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Class Twelve Mathematics

Matrices-NCERT Exemplar Problems & Solution

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Short Answer Question:part-3

Question(32) Let `A= [(1, 2), (-1, 3)], B= [(4, 0), (1, 5)],`

`C=[(1, 0), (1, -2)]` and `a = 4, b =-2`

Show that

(a)`A+(B+C)+(A+B)+C`

Solution:

Here `A= [(1, 2), (-1, 3)]`

`B= [(4, 0), (1, 5)]`

`C= [(2, 0), (1, -2)]`

and `a= 4, b= -2`

(a)`A+(B+C)+(A+B)+C`

Now, `B+C = [(4, 0), (1, 5)]+[(2, 0), (1, -2)]= [(6, 0), (2, 3)]`

`:. A + (B+C)= [(1, 2), (-1, 3)]+[(6, 0), (2, 3)]`

`= [(7, 2), (1, 6)]`----(i)

`A+B = [(1, 2), (-1, 3)]+ [(4, 0), (1, 5)]`

` = [(5, 2), (0, 8)]`

`:. (A+B)+ C= [(5, 2), (0, 8)]+[(2, 0), (1, -2)]`

` =[(7, 2), (1, 6)]` ------(ii)

From (i) and (ii).

`A+(B+C) = (A + B)+ C ` Proved

(b)`A (BC)= (AB)C`

Solution:

`BC = [(4, 0), (1, 5)][(2, 0) (1, -2)]`

`=[(8, 0), (7, -10)]`

`:. A(BC)= [(1, 2), (-1, 3)][(8, 0), (7, -10)]`

`= [(22, -20), (13, -30)]` ----(iii)

`AB= [(1, 2), (-1, 3)][(4, 0), (1, 5)]= [(6, 10), (-1, 15)]`

`:. (AB)C = [(6, 10), (-1, 15)][(2, 0), (1, -2)]`

`= [(22, -20), (13, -30)]`----(iv)

From (iii)and (iv),

`A(BC)= (AB)C` Proved

(c) `(a+ b) B = aB+ bB, a= 4, b=-2`

Solution:

`(a+b)B = (4-2)[(4, 0), (1, 5)]`

`=2[(4, 0), (1, 5)]=[(8, 0), (2, 10)]`----(v)

`aB+bB = 4 [(4, 0), (1, 5)]+ (-2)[(4, 0), (1, 5)]`

`=[(16, 0), (4, 20)]- [(8, 0), (2, 10)]`

`= [(8, 0), (2, 10)]` -----(vi)

From (v)and (vi), we get

`(a+b) B = aB + bB` Proved.

(d)`a(C-A)=aC- aA, a=4`

Solution:

`:.a(C-A)=4{[(2, 0), (1, -2)]-[(1, 2), (-1, 3)]}`

`=4 [(1, -2), (2, -5)]`

`=[(4, -8), (8, -20)]`----(vii)

`aC - aA = 4[(2, 0), (1, -2)]-4[(1, 2), (-1, 3)]`

`=[(8, 0), (4, -8)]-[(4, 8), (-4, 12)]`

`[(4, -8), (8, -20)]`

From (vii)and (viii),

`a(C-A)= aC-aA` Proved

(e)`(A^T)^T = A`

Solution:

`A = [(1, 2), (-1, 3)]`

`A^T = [(1, -1), (2, 3)]`

`(A^T)^T = [(1, 2), (-1, 3)]=A` Proved.

