Matrices

NCERT Solution

Exercise 1 Part 2

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

class 12 math matrices ncert exercise 3.1-8

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

Answer:

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.



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