Matrices | ML Aggarwal | Class 12th Solutions | ISC Board

Matrices

Exercise 3.1

1.If A is a matrix of type p×q and R is a row of A, then what is the type of R as a matrix?

Ans. Since, A is a matrix of type p×q therefore, each row of A contains q elements.

Hence, R is a row matrix of type 1× q.

2.If A is a column matrix with 5 rows, then what type of matrix is a row of A?

Ans. 1× 1

3. If a matrix has 18 elements, what the possible orders it can have? What , if it has 11 elements?

Ans. We know that if a matrix is of order m×n, then it have mn elements. Thus to find all possible orders of a matrix with 18 elements, we shall find all ordered pairs(of natural numbers) whose product is 18.

All possible ordered pairs are-

(1,18) , (18,1) , (2,9) , (9,2) , (3, 6) , (6,3)

Hence, the possible orders are-

(1×18) , (18×1) , (2×9) , (9×2) , (3×6) , (6× 3).

Further, if it has 11 elements, then the possible orders are (1×11) , (11×1).

4. If A is a 3×3 matrix whose elements are given by a ij =|-3i+j|, write the value of a23.

Ans. $$ The value of \begin{matrix}a_{23}\end{matrix}=$$

$$\begin{matrix}a_{23}\end{matrix}= 1/3|-3×2 +3| $$

$$\begin{matrix}a_{23}\end{matrix}= 1/3|-6 +3| $$

$$\begin{matrix}a_{23}\end{matrix}= 1/3|3| $$

$$\begin{matrix}a_{23}\end{matrix}= 1/3(3) $$

$$\begin{matrix}a_{23}\end{matrix}= 1 $$

5. Construct a 2×2 matrix A= [a ij] whose elements aij are given by:

$$(i)\begin{matrix}a_{ij}\end{matrix}= 2i-j$$ $$(ii)\begin{matrix}a_{ij}\end{matrix}=|2i- 3j|.$$

