Matrices-NCERT Exemplar Problems & Solution

Short Answer Type Questions:Part-2

Question(17)Given `A= [(2, 4, 0), (3, 9, 6)]` and `B = [(1, 4), (2, 8), (1, 3)]` is `(AB) =B' A'`?

Solution:

Here

`A=[(2, 4, 0), (3, 9, 6)]`

and `B = [(1, 4), (2, 8), (1, 3)]`

`AB= [(2, 4, 0), (3, 9, 6)][(1, 4), (2, 8), (1, 3)]`

`= [(10, 40), (27, 102)]`

`(AB)' = [(10, 27), (40, 102)]`

`B' = [(1, 2, 1), (4, 8, 3)]`

`A' = [(2, 3), (4, 9), (0, 6)]`

`BA=[(1, 2, 1), (4, 8, 3)][(2, 3), (4, 9), (0, 6)]`

`=[(10, 27), (40, 102)]`----(ii)

From (i) and (ii),

`(AB)' = B'A'`

`:. yes (AB)' =B'A'`

Question(18) Solve for `x` and `y`

`x[(2), (1)]+ y [(3), (5)]+ [(-8), (-11)]=0`

Solution:

`x[(2), (1)]+ y [(3), (5)]+ [(-8), (-11)]=0`

` => [(2x, x), (3y, 5y)]+ [(-8), (-11)]= [(0), (0)]`

`=> [(2x+3y-8), (x+5y-11)]=[(0), (0)]`

`2x+3y-8=0`

`x+5y-11=0`

Solving equations(i)and (ii), we get

` x=1, y = 2`

Question(19)If `X` and `Y` are 2xx2 matrices, then solved the following matrix equation for `X` and `Y`

`2X+3Y=[(2, 3), (4, 0)]`

`3X+2Y= [(-2, 2), (1, -5)]`

Solution:

Here `2X+3Y=[(2, 3), (4, 0)]`

and `3X+2Y= [(-2, 2), (1, -5)]`

`:. 4X+6Y= 2[(2, 3), (4, 0)]`

`9X+6Y= 3[(-2, 2), (1, -5)]`
----------------------------

`-5X= [(4, 6), (8, 0)]=-[(-6, 6), (3, -15)]`

`= [(4+6, 6-6), (8-3, 0+15)]`

`=[(10, 0), (5, 15)]`

`=> X = -1/5 [(10, 0), (5, 15)]= [(-2, 0), (-1, -3)]`

Substituting the value of `X` in the first equation, we get

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

`=>[(-4, 0), (-2, -6)]+3y = [(2, 3), (4, 0)]`

`=> 3y= [(2, 3), (4, 0)]- [(-4, 0), (-2, -6)]`

`=> 3Y= [(2+4, 3-0), (4+2, 0+6)]`

`=> 3Y= [(6, 3), (6, 6)]`

`y=[(2, 1), (2, 2)]`

`:. X= [(-2, 0), (-1, -3)]`

and `y = [(2, 1), (2, 2)]`

Question(20)If `A=[(3, 5)],B=[(7, 3)]` then find a non-zero matrix `C` such that `AC =BC`.

Solution:

Since `A= [(3, 5)]` and `B= [(7, 3)]`

Hence `C` has to be a matrix of order 2xxn where n can be any natural number.

Now, Let `C= [(a), (b)]`

`:. AC=BC`

`=>[(3, 5)][(a), (b)]=[(7, 3)][(a), (b)]`

`=>[(3a+5b)]=[(7a+3b)]`

`=> 3a+5b= 7a+3b`(using equality of two matrices)

`=> 4a= 2ab`

`2a= b`

`:. a =k, b = 2

`:.` The required matrices can be `[(k), (2K)],[(k, k), (2K, 2K)]`etc.Where k is the real number.