(f)`(bA)T=bA^T, b=-2`

Solution:

` bA =-2[(1, 2), (-1, 3)]`

`=[(-2, -4), (2, -6)]`

`:. (bA)^T= [(-2, -2), (4, -6)]` ----(ix)

`A^T = [(1, -1), (2, 3)]`

`:. bA^T= -2 [(1, -1), (2, 3)]`

`=[(-2, 2), (-4, -6)]`

From (ix and (x), we get

`(bA)^T = bA^T`

(g) `(AB)^T= B^TA^T`

Solution:

`AB= [(1, 2), (-1, 3)][(4, 0), (1, 5)]= [(6, 10), (-1, 15)]`

`:. (AB)^T = [(6, -1), (10, 15)]`-----(ix)

`A^T= [(1, 1), (1, -1), (2, 3)]`

`B^T = [(4, 1), (0, 5)]`

`B^TA^T= [(4, 1), (0, 5)][(1, -1), (2, 3)]`

`= [(6, -1), (10, 15)]`------(xii)

From (xi) and (xii), we get

`(AB)^T= B^TA^T` Proved

(h) `(A-B)C= AC-BC`

Solution:

`A-B= [(1, 2), (-1, 3)]-[(4, 0), (1, 5)]`

`= [(-3, 2), (-2, -2) ]`

`:. (A-B)C= [(-3, 2), (-2, -2)][(2, 0), (2, 0), (1, -2)]`

`=[(-4, -4), (-6, 4)]`-----(xiii)

`AC= [(1, 2), (-1, 3)][(2, 0), (1, -2)]`

`= [(4, -4), (1, 5)]`

`BC= [(4, 0), (1, 3)][(2, 0), (1, -2)]`

`= [(8, 0), (7, -10)]`

`AC - BC= [(4, -4), (1, -6)]-[(8, 0), (7, -10)]`

`= [(4, -4), (-6, 4)]`-------(xiv)

`(a -B)C = AC-BC` Proved

(i)`(A- B)^T = A^T- B^T`

Solution:

`A-B= [(1, 2), (-1, 3)]-[(4, 0), (1, 5)]`

`= [(-3, 2), (-2, -2)]`

`:. (A-B)^T = [(-3, 2), (-2, -2)]`----(xv)

`A^T=[(1, -1), (2, 3)]`

`B^T= [(4, 1), (0, 5)]`

`A^T-B^T = [(1, -1), (2, 3)]-[(4, 1), (0, 5)]`

`= [(-3, -2), (2, -2)]`----(xvi)

From (xv)and (xvi), we get

`(A-B)^T = A^T - B^T` Proved

Question (33) If `A = [(cos theta, sin 2theta), (-sin theta , cos theta)]`, then show that

`A^2 = [(cos 2theta, sin 2theta), (-sin 2theta , cos 2theta)]`

Solution:

Here `A[(cos theta, sin 2theta), (-sin theta , cos theta)]`

`A^2= [(cos theta, sin 2theta), (-sin theta , cos theta)][(cos theta, sin 2theta), (-sin theta , cos theta)]`

`=[(cos^2theta - sin^2 theta, cos theta sin theta + sin theta cos theta), (-sin theta cos theta - cos theta sin theta, -sin^2 theta + cos^2 theta)]`

`= [ (cos 2theta, 2sin theta cos theta), (-2sin theta cos theta, cos 2theta)]`

`(because cos^2theta - sin^2theta = cos 2theta)`

`= [(cos 2theta, sin 2theta), (-sin 2theta, cos 2theta)]`

`(because 2 sin theta cos theta = sin 2theta)` Proved

Question (34) If `A= [(0, -x), (x, 0)]-[(0, 1), (1, 0)]` and ` x^2=-1`, then show that `(A+B)^2 = A^2 + B^2` .

Solution:

Here `A= [(0, -x), (x, 0)]`

`B= [(0, 1), (1, 0)]`

and `x^2 =-1`

`(A+B)= [(0, -x), (x, 0)]+[(0, 1), (1, 0)]`

`=[(0, 1-x), (1+x, 0)]`

`(A+B)^2= [(0, 1-x), (1+x, 0)][(0, 1-x), (1+x, 0)]`

`[(1-x^2, 0), (0, 1-x^2)]`

`=[(1+1, 0), (0, 1+1)](because x^2=-1)`

`[(2, 0), (0, 2)]`

`A^2= [(0, -x), (x, 0)][(0, -x), (x, 0)]`

`= [(-x^2, 0), (0, -x^2)]`

`B^2 = [(0, 1), (1, 0)][(0, 1), (1, 0)]=[(1, 0), (0, 1)]`

`A^2+B^2 = [(-x^2, 0), (0, -x^2)]+[(1, 0), (0, 1)]`

`[(-x^2+1, 0), (0, -x^2+1)]`

`[(1+1, 0), (0, 1+1)]`

`

(because x^2 =-1)`

`[(2, 0), (0, 2)]` ----(ii)

