**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,

a

_{11 }: x-2 = y ………….(eq.1)a

_{12 }: z = 3 ………….(eq.2)a

_{13 }: 2z = 6 ………….(eq.3)a

_{21 }: 18z = 6y ……….(eq.4)a

_{22 }: y+2 = x ………..(eq.5)a

_{23 }: 6z = 2y …………(eq.6)From (eq.2) and (eq.3),

=>

z = 3From (eq.4) and (eq.6),

=> y = 3z

=> y = 3(3)

=>

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

=> x-2 = 9

=>

x = 11Thus 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,

a

_{11 }: x = 3 ………….(eq.1)a

_{12 }: 3x-y = 2 ………….(eq.2)a

_{21 }: 2x+z = 4 ………..(eq.3)a

_{22 }: 3y-w = 7 ……….(eq.4)From (eq.1) ,

=>

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

=> 3(3)-y = 2

=> y = 3(3) – 2

=>

y = 7Substitute ( x=3 ) in (eq.3),

=> 2(3)+z = 4

=>z = 4-6

=>

z = -2Substitute (y=7) in (eq.4),

=>3(7) – w = 7

=>w = 21 – 7

=>

w = 14Thus 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,

a

_{11 }: x-y = -1 ………….(eq.1)a

_{12 }: z = 4 ………….(eq.2)a

_{21 }: 2x-y = 0 ………..(eq.3)a

_{22 }: w = 5 …………….(eq.4)From (eq.2),

=>

z = 4And from (eq.4),

=>

w = 5Now, (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 = 1Substitute (x=1) in (eq.1),

=> 1-y = -1

=>

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

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

**Solution:**

Thus,

a

_{11 }: x+3 = 0 ……………..(eq.1)a

_{12 }: z+4 = 6 ………….(eq.2)a

_{13 }: 2y-7 = 3y-2 ……..(eq.3)a

_{21 }: 4x+6 = 2x ………….(eq.4)a

_{22 }: a-1 = -3 ……………..(eq.5)a

_{23 }: 0 = 2c+2 …………….(eq.6)a

_{31 }: b-3 = 2b+4…………(eq.7)a

_{32 }: 3b = -21………………(eq.8)a

_{33 }: z+2c = 0………………(eq.9)From (eq.1) and (eq.4),

=>

x = -3From (eq.2) ,

=> z+4 = 6

=>

z = 2From (eq.3),

=> 2y-7 = 3y-2

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

=>

y = -5From (eq.5),

=> a = -3+1

=>

a = -2From (eq.8),

=> 3b = -21

=>

b = -7Substitute (z=2) in (eq.9),

=> 2+ 2c = 0

=>

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

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

**Solution:**

Thus,

a

_{11 }: 2x+1 = x+3………..(eq.1)a

_{12 }: 5x = 10 ………………(eq.2)a

_{21 }: 0 = 0 …………………(eq.3)a

_{22 }: y^{2}+1 = 26 …………(eq.4)From (eq.1) and (eq.2),

=> 2x+1 = x+3

=>

x=2From (eq.4),

=> y

^{2}+1 = 26=> y

^{2}= 25=>

y = ± 5Thus if y = +5,

=>

x+y = 7And if y = -5,

=>

x+y = -3

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

**Solution:**

Thus,

a

_{11 }: xy = 8 ……………..(eq.1)a

_{12 }: 4 = w ………………(eq.2)a

_{21 }: z+6 = 0 …………..(eq.3)a

_{22 }: x+y = 6 …………..(eq.4)From (eq.2),

=>

w = 4And from (eq.3),

=>

z = -6Now, 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

=> y

^{2}– 6y +8 = 0Solving the above equation,

=> y

^{2}– 4y – 2y + 8 = 0=> y( y-4 ) -2( y-4 ) = 0

=> (y – 2)(y – 4) = 0

=>

y = 2 or 4Substitute in (eq.1):

=> when

x=2, y=4and whenx=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

(1×1).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 2×3 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,

DeluxePremiumStandardDealer A5 3 4 Dealer B7 2 3 For January to February,

DeluxePremiumStandardDealer A8 7 6 Dealer B10 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:**

Thus,

a

_{11 }: 2x+1 = x+3 …………(eq.1)a

_{12 }: 2y = y^{2}+2 ……………(eq.2)a

_{21 }: 0 = 0 …………………….(eq.3)a

_{22 }: y^{2}-5y = -6 …………..(eq.4)From (eq.1),

=> 2x-x = 3-1

=>

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

=> y

^{2}-2y+2 = 0=> y =

=>

y = -1 ± i( No real solutions )Taking (eq.4), it can be re-written as,

=> y

^{2}– 5y +6 = 0Solving the equation,

=> y

^{2}– 2y – 3y +6 = 0=> y( y-2 ) -3( y-2 ) = 0

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

=>

y = 2 or 3As 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:**

Thus,

a

_{11 }: x+10 = 3x+4 …………(eq.1)a

_{12 }: y^{2}+2y = 3 ……………(eq.2)a

_{21 }: 0 = 0 ……………………..(eq.3)a

_{22 }: y^{2}-5y = -4 …………..(eq.4)From (eq.1),

=> 2x = 6

=>

x = 3Taking (eq.2), it can be re-written as,

=> y

^{2}+2y-3 = 0=> y

^{2 }+ 3y -y -3 = 0=> y( y+3 ) -1( y+3 ) = 0

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

=> y

= -3 or 1Taking (eq.3), it can be re-written as,

=> y

^{2}-5y+4 = 0=> y

^{2}-4y -y +4 =0=> y( y-4 ) -1( y-4 ) = 0

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

=>

y = 4 or 1The value of y that can satisfy both (eq.2) and (eq.3) is 1.

=>

y=1Thus, x=3 and y=1.

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

**Solution:**

Thus,

a

_{11 }: a+4 = 2a+2 …………..(eq.1)a

_{12 }: 3b = b^{2}+2 …………..(eq.2)a

_{21 }: 8 = 8 ……………………..(eq.3)a

_{22 }: -6 = b^{2}-5b ………….(eq.4)From (eq.1),

=>

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

=> b

^{2}-3b+2=0=> b

^{2}– 2b -b +2 = 0=> b(b-2) -1(b-2) = 0

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

=>

b=1 or 2Taking (eq.4) it can be re-written as,

=> b

^{2}-5b +6 = 0=> b

^{2}– 3b -2b + 6 = 0=> b( b-3 ) -2( b-3 ) = 0

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

=>

b = 2 or 3Thus, b=2 can satisfy both (eq.2) and (eq.4).

=>

b = 2Thus, a=2 and b=2.