Question(21)Given an example of matrices A, B and C such that AB = AC, Where A is nonzero matrix , but `B !=C`

Solution:

Let `A= [(1, 0), (0, 0)]`

`B= [(0, 0), (2, 3)]`

`C= [(0, 0), (1, 5)]`

Clearly `A` is a non - Zero matrix and `B!=C`

But `AB = [(1, 0), (0, 0)] [(0, 0), (2, 3)]`

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

`AC= [(1, 0), (0, 0)][(0, 0), (1, 5)]`

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

i.e.,`AB= AC`

Question(22)If `A [(1, 2), (-2, 1)] , B = [(2, 3), (3, -4)]` and `C=[(1, 0), (-1, 0)]`

Verify ;(i)`(AB)C = A(BC)`

(ii)`A(B+C)=AB+AC`

Solution:

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

`B=[(2, 3), (3, -4)]`

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

WE have to verify

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

(ii)`A(B+C)= AB+AC`

For (i) `(AB)C =A(BC)`

Take L.H.S `= (AB)C`

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

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

`= [(13, 0), (9, 0)]`

R.H.S.= `A(BC)`

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

`= [(1, 2), (-2, 1)][(-1, 0), (7, 0)]`

`=[(13, 0), (9, 0)]`

(AB)C = A(BC)verified

For (ii) `A(B+C)= AB + AC`

Take L.H.S `= A(B+C)`

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

`=[(8, -5), (-1, -10)]+[(-1, 0), (-3, 0)]`

`= [(7, -5), (-4, -10)]`

`:. A(B+C) = AB+AC`

Question(23)If `P =[(x, 0, 0), (0, y, 0), (0, 0, z)]`

and `Q= [(a, 0, 0), (0, b, 0), (0, 0, c)]`

Prove that `PQ =[(xa, 0, 0), (0, yb, 0), (0, 0, zc)]=QP`

Solution:

Here ` p=[(x, 0, 0), (0, y, 0), (0, 0, z)]`

and Q= `[(a, 0, 0), (0, b, 0), (0, 0, c)]`

`:. PQ = [(x, 0, 0), (0, y, 0), (0, 0, z)][(a, 0, 0), (0, b, 0), (0, 0, c)]`

`=[(xa, 0, 0), (0, yb, 0), (0, 0, zc)]`-----(i)

`Qp =[(a, 0, 0), (0, b, 0), (0, 0, c)][(x, 0, 0), (0, y, 0), (0, 0, z)]`

`= [(ax, 0, 0), (0, by, 0), (0, 0, cz)]`

`=[(xa, 0, 0), (0, yb, 0), (0, 0, zc) ]`-------(ii)

From(i) and (ii), we have

`PQ = [(xa, 0, 0), (0, yb, 0), (0, 0, zc )]`

`= QP . Proved .

Question(24)If `[(2, 1, 3)][(-1, 0, -1), (-1, 1, 0), (0, 1, 1)][(1), (0), (-1)]=A`find A.

Solution:

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

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

`=> [(-4)]=A`

`:.A= [-4]`Answer

Question(25)If `A = [(2, 1)], B= [(5, 3, 4), (8, 7, 6)]and C = [(-1, 2, 1), (1, 0, 2)], verity that A(B+C)= (AB+AC).`

Solution:

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

`B = [(5, 3, 4)], [(8, 7, 6)]`

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

To veryfy `A(B+C)= (AB + AC)`

Taking L.H. S. `= A(B+C) `

` [(2, 1)]{[(5, 3, 4), (8, 7, 6)]}+ [(-1, 2, 1), (1, 0, 2)]}`

`=[(2, 1)][(4, 5, 5), (9, 7, 8)]`

`=[(17, 17, 18)]`

R.H.S.= `AB+AC`

`= [(2, 1)][(5, 3, 4), (8, 7, 6)]+[(2, 1)][(-1, 2, 1), (1, 0, 2)]`

`[(18, 13, 14)]+[(-1, 4, 4)]`

[(17, 17, 18)]`

`:.` L. H. S. = R.H.S.

Question(26)If `A= [(1, 0, -1), (2, 1, 3), (0, 1, 1)]`, then verify that `A^2+A = A(A+1)`, Where 1 is `3xx3` unit matrix.