From (i)and (ii), we have

`(A+B)^2 = A^2 +B^2` Proved

Question (35)Verify that `A^2 =1` when `A = [(0, 1, -1), (4, -3, 4), (3, -3, 4)]`

Solution:

Here , `A = [(0, 1, -1), (4, -3, 4), (3, -3, 4)]`

`A^2 = [(0, 1, -1), (4, -3, 4), (3, -3, 4)] [(0, 1, -1), (4, -3, 4), (3, -3, 4)]`

` = [(1, 0, -0), (0, -1, 0), (0, 0, 1)]`=1 Proved

Question (36)Prove by Mathematical Induction that `(A^1)^n= (A^n)^1`, Where `n in N` for any square matrix `A`

Solution:

Let `p(n) :(A')^n =(A^n)^1`

`=> A^1= A^1`, Which is true.

Let it be true for `n=k`

`:. P(K): (A^1)^k= (A^k)^1`

Now for `n= k+1`

` p(k+1): (A^1)^(k+1)= (A^1)^k A^1 = (A^k)^1 A^1`

` (A^k)^1 (because B^1A^1 = (AB)^1)`

`(A^(k+1))^1`,

Which is true whenever.

`P(k) ` is true

`:. ` By PMI it is true for all `n=inN`

Question (37) Find inverse, by elementary row operations (if possible), of the following matrices: (i)`[(1, 3), (-5, 7)]` (ii)`[(1,- 3), (-2, 6)]`

Solution:

(i) Let `A=[(1, 3), (-5, 7)]`

Now,` A= IA`

`=> [(1, 3), (-5, 7)]=[(1, 0), (0, 1)]A`

`R_2-> R_2 + 5R_1`

`[(1, 3), (0, 22)]=[(1, 0), (5, 1)]A`

`R_2 -> 1/22 R_2`

`[(1, 3), (0, 1)]= [(1, 0), (5/22, 1/22)]A`

`R_1 -> R_1 - 3 R_2`

`[(1, 0), (0, 1)]= [(1 - \15/22, 0 - 3/22), (5/22, 1/22)]A`

`[(1, 0), (0, 1)]= [(7/22, 3/22), (5/22, 1/22)]A`

`:. A^-1= [(7/22, 3/22), (5/22, 1/22)]A` Answer.

(ii) Let `A= [(1, -3), (-2, 6)]`

Now, A=IA

`=> [(1, -3), (-2, 6)]=[(1, 0), (0, 1)]A`

`R_2->R_2 +2R_1`

`[(1, -3), (0, 0)]=[(1, 0), (2, 1)]A`

Since we have all zeroes in `R_2`

Hence inverse for the given matrix does not exist.

Question (38)If `[(xy, 4) (z+6, x+y)]= [(8, w), (0, 6)]`,then find values of `x, y, z` and `w`.

Solution:

Here,`[(xy, 4) (z+6, x+y)]= [(8, w), (0, 6)]`

`=> xy = 8, w =4`(using equality of two matrices)

` z+6 = 4, x+y=6`

` =>w = 4, z = -6`

`y= 6-x`

`x(6-x)=8`

`=>x^2- 6x+8= 0`

`=>x^2-4x-2x+8=0`

`(x-4(x-2)=0`

`x=4, 2`

`:. y=2, 4`

`:. x=4,y= 2,z=-6, w=4`

or `x=2,y= 4,z=-6, w=4`

Question (39)If`A =[(1, 5), (7, 12)]` and `B=[(9, 1), (7, 8)]`, find a matrix C such that `3A +5B + 2 is a null matrix.

