Matrices-NCERT Exemplar Problems & Solution
Short Answer Type Questions
Question: (1)If a matrix has `28` elements, what are the possible orders it can have? What if it has 13 elements?
Solution:
No of elements = `28`
`:.`Possible orders of the matrix
` = 1 xx 28, 2 xx 14, 4 xx 7, 7 xx 4, 14 xx 2, 28 xx 1`
If the matrix has 13 elements, then the possible orders are `1 xx 13` and `13 xx 1`.
Question: (2)In the matrix `A = [(a, 1, x), (2, sqrt3, x^2-y), (theta, 5, - 2/5)]`
Write
(i)The order of the matrix `A`
(ii)The number of elements
(iii)Write elements `a_23 a_3p a_12`
Solution:
Here `A = [(a, 1, x), (2, sqrt3, x^2-y), (theta, 5, - 2/5)]`
(i) Order oof the matrix `A= 3 xx3`
(ii)The numbers of elements`=9`
(iii)`a_23= x^2- y, a_31 = 0, a_12 =1`
Question: (3)Construct `a_(2xx2)` maatrix where;
(i)`a_(ij)=(i - 2j)^2/2`
Question: (ii)`a_(ij)=|-2i+3j|`
Solution:
(i)`a_(ij)=(i - 2j)^2/2`
`:. a_11= (1-2 xx1)^2/2`
`=(1-2)^2/2 =1/2`
`a_12 = (1-2 xx1)^2/2 = 9/2`
`a_21 = (1-2 xx1)^2/2=0`
`a_22 = (2-2xx2)^2/2`
`(2-4)^2/2 = 4/2=2`
`:.` Required matrix is `[(1/2, 9/2), (0, 2)]`
(ii) `a_ij = |-2i+3j|`
` a_11 = |-2 xx 1 + 3 xx 1|`
`= |-2+3|=1`
`:. a_12 = |-2 xx 1 + 3 xx 2|`
`= |-2 + 6|= 4`
`a_21 = | -2 xx2 + 3 xx 1|`
`=|-4 + 3| = |-1| =1`
`a_22 |-2 xx2 + 3 xx 2|`
`|-4 + 6 |=2`
Required matrix is `[(1, 4), (1, 2)]`
Question: (4)Construct a `3 xx 2` matrix whose elements are given by `a_ij=e^ix sin jx`.
Solution:
Here, we have to construct `a 3xx2` matrix whose elements are given by
`a_(ij)= e^(ix)sin jx`
`:. a_11 = e^x sin x`
` a_12 = e^x sin 2x`
` a_21 = e^2x sin x`
` a_22 = e^2x sin 2x`
` a_31 = e^3x sin x`
` a_32 = e^3x sin 2x`
Required matrix is `[(e^x sin x, e^x sin 2x), (e^2x sin x, e^2x sin2x),(e^3x sin x, e3x sin 2x)]`
Question: (5)Find the values of `a` and `b` if `A=B`, where
`A = [(a+4, 3b), (8, -6)]`
`B = [(2a+2, b^2+2), (8, b^2-5b)]`
Solution:
Here , `A = [(a+4, 3b), (8, -6)]`
and `B = [(2a+2, b^2+2), (8, b^2-5b)]`
and `A=B`
=>` [(a+4, 3b), (8, -6)] = [(2a+2, b^2+2), (8, b^2-5b)]`
`=> a + 4 =2a +2`
`3b= b^2+2`
`b^2-5b =-6`
`=> a = 2`
`b^2-3b + 2 = 0`
`b^2 - 5b + 6 = 0`
`=> a = 2`
`b^2 - 2b - b + 2 = 0`
`b^2 -3b -2b +6 =0`
`=> a = 2`
` b(b-2) - (b-2) = 0`
` b(b-3) - 2(b-3) = 0`
`=> a = 0`
`(b-2) - (b-1) = 0`
` (b-3) - (b-2) = 0`
`=> a = 2`
` b = 2 or 1`
`b = 3 or 2`
` a = 2`
`b = 2` Answer
Question: (6)If possible, find the sum of the matriees `A` and `B` where
`A = [(sqrt3, 1), (2, 3)]` and `B= [(x, y, z), (a, b, 6)] `.