Solution:

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

To verify `A^2 + A = A(A+1)`

Taking L.H.S.= `A^2+A`

`[(2, -1, -3), (6, 5, 7), (2, 3, 5)]`

Question(27)If `A= [(0, -1, 2), (4, 3, -4)]` and `B= [(4, 6), (1, 3), (2, 6)]`

then verify that (i)`(A)'= A`

(ii)`(AB)'= BA'`

(iii) `(KA)' = (KA')`

Solution:

Here `A= [(0, -1, 2), (4, 3, -4)] `and `B= [(4, 0), (1, 3), (2, 6)]`

To verify

(i)`(A)'= A`

(ii)`(AB)'= BA'`

(iii) `(KA)' = (KA')`

for (i)`(A')'= A`

L.H.S.`= (A')'= [(0, 4), (-1, 3), (2, -4)]`

`=[(0, -1, 2), (4, 3,-4)]= A` = R.H.S.

(ii)`(AB')'= {[(0, -1, 2), (4, 3, -4)]` `[(4, 0), (1, 3), (2, 6)]}`

`[(3, 9), (11, -15)]` `=[(3, 11), (9, 15)]`

`B'=[(4, 1, 2), (0, 3, 6)]`

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

`:. B'A'=[(4, 1, 2), (0, 3, 6)][(0, 4), (-1, 3), (2, -4)]`

`= [(3, 11), (9, -15)]`

`:. (AB)'=B'A' `

(iii) `(KA) = KA`

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

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

`:.(KA)'= [(0, 4k), (-k, 3k), (2k, -4k)]`---(i)

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

`:. KA' = [(0, 4k), (-k, 3k),(2k, -4k)]`

From (i) and (ii),

`(KA)'= (KA').

Question(28)If `A= [(1, 2), (4, 1), (5, 6)]` `' B= [(1, 2), (6, 4), (7, 3)]`, then verify that:

(i)`(2A+B)'= 2A'+B'`

(ii) `(A-B)' = A' - B' `

Solution:

Here, `A = [(1, 2), (4, 1), (5, 6)]' B =[(1, 2), (6, 4), (7, 3)]`

To verify:

(i) `(2A+ By= 2A' +B)'`

(ii)`(A-B)' = A'-B'`.

For(i)`(2A+B)' = 2A' +B'`

`2A+B -2 [(1, 2), (4, 1), (5, 6)]+ [(1, 2), (6, 4), (7, 3)]`

`[(2, 4), (8, 2), (10, 12)]+ [(1, 2), (6, 4), (7, 3)]`

` =[(3, 6), (14, 6), (17, 15)]`

` (2A+B)' = [(3, 14, 17), (6, 6, 15)]`----- (i)

`2A' + B' =2[(1, 2), (4, 1), (5, 6)]+ [(1, 2), (6, 4), (7, 3)]`

`= [(2, 4), (8, 2), (10, 12)]+[(1, 2), (6, 4), (7, 3)]`

`[(2, 8, 10), (4, 2, 12)]+ [(1, 6, 7), (2, 4, 3)]`

`= [(3, 14, 17), (6, 6, 15)]`-----(ii)

From (i) and (ii),

`(2A+B)' = 2A' +B' `

For (ii)`(A-B)' = A' - B`

Verified.

Question(29) Show that `A'A` and `A'A` are both symmetric matrices for any matrix `A`.

Solution:

Here (A'A)' = A' (A')`

`(:. (AB)' = B' A' )`

`=A'A (:.(A')' =A')`

∴ A A is symmetric.

`=> A'A and A'A are both symmetric matrices for any matrix A`

Question(30) Let A and B be square matrices of order `3xx3`. Is `(Ab)^2 = A^2B^2` ? Give reasons.

Solution:

Let`A` and `B` be square matrices of order`3xx3`

Now, `(AB)^2 = (AB)(AB)`

`=BAAB (if AB=BA)`

`:. (AB)^2 = A^2B^2 if AB=BA.`

Question(31)Show that if `A` and `B` are square matrices such that `AB= BA,` then

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

Solution:

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

`=A^2 +AB +AB + B^2`