Solution:

Here , `A =[(1, 5), (7, 12)]`and `B=[(9, 1), (7, 8)]` We find a matrix C such that `3A+5B+2C is a null matrix.

`:. 3A + 5B +2C =0`

`3A+5B = -2C`

`=3[(1, 5), (7, 12)]+5[(9, 1), (7, 8)]=-2C`

`[(3, 15), (21, 36)]+ [(45, 5), (35, 40)]=-2C`

`=>[(48, 20), (56, 76)]=-2C`

`=> C = - 1/2[(48, 20), (56, 76)]`

` => C= [(-24, -10), (-28, -38))]`Answer

Question (40) If `A= [(3, -5), (-4, 2)]`, then find `A^2 - 5A-141`. Hence, obtain `A^3`.

Solution:

Here ,`A= [(3, -5), (-4, 2)]`

To find `A^2-5A-141`

`:.A^2- 5A-141`

`= [(3, -5), (-4, 2)][(3, -5), (-4, 2)]` `-5[(3, -5), (-4, 2)]-14[(1, 0), (0, 1)]`

`= [(29, -25), (-20, 24)]-[(15, -25), (-20, 10)]` `-[(14, 0), (0, 14)]`

`= [(0, 0), (0, 0)]`

To get `A^3`, pre-multiply both sides of

`A^2-5A-141=0`, we get

`AA^2 - 5AA-14AI =0`

`=>A^3 - 5A^2 - 14 A= 0`

`(because AI=A) `

`=> A^3 = 5A^2 + 14A`

`= 5[(29, -25), (-20, 24)]+14 [(3, -5), (-4, 2)]`

`[(145, -125), (-100, 120)]+[(42, -70), (-56, 28)]`

`= [(187, -195), (-156, 148)]`Ans.

Question (41)Find the values of `a,b,c` and `d` if `2[(a, b), (c, d)]= [(a, 6), (1, 2d)]=[(4, a+b), (c+d , 3)]`

Solution:

Here, `3[(a, b), (c, d)]= [(a, 6), (-1, 2d)]+[(4, a+b), (c+d , 3)]`

`[(3a, 3b), (3c, 3d)]= [(a+4, 6+a+b), (-1+c+d, 2d+3)]`

` 3a = a+4`

`3b= 6+a+b`

`3c= -1+c+d`

`3d= 2d+3`

`2a= 4`

`2b= 6+a`

`2c-d=-1`

`=>d=3`

`a=2`

`2b = 6 +2`

`2c-3 =-1`

`d=3`

`=> a=2`

`b=4`

`c=1`

` d=3`

Question (42)Find the matrix `A` such that `[(2, -1), (1, 0), (-3, 4)]A= [(-1, -8, -10), (1, -2, -5), (9, 22, 15)]`

Solution:

Here `[(2, -1), (1, 0), (-3, 4)]A= [(-1, -8, -10), (1, -2, -5), (9, 22, 15)]`

Since `[(2, -1), (1, 0), (-3, 4)]` is of order ` 3 xx 2 a d `the

matrix `[(-1, -8, -10), (1, -2, -5), (9, 22, 15)]` is of order `3xx3`.

A must be of order `2xx3`

Let `A= [(a, b, c), (d, e, f)]`

`[(2, -1), (1, 0), (-3, 4)][(a, b, c), (d, e, f)]=[(-1, -8, -10), (1, -2, -5), (9, 22, 15)]`

`=> [(2a-d, 2b-c, 2c-f), (a, b, c), (-3a+4d, -3b+4c, -3c+4f)]= [(-1, -8, -10), (1, -2, -5), (9, 22, 15)]`

`=>2a- d=-1`

`2b-e=-8`

`2c-f=-10`

`a=1`

`b=-2`

`c=-5`

`=> d = 3`

`e = 4`

`f = 0`

`:.A= [(1, -2, -5), (3, 4, 0)]`Ans.