Solution:
Here `A = [(sqrt3, 1), (2, 3)]`
and `B = [(x, y, z), (a, b, 6)]`.
`:.` Order of `A= 2 xx 2` and order of `B= 2 xx 3`
`:.` Order of `A !=` Order of `B`
`:.` Sum of matrices `A` and `B` is not possible.
Question: (7)If `X = [(3, 1, -1), (5, -2, -3)]` and `y= [(2, 1, -1), (7, 2, 4)]` find
(i) ` X+Y`
(ii)`2X- 3Y`
(iii)A matrix `Z` such that `X + Y + Z ` is a zero matrix.
Solution:
Here, `X = [(3, 1, -1), (5, -2, -3)]`
and `Y = [(2, 1, -1), (7, 2, 4)]`
(i) `X+Y=[(3, 1, -1), (5, -2, -3)]+[(2, 1, -1), (7, 2, 4)]`
`[(3+2, 1+1, -1+(-1)), ( 5+7, -2+2, -3+4)]`
`[(5, 2, -2), (12, 0, 1)]`
(ii)`2X-3Y = 2[(3, 1, -1), (5, -2, -3)]-3 [(2, 1, -1), (7, 2, 4)]`
`= [(6, 2, -2), (10, -4, -6)]- [(6, 3, -3), (21, 6, 12)]`
`= [(0, -1, 1), (-11, -10, -8)]`
(iii) Let `Z = [(a, b, c), (d, e, f)]`
`because X+Y+Z=0`
`=>[(3, 1, -1), (5, -2, -3)]+ [(2, 1, -1), (7, 2, 4)]` `+[(a, b, c), (d, e, f)]`
`=[(0, 0, 0), (0, 0, 0)]`
`=> [(5, 2, -2), (12, 0, 1)]+ [(a, b, c), (d, e, f)]`
`=[(0, 0, 0), (0, 0, 0)]`
`=> [(5, 2, -2), (12, 0, 1)]=[(-a, -b, -), (-d, -e, -f)]`
`=> a=-5`
` b=-2`
`c=2`
`d=-12`
`e=0`
`f=-1`
`:. Z= [(-5, -2, 2), (-12, 0, -1)]`
Question: (8) Find non- zero values of `x` satisfying the matrix equation:
`x[(2x, 2), (3, x)]+2 [(8, 5x), (4, 4x)]`
`=2[(x^2+8, 24), (10, 6x)]`
Solution:
Here ,`x[(2x, 2), (3, x)]+2 [(8, 5x), (4, 4x)]`
`=2[(x^2+8, 24), (10, 6x)]`
`[(2x^2, 2x), (3x, x^2)]+2 [(1, 10x), (8, 8x)]`
`= [(x^2+16, 48), (20, 12x)]`
`=[(2x^2+16, 2x+10x), (3x+8, x^2+8x)]`
`= [(2x^2+16, 48), (20, 12x)]`
`=[(2x^2+16, 12x), (3x+8, x^2+8x)]`
`= [(2x^2+16, 48), (20, 12x)]`
Using equality of two matrices, we have
`2x^2+ 16 = 2x^2 +16;`
`12x=48;`
`3x+8=20:
and `x^2 +8x = 12x`
`=> x = 4, x = 4, x = 0 ` or `4`
`:. x = 4`
Question: (9)If `A= [(0, 1), (1, 1)]` and `B =[(0, -1), (1, 0)]`, show that `(A+B)(A-B)!= A^2 -B^2`
Solution:
Here, `A= [(0, 1), (1, 1)]` and `B =[(0, -1), (1, 0)]`
`:. (A+B)(A -B)`
`{[(0, 1), (1, 1)] + [(0, -1), (1, 0)]}{[(0, 1), (1, 1)] -[(0, -1), (1, 0)]}`
`{[(0, 0), (2, 1)]}{[(0, 2), (0, 1)]}`
`[(0, 0), (2, 1)][(0, 2), (0, 1)]`
`[(0, 0), (0, 5)]`
`A^2 = [(0, 1), (1, 1)][(0, 1), (1, 1)]`
` = [(1, 1), (1, 2)]`
`B^2 = [(0, -1), (1, 0)][(0, -1), (1, 0)]`
` = [(-1, 0), (0, -1)]`
`:. A^2- B^2 = [(1, 1)(1, 2)]-[(-1, 0), (0, -1)]`
`[(2, 1), (1, 3)]` -------(ii)
From (i) and (ii), we have `(A+B)(A-B)!= A^2 - B^2`, Proved.
