Class 12 Maths

# Matrices

## NCERT Solution

### Exercise 1 Part 2

Question 5: Construct a 3 x 4 matrix, whose elements are given by

(i) a_(ij)=1/2|-3i+j| (ii) a_(ij)=2i-j

Let [A]_(3xx4)=\begin{bmatrix} a_(11)& a_(12)& a_(13)& a_(14) \\\ a_(21)& a_(22)& a_(23)& a_(24)\\\ a_(31)& a_(32) & a_(33)& a(34)\end{bmatrix}

(i)a_(ij)=1/2|-3i+j|

So, a_(11)=1/2|-3xx1+1|=1/2|-3+1|=1/2xx|-2|=1/2xx2=1

a_(12)=1/2|-3xx1+2|=1/2|-3+2|=1/2|-1|=1/2xx1=1/2

a_(13)=1/2|-3xx1+3|1/2|-3+3|=1/2|0|=1/2xx0=0

a_(14)=1/2|-3xx1+4|=1/2|-3+4|=1/2|1|=1/2xx1=1/2

a_(21)=1/2|-3xx2+1|=1/2|-6+1|=1/2xx|-5|=1/2xx5=5/2

a_(22)=1/2|-3xx2+2|=1/2|-6+2|=1/2xx|-4|=1/2xx4=2

a_(23)=1/2|-3xx2+3|=1/2|-6+3|=1/2xx|-3|=1/2xx3=3/2

a_(24)=1/2|-3xx2+4|=1/2|-6+4|=1/2xx|-2|=1/2xx2=1

a_(31)=1/2|-3xx3+1|=1/2|-9+1|=1/2xx|-8|=1/2xx8=4

a_(32)=1/2|-3xx3+2|=1/2|-9+2|=1/2xx|-7|=1/2xx7=7/2

a_(33)=1/2|-3xx3+3|=1/2|-9+3|=1/2xx|-6|=1/2xx6=3

a_(34)=1/2|-3xx3+4|=1/2|-9+4|=1/2xx|-5|=1/2xx5=5/2

So, A=\begin{bmatrix}1 & 1/2 & 0 &1/2\\5/2 & 2 & 3/2 & 1\\4 & 7/2 & 3 & 5/2\end{bmatrix}

(ii) a_(ij)=2i-j

So, a_(11)=2xx1-1=2-1=1

a_(12)=2xx1-2=2-2=0

a_(13)=2xx1-3=2-3=-1

a_(14)=2xx1-4=2-4=-2

a_(21)=2xx2-1=4-1=3

a_(22)=2xx2-2=4-2=2

a_(23)=2xx2-3=4-3=1

a_(24)=2xx2-4=4-4=0

a_(31)=2xx3-1=6-1=5

a_(32)=2xx3-2=6-2=4

a_(33)=2xx3-3=6-3=3

a_(34)=2xx3-4=6-4=2

Hence, A = \begin{bmatrix}1 & 0 & -1 & -2\\3 & 2 & 1& 0\\5&4&3&2\end{bmatrix}

Question: 6 – Find the values of x, y and z from the following equations:

(i) \begin{bmatrix}4&3\\x&5\end{bmatrix}=\begin{bmatrix}y& z\\1&5\end{bmatrix}

Answer: Here, it is clear that, x=1, y=4 and z=3

(ii) \begin{bmatrix}x+y&2\\5+z& xy\end{bmatrix}=\begin{bmatrix}6&2\\5&8\end{bmatrix}

Solution: Here, x+y=6
Hence, x=6-y
5+z=5
Hence, z=0
Again xy=8
So, (6-y)y=8 (Because x=6-y)
Or, 6y-y^2-8=0
Or, y^2-6y+8=0
Or, y^2-4y-2y+8=0
Or, y(y-4)-2(y-4)=0
Or, (y-2)(y-4)=0
Or, y=2 or y=4
Now, if y=2
Then, x=6-2=4
i.e. x=4, y=2 and z=0
And, if y=4
Then, x=6-4=2
Thus, x=2, y=4 and z=0

(iii) \begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}

Solution: Here, x+y+z=9 ……(i)
x+z=5 …….(II)
y+z=7 ….(iii)
From equations (i) and (ii) we get
y+5=9
Or, y=4
After substituting the value of y in equation (iii) we get
4+z=7
Or, z=3
After substituting the values of z in equation (ii) we get
x+3=5
Or, x=2
Thus, x=2, y=4 and z=3

Question 7: Find the values of a, b, c and d from the equations:

\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5\\0 & 13\end{bmatrix}

Solution: Here, a-b=-1 ……………(i)
2a-b=0 ………………..(ii)
2a+c=5 ……………….(iii)
3c+d=13……………….(iv)

From equations (i) and (ii) we get
a-b-(2a-b)=-1-0
Or, a-b-2a+b=-1
Or, -a=-1 or a=1

After substituting the value of a in equation (i) we get
1-b=-1
Or, -b=-1-1=-2
Or, b=2

After substituting the value of a in equation (iii) we get
2xx1+c=5
Or, c=5-2=3

After substituting the value of c in equation (iv) we get
3xx3+d=13
Or, 9+d=13
Or, d=13-9=4
Thus, a=1, b=2, c=3 and d=4

Question 8: A=[a_(ij)]_(m\xx\n) is a square matrix if:

(A) m < n

(B) n > n

(C) m = n

(D) None of these

Solution: (C) m = n

Explanation: When the number of column is equal to number of row, then matrix is a square matrix. Therefore, option (C) m = n is correct.

Question 9: Which of the given values of x and y make the following pair of matrices equal?

\begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix} \begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}

1. x=-1/3, y=7
2. Not possible to find
3. y=7, x=-2/3
4. x=-1/3, y=-2/3

Solution: (b) Not possible to find

Explanation:

Let \begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix}

Is equal to \begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}

Thus, 3x+7=0
Or, 3x=-7
Or, x=-7/3
And, 2-3x=4
Or, -3x=4-2=2
Or, x=-2/3

Here, x has two values which is not possible, because x can have only one value at a time. Thus, given matrices can never be equal. Thus, option ‘(B) Not possible to find’ is correct.

Question 10: The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512

Solution: (D) 512

Explanation: Given matrices is of 3 x 3 order.

This means the given matrices has 3 x 3 = 9 elements.

As given, there are two options for entry, i.e. 0 and 1.

Thus, number of possible matrices = (2)9 = 512

Hence, option ‘(D) 512’ is correct.