Question (43) If `A= [(1, 2), (4, 1)]`, find `A^2 + 2A +71`,

Solution:

Here, `A = [(1, 2), (4, 1)]`

To find `A^2 + 2A+71`

Now, `A^2+2A+71`

`=[(1, 2), (4, 1)][(1, 2), (4, 1)]+2[(1, 2), (4, 1)]+7[(1, 0), (0, 1)]`

` [(9, 4), (8, 9]+ [(2, 4), (8, 2)]+[(7, 0), (0, 7)]`

`[(18, 8), (16, 18)]`Ans.

Question (44)If`A= [(cos alpha, sin alpha), (-sin alfa, cos alpha)]`and `A^-1 = A^1` find value of `alpha`.

Solution:

Here `A= [(cos alpha, sin alpha), (-sin alpha, cos alpha)]`

`M_11 = cos alpha, C_11 = cos alpha`

`M_12 = -sin alpha, C_12 = sin alpha`

`M_21 = sin alpha, C_21 =- sin alpha`

`M_22 = cos alpha, C_22 = cos alpha`

Adj.` A = [(cos alpha, sin alpha), (-sin alpha, cos alpha)]`

`=[(cos alpha, -sin alpha), (sin alpha, cos alpha)]`

`|A|[(cos alpha, sin alpha), (-sin alfa, cos alpha)]`

`= cos^2alpha + sin^2 alpha=1`

`A^-1 =(adj.A)/|A| = [(cos alpha, -sin alpha), (sin alpha, cos alpha)]`

`A^1 = [(cos alpha, sin alpha), (-sin alpha, cos alpha)]`

Since, `A^1=A^1`

`=>[(cos alpha, -sin alpha), (sin alpa, cos alpha)]`

`[(cos alpha, -sin alpha), (sin alpha, cos alpha)]` , Which is true for all real values of `alpha`

Question (45)If the matrix `[(0, a, 3), (2, b, -1), (c, 1, 0)]` is a akewsymmetric matrix, find the values of a, b and c.

Solution:

Let `A=[(0, a, 3), (2, b, -1), (c, 1, 0)]`

`because A ` is a akew- symmetric matrix.

`:, A^1 = -A`

`=>[(0, 2, c), (a, b, 1), (3, -1, 0)]`

`=-[(0, a, 3), (2, b, -1), (c, 1, 0)]`

`=>[(0, 2, c), (a, b, 1), (3, -1, 0)]`

`-[(0, -a, -3), (-2, -b, 1), (-c, -1, 0)]`

`=> `a= -2, b=-b, c= -3`

`=> a=-2, 2b= , c=-3`

`=> a=-2, b=0, c=-3`

Answer.

Question (46)If `P (x)= [(cos x, sin x), (-sin x, cos x)]`, then show that `P(x).P(y) = P(x+y) = P(y). P(x)`

Solution:

Here, `P(x) = [(cos x, sin x), (-sin x, cos x)]`

`:. P(y)= [(cos y, sin y), (-sin y, cos y)]`

Now, `P(x).P(y)=[(cos x, sin x), (-sin x, cos x)][(cos y, sin y), (-sin y, cos y)]`

`[(cos x cos y- sin x sin y, cos x sin y+sin x cos y), (-sin x cos y - cos x sin y - sin x sin y + cos x cos y)]`

`=[(cos(x+y), sin(x+y), (-sin(x+y), cos(x+y))]`

`(because cos x cos y- sin x sin y = cos (x+y), sin x cos y + cos x sin y = sin (x+y)= P(x+y)` --- (i)

Also, `P(y), P(x) = [(cosy, siny), (-siny, cos y)][(cos x, sin x), (-sinx, cos x)]`

`[(cos y cos x- sin y sin x, cos y sin x+sin y cos x), (-sin y cos x - cos y sin x - sin y sin x + cos y cos x)]`

`=[(cos (y+x), sin(y+x), (-sin(y+x), cos (y+x)]`

`=[(cos (x+y), sin(x+y), (-sin(x+y), cos (x+y)]`-----(ii)

From (i) and (ii), we get

`P(x). P(Y)= P(x+y)=P(y). P(x)`. Proved.

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