Question: (10) Find the value of `x` if
`[(1, x, 1)][(1, 3, 2), (2, 5, 1), (15, 3, 2)][(1), (2), (x)]=0`
Solution:
Here `[(1, x, 1)][(1, 3, 2), (2, 5, 1), (15, 3, 2)][(1), (2), (x)]=0`
`=> [(1+2x+15, 3+5x+3, 2+x+2)] [(1), (2), (x)]=0`
`[2x + 16 + 12 + 10x +4x + x^2]=0`
` => x^2+16x+28 =0`
`=> x^2+ 14 x + 2x + 28 =0`
`=> x(x+1)+2(x+14)=0`
`=> (x+14)(x+2)=0`
`=>(x+14)(x+2)=0`
`x=-14` or -2` Answer
Question: (11)Show that `A= [(5, 3), (-1, -2)]` satisfies the equation `A^2-3A-71=0` and hence find`A^-1`.
Solution:
Here , `A = [(5, 3), (-1, -2)]` and given equation is
`A^2=3A-71 =0 `
Now` A^2= [(5, 3), (-1, -2)][(5, 3), (-1, -2)]`
`[(22, 9), (-3, 1)]`
`3A= 3[(5, 3), (-1, -2)]=[(15, 9), (-3, -6)]`
`71 =7[(1, 0), (0, 1)]=[(7, 0), (0, 7)]`
`:.A^2-3A-71`
` = [(22, 9), (-3, 1)]- [(15, 9), (-3, -6)]-[(7, 0), (0, 7)]`
`= [(22-15-7, 9-9-0), (-3+3-0, 1+6-7)]`
`=[(0, 0), (0, 0)]=0`
This show that `A` satisfies the given equation
`A^2 - 3A-71 = 0`
Pre- multiplying both sides by `A^-1`, we get `A^-1A^2-3A^-1A-7A^-1 1 = A^-1 0`
`=> A-31-7A^-1 =0`
`[:. A^-1A =1` and `A^-1 1 =A^-1]`
`=>7A^-1 = A-31`
`=[(5, 3), (-1, -2)]- [(3, 0), (0, 3)]`
`= [(2, 3), (-1, -5)]`
`=> A^-1 = 1/7 [(2, 3), (-1, -5)] = [(2/7, 3/7), (-1/7, -5/7)]`
Question: (12)Find the matrix A satisfying the matrix equation;
`[(2, 1), (3, 2)]A[(-3, 2), (5, -3)]`
`=[(1, 0), (0, 1)]`
Solution:
Here , `[(2, 1), (3, 2)]A[(-3, 2), (5, -3)]`
`= [(1, 0), (0, 1)]`
Let ` A = [(x, y), (z, u)]`
Now, `[(2, 1), (3, 2)][(x, y), (z, u)][(-3, 2), (5, -3)]`
`= [(1, 0), (0, 1)]`
` => [(2x+z, 2y+u), (3x+2z, 3y+2u)][(-3, 2), (5, -3)]`
`=[(1, 0), (0, 1)]`
`=>[(-6x-3z+10y+5u, 4x+2z-6y-3u), (-9x-6z+15y+10u, 6x+4z-9y-6u)]`
`=[(1, 0), (0, 1)]`
`=> -6x-3z+10y+5u=1` ---(i)
` =>4x +2z-6y-3u=0` -----(ii)
`=> -9x-6z+15y+10u=0` ----(iii)
`=> 6x+4x-9y-6u=1`
------(iv)Solving (i), (ii), (iii) and (iv) , we get
