Solution:

 As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : x = 3      .............(eq.1)

a₁₂ : 3x-y = 2 .............(eq.2)

a₂₁ : 2x+z = 4 ...........(eq.3)

a₂₂ : 3y-w = 7 ..........(eq.4)

From (eq.1) ,

=> x = 3

Substitute (x=3) in (eq.2),

=> 3(3)-y = 2

=> y = 3(3) - 2

=> y = 7

Substitute ( x=3 ) in (eq.3),

=> 2(3)+z = 4

=>z = 4-6

=>z = -2

Substitute (y=7) in (eq.4),

=>3(7) - w = 7

=>w = 21 - 7

=>w = 14

Thus x=3, y=7, z=-2 and w=14.

Question 13: If , find x, y, z and w.

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : x-y = -1 .............(eq.1)

a₁₂ : z = 4     .............(eq.2)

a₂₁ : 2x-y = 0 ...........(eq.3)

a₂₂ : w = 5 ................(eq.4)

From (eq.2),

=> z = 4

And from (eq.4),

=> w = 5

Now, (eq.1) and (eq.3) form a system of equations comprising of variables x and y. 

Thus, (eq.1) - (eq.2),

=> (x-2x) +(-y+y) = -1 - 0

=> -x = -1

=> x = 1

Substitute (x=1) in (eq.1),

=> 1-y = -1

=>y = 2

Thus, x=1, y=2, z=4 and w=5.

Question 14: If . Obtain the values of a, b, c, x, y and z.

Solution:

 As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : x+3 = 0 .................(eq.1)

a₁₂ : z+4 = 6    .............(eq.2)

a₁₃ : 2y-7 = 3y-2  ........(eq.3)

a₂₁ : 4x+6 = 2x .............(eq.4)

a₂₂ : a-1 = -3 .................(eq.5)

a₂₃ : 0 = 2c+2 ................(eq.6)

a₃₁ : b-3 = 2b+4............(eq.7)

a₃₂ : 3b = -21..................(eq.8)

a₃₃ : z+2c = 0..................(eq.9)

From (eq.1) and (eq.4),

=> x = -3

From (eq.2) ,

=> z+4 = 6

=> z = 2

From (eq.3),

=> 2y-7 = 3y-2

=> 3y-2y = -7+2

=> y = -5

From (eq.5),

=> a = -3+1

=> a = -2

From (eq.8),

=> 3b = -21

=> b = -7

Substitute (z=2) in (eq.9),

=> 2+ 2c = 0

=> c = -1

Thus, x=-3, y=-5, z=2, a=-2, b=-7 and c=-1.

Question 15: If , find the value of (x+y).

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : 2x+1 = x+3...........(eq.1)

a₁₂ : 5x = 10 ..................(eq.2)

a₂₁ : 0 = 0  .....................(eq.3)

a₂₂ : y²+1 = 26 ............(eq.4)

From (eq.1) and (eq.2),

=> 2x+1 = x+3

=> x=2

From (eq.4),

=> y²+1 = 26

=> y² = 25

=> y = ± 5

Thus if y = +5,

=> x+y = 7

And if y = -5,

=> x+y = -3

Question 16: If , then find the values of x, y, z and w.

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : xy = 8 .................(eq.1)

a₁₂ : 4 = w ..................(eq.2)

a₂₁ : z+6 = 0 ..............(eq.3)

a₂₂ : x+y = 6 ..............(eq.4)

From (eq.2),

=> w = 4

And from (eq.3),

=> z = -6

Now, we can see that (eq.1) and (eq.4) form a system of equations comprising of variables x and y.

From (eq.1), 

=> x = 8/y

Substitute (x=8/y) in (eq.4):

=> y + (8/y) = 6

=> y² - 6y +8 = 0

Solving the above equation,

=> y² - 4y - 2y + 8 = 0

=> y( y-4 ) -2( y-4 ) = 0

=> (y - 2)(y - 4) = 0

=> y = 2 or 4

Substitute in (eq.1):

=> when x=2, y=4 and when x=4, y=2.

Thus, (x,y) = (2,4) or (4,2) and z = -6 and w = 4.

Question 17(i): Give an example of a row matrix which is also a column matrix.

Solution:

We know the order of a row matrix can be written as 1xn (1 row with n elements).

And similarly, the order of a column matrix is mx1.

So, a row matrix which is also a column matrix must be of the order (1x1).

As an example, we can take the matrix : 

Question 17(ii): Give an example of a diagonal matrix which is not scalar.

 Solution:

In a diagonal matrix, only the diagonal elements possess non-zero values. Thus, for a nxn diagonal matrix, a_ii ≠ 0, for 1≤ i ≤ n.

And a scalar matrix is a diagonal matrix, such that all the diagonal elements are equal.

Thus, a matrix which is diagonal but not scalar is:

Question 17(iii): Give an example of a triangular matrix.

Solution:

A triangular matrix is a square matrix, and it is filled in such a way that, either the triangle above the main-diagonal is non-zero (upper-triangular) or the triangle below the diagonal is non-zero (lower-triangular).

Thus, an example would be : 

Question 18: The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the month period of January- February revealed that dealer A sold 8 deluxe, 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2x3 matrices summarizing sales data for January and 2-month period for each dealer.

Solution:

The above data can be represented in the form of tables:

For January 2013,

Deluxe	Premium	Standard
Dealer A	5	3	4
Dealer B	7	2	3

For January to February,

Deluxe	Premium	Standard
Dealer A	8	7	6
Dealer B	10	5	7

Thus, the two matrices are :  and 

Question 19: For what value of x and y are the following matrices equal?

A= and B =

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : 2x+1 = x+3 ............(eq.1)

a₁₂ : 2y = y²+2 ...............(eq.2)

a₂₁ : 0 = 0 .........................(eq.3)

a₂₂ : y²-5y = -6 ..............(eq.4)

From (eq.1),

=> 2x-x = 3-1

=> x=2

Taking (eq.2), it can be re-written as,

=> y²-2y+2 = 0

=> y = 

=> y = -1 ± i ( No real solutions )

Taking (eq.4), it can be re-written as,

=> y² - 5y +6 = 0

Solving the equation,

=> y² - 2y - 3y +6 = 0

=> y( y-2 ) -3( y-2 ) = 0

=> ( y-2 )( y-3 ) = 0

=> y = 2 or 3

As values of y are inconsistent, we can say that the above matrices are not equal for any (x,y) pair.

Question 20. Find the values of x and y if  .

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : x+10 = 3x+4 ............(eq.1)

a₁₂ : y²+2y = 3   ...............(eq.2)

a₂₁ : 0 = 0  ..........................(eq.3)

a₂₂ : y²-5y = -4   ..............(eq.4)

From (eq.1),

=> 2x = 6

=> x = 3

Taking (eq.2), it can be re-written as,

=> y²+2y-3 = 0

=> y² + 3y -y -3 = 0

=> y( y+3 ) -1( y+3 ) = 0

=> ( y+3 )( y-1 ) = 0

=> y = -3 or 1

Taking (eq.3), it can be re-written as,

=> y²-5y+4 = 0

=> y² -4y -y +4 =0

=> y( y-4 ) -1( y-4 ) = 0

=> ( y-4 )( y-1 ) = 0

=> y = 4 or 1

The value of y that can satisfy both (eq.2) and (eq.3) is 1.

=> y=1

Thus, x=3 and y=1.

Question 21. Find the values of a and b if A = B , where A= , B= .

Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).

Thus,

a₁₁ : a+4 = 2a+2 ..............(eq.1)

a₁₂ : 3b = b²+2   ..............(eq.2)

a₂₁ : 8 = 8  ..........................(eq.3)

a₂₂ : -6 = b²-5b   .............(eq.4)

From (eq.1),

=> a = 2

Taking (eq.2), it can be re-written as,

=> b²-3b+2=0

=> b² - 2b -b +2 = 0

=> b(b-2) -1(b-2) = 0

=> (b-2)(b-1) = 0

=> b=1 or 2

Taking (eq.4) it can be re-written as,

=> b² -5b +6 = 0

=> b² - 3b -2b + 6 = 0

=> b( b-3 ) -2( b-3 ) = 0

=> ( b-3 )( b-2 ) = 0

=> b = 2 or 3

Thus, b=2 can satisfy both (eq.2) and (eq.4).

=> b = 2

Thus, a=2 and b=2.

Summary

Exercise 5.1 Set 2 in Chapter 5 expands on basic matrix concepts, focusing on more advanced operations and properties. It likely covers topics such as matrix equality, additive inverse of matrices, zero matrices, and special types of matrices like skew-symmetric matrices. This set helps students develop a deeper understanding of matrix algebra, preparing them for more complex applications in later chapters and in higher mathematics.

FAQs on Algebra of Matrices

What is a zero matrix?

A zero matrix is a matrix in which all elements are zero. It acts as the additive identity in matrix addition.

How do you find the additive inverse of a matrix?

The additive inverse of a matrix A is -A, where each element of A is multiplied by -1.

What is a skew-symmetric matrix?

A skew-symmetric matrix is a square matrix A where A^T = -A, or equivalently, a_ij = -a_ji for all i ≠ j, and a_ii = 0.

Can matrices of different orders be added or subtracted?

No, only matrices of the same order can be added or subtracted.

What is meant by the equality of two matrices?

Two matrices are equal if they have the same order and corresponding elements are equal.