Ans. $$(i)\begin{matrix}a_{ij}\end{matrix}=2i-j$$

$$\begin{matrix}a_{11}\end{matrix}=2×1-1=1$$

$$\begin{matrix}a_{12}\end{matrix}=2×1-2=0$$

$$\begin{matrix}a_{21}\end{matrix}=2×2-1=3$$

$$\begin{matrix}a_{22}\end{matrix}=2×2-2=2$$

The matrix of order 2×2 is

$$\begin{bmatrix}1&0\\3&2\end{bmatrix}$$

$$(ii)\begin{matrix}a_{ij}\end{matrix}=|2i- 3j|.$$

$$\begin{matrix}a_{11}\end{matrix}=|2×1 – 3×1| =|2-3| =|-1| =1.$$

$$\begin{matrix}a_{12}\end{matrix}=|2×1 – 3×2| =|2-6| =|-4| =4.$$

$$\begin{matrix}a_{21}\end{matrix}=|2×2 – 3×1| =|4-3| =|-1| =1.$$

$$\begin{matrix}a_{22}\end{matrix}=|2×2 – 3×2| =|4-6| =|-2| =2.$$

The matrix of order 2×2 is

$$\begin{bmatrix}1&4\\1&2\end{bmatrix}$$

6. Construct a 2×3 matrix B=[b ij] whose elements bij are given by:

$$(i)\begin{matrix}b_{ij}\end{matrix}=i-3j$$

$$(ii)\begin{matrix}b_{ij}\end{matrix}=(i+2j)^2.$$

Ans. $$(i)\begin{matrix}b_{ij}\end{matrix}=i-3j$$

$$\begin{matrix}b_{11}\end{matrix}=1-3×1 = -2$$

$$\begin{matrix}b_{12}\end{matrix}=1-3×2 = -5$$

$$\begin{matrix}b_{13}\end{matrix}=1-3×3 = -8$$

$$\begin{matrix}b_{21}\end{matrix}=2-3×1 = -1$$

$$\begin{matrix}b_{22}\end{matrix}=2-3×2 = -4$$

$$\begin{matrix}b_{23}\end{matrix}=2-3×3 = -7$$

The matrix of order 2 × 3 is

$$\begin{bmatrix}-2&-5&-8\\-1&-4&-7\end{bmatrix}$$

$$(ii)\begin{matrix}b_{ij}\end{matrix}=(i+2j)^2.$$

$$\begin{matrix}b_{11}\end{matrix}=(1+2×1)^2=3^2=9$$

$$\begin{matrix}b_{11}\end{matrix}=(1+2×1)^2=3^2=9$$

$$\begin{matrix}b_{12}\end{matrix}=(1+2×2)^2=5^2=25$$

$$\begin{matrix}b_{13}\end{matrix}=(1+2×3)^2=7^2=49$$

$$\begin{matrix}b_{21}\end{matrix}=(2+2×1)^2=4^2=16$$

$$\begin{matrix}b_{22}\end{matrix}=(2+2×2)^2=6^2=36$$

$$\begin{matrix}b_{23}\end{matrix}=(2+2×3)^2=8^2=64$$

The matrix of order 2×3 is

$$\begin{bmatrix}9&25&49\\16&36&64\end{bmatrix}$$

$$7. (i)If \begin{bmatrix}x+2y&-y\\3x&4\end{bmatrix}=\begin{bmatrix}-4&3\\6&4\end{bmatrix}$$,find the values of x and y.

$$(ii) If \begin{bmatrix}x+3y&y\\7-x&4\end{bmatrix}=\begin{bmatrix}-4&-1\\0&4\end{bmatrix}$$,find the values of x and y.

(iii)Find the value of y,$$if\begin{bmatrix}x-y&2\\x&5\end{bmatrix}=\begin{bmatrix}2&2\\3&5\end{bmatrix}$$

(iv) Find the value of x, $$if\begin{bmatrix}3x+3y&-y\\2y-x&3\end{bmatrix}=\begin{bmatrix}-1&2\\-5&3\end{bmatrix}.$$

$$(v)If \begin{bmatrix}x+2y&3y\\4x&2\end{bmatrix}=\begin{bmatrix}0&-3\\8&2\end{bmatrix}$$,find the values of (x-y).

$$(vi)If \begin{bmatrix}x-y&z\\2x-y&w\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}$$,find the values of (x+y).

$$(vii)If \begin{bmatrix}x+y&x+2\\2x-y&16\end{bmatrix}=\begin{bmatrix}8&5\\1&3y+1\end{bmatrix}$$,find the values of (y-x).

Ans.

7(i). $$\begin{bmatrix}x+2y&-y\\3x&4\end{bmatrix}=\begin{bmatrix}-4&3\\6&4\end{bmatrix}$$

=>x+2y=-4

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

=>x-6=-4

=>x=2

And,

=>-y=3

So, y=-3

$$7(ii).\begin{bmatrix}x+3y&y\\7-x&4\end{bmatrix}=\begin{bmatrix}-4&-1\\0&4\end{bmatrix}$$

=>y=-1

=>7-x=0

=>x=7

$$7(iii).\begin{bmatrix}x-y&2\\x&5\end{bmatrix}=\begin{bmatrix}2&2\\3&5\end{bmatrix}$$

x=3……….(eq 1)

x-y=2………(eq 2)

putting the value of x in eq 2

=>3-y=2

=>-y=2-3

=>-y=-1

=>y=1

So, the value of y=1.

$$7(iv).\begin{bmatrix}3x+3y&-y\\2y-x&3\end{bmatrix}=\begin{bmatrix}-1&2\\-5&3\end{bmatrix}.$$

-y=2

y=-2……eq 1

3x+y=1……eq 2

Putting the value of y in eq 2

3x-2=1

3x=1+2

3x=3

x=1

So, the value of x=1.

$$7(v).\begin{bmatrix}x+2y&3y\\4x&2\end{bmatrix}=\begin{bmatrix}0&-3\\8&2\end{bmatrix}$$

4x=8

x=2……eq 1

x+2y=0……..eq 2

Putting the value of x in eq 2

2+2y=0

2y=-2

y=-1……eq 3

So, the value of x – y=2-(-1)= 3.

$$7(vi).\begin{bmatrix}x-y&z\\2x-y&w\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}$$

x-y=-1

x=y-1…………..eq 1

2x-y=0…………eq 2

Putting the value of x in eq 2

2(y-1)-y=0

2y-2-y=0

y-2=0

y=2……………eq 3

Putting the value of y in eq 1

x=y-1

x=2-1

x=1

Hence, the value of x+y=1+2=3

$$7(vii).\begin{bmatrix}x+y&x+2\\2x-y&16\end{bmatrix}=\begin{bmatrix}8&5\\1&3y+1\end{bmatrix}$$

x+2=5

x=5-2

x=3……..eq 1

x+y=8…………eq 2

Putting the value of x in eq 2

3+y=8

y=5

So, the value of(y-x)=5-3=2

$$8. If \begin{bmatrix}x&3x-y\\2x+z&3y-w\end{bmatrix}=\begin{bmatrix}3&2\\4&7\end{bmatrix}$$, find x, y, z and w.

Ans. x=3…………eq 1

3x-y=2……………..eq 2

Putting the value of x in eq 2

3×3 – y=2

9-y=2

-y=2-9

-y=-7

y = 7………..eq 3

2x+z=4……….eq 4

Putting the value of x in eq 4

2×3+z=4

6+z=4

z=4-6

z=-2

3y-w=7…………..eq 5

Putting the value of y in eq 5

3×7-w=7

21-7=y

y=14

So, the value of x, y , z and w are 3, 7,-2 and 14.

9.Find the value of a, b,c and d from the following equations: $$\begin{bmatrix}2a+b&a -2\\5c-d&4c+3d\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix}$$

=> 2a+b=4……………eq 1

(a- 2b=-3)×2

=> 2a+b=4

2a-4b=-6

—————–

– + +

5b=10

b=2……….eq 2

Putting the value of b in eq 1

2a+2=4

a=1

=> 5c-d=11

d=5c-11……..eq 3

4c+3d=24………..eq 4

Now, putting the value of d in eq 4

4c + 3(5c- 11)=24

4c + 15c -33 =24

19c=24+33

19 c=57

c=3

Now, put the value of c in eq 3

d=5c-11

d=5×3-11

d=15-11

d=4

10. Find x, y, a and b $$If \begin{bmatrix}2x-3y&a-b&3\\1&x+4y&3a+4b\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$$

Ans. x+4y=6

x=6-4y…………eq 1

2x-3y=1…………..eq 2

Now, putting the value of x in eq 2

2(6-4y)-3y=1

12-8y-3y=1

12-11y=1

11y = 11

y =1

Now, putting the value of y in eq 1

x=6-4y

x=6-4×1

x=2

a-b=2

a=2+b……………eq 3

3a + 4b=29………..eq 4

Now, putting the value of a in eq 4

3(2+b)+4b=29

6+3b+4b=29

6+7b=29

7b=21

b=3

Now put the value of b in eq 3

a=2+b

a=2+3

a=5

Hence, the value of x,y a and b are 2 , 1, 3 and 5

11. (i) Write the numbers of all possible matrices of order 2×3 with each entry 1 or 2

Ans. Since, the number of elements in matrix of order 2 × 3is 1 or 2 i.e. 2 elements.

So, the number of possible matrices of order 2 × 3 are $$\begin{matrix}2^6=64\end{matrix}$$

(ii) Write the numbers of all possible matrices of order 2×2 with each entry 1, 2 or 3

Ans. Since, the number of elements in matrix of order 2 × 2is 1,2 or 3 i.e. 3 elements.

So, the number of possible matrices of order 2 × 2 are $$\begin {matrix}3^4=81\end{matrix}$$