`x=1,y = 1, z= 1, u =0`
`:. ` The matrix `A = [(1, 1), (1, 0)]`
Altier : `[(2, 1), (3, 2)]A[(-3, 2), (5, -3)]`
` = [(1, 0), (0, 1)]`
`=>A [(-3, 2), (5, -3)]`
`=>[(2, 1), (3, 2)]^-1[(1, 0), (0, 1)]`
``= [(2, -1), (-3, 2)][(1, 0), (0, 1)]`
`=[(2, -1), (-3, 2)]`
`=> A = [(2, -1), (-3, 2)][(-3, 2), (5, -3)]^-1`
`= [(2, -1), (-3, 2)][(3, 2), (5, 3)]`
`[(1, 1), (1, 0)]`
Question: (13)Find A, if `[(4), (1), (3)]A = [(-4, 8, 4), (-1, 2, 1), (-3, 6, 3)]`
Solution:
Since,`[(4), (1), (3)]A = [(-4, 8, 4), (-1, 2, 1), (-3, 6, 3)]`
`:.` Order of A should be `1xx3`
Let `A=[(x, y, z)]`
Now , `[(4), (1), (3)][(x, y, z)]= [(-4, 8, 4), (-1, 2, 1), (-3, 6, 3)]`
`=> [(4x, 4y, 4z), (x, y, z), (3x, 3y, 3z)] = [(-4, 8, 4), (-1, 2, 1), (-3, 6, 3)]`
`=> 4X = -4, 4Y = 8, 4Z = 4`
`:. A= [(-1, 2, 1)]`
Question: (14) If `A = [(3, -4), (1, 1), (2, 0)]` and `B = [(2, 1, 2), (1, 2, 4)]`
then verify `(BA)^2 != B^2 A^2`
Solution:
Here `A = [(3, -4), (1, 1), (2, 0)]`
`B = [(2, 1, 2), (1, 2, 4)]`
`BA = [(2, 1, 2), (1, 2, 4)][(3, -4), (1, 1), (2, 0)]`
`= [(11, -3), (13, -2)]`
`(BA)^2 =[(11, 7), (13, -2)][(11, -7), (13, -2)]`
`= [(121-91, -77+14), (143-26, -91+4)]`
`=[(30, -63), (117, -87)]`----(i)
` B^2 = [(2, 1, 2), (1, 2, 4)][(2, 1, 2), (1, 2, 4)]`
Question: (15)If possible, find `BA` and `AB`, where
`A= [(2, 1, 2), (1, 2, 4)]` and ` B= [(4, 1), (2, 3), (1, 2)]`
Solution:
Here `A= [(2, 1, 2), (1, 2, 4)]`
and ` B= [(4, 1), (2, 3), (1, 2)]`
`BA= [(4, 1), (2, 3), (1, 2)][(2, 1, 2), (1, 2, 4)]`
` = [(9, 6, 12), (7, 8, 16), (4, 5, 10)]`
And `AB= [(2, 1, 2), (1, 2, 4)][(4, 1), (2, 3), (1, 2)]`
`=[(12, 9), (12, 15)]`
Question(16)Show by an example that for `A!= 0,B!=0,AB =0`
Solution:
Let `A = [(1, 0), (0, 0)]=0`
` B = [(0, 0), (4, 100)] = 0`
`AB= [(1, 0), (0, 0)][(0, 0), (4, 100)]`
`= [(0, 0), (0, 0)]=0` Proved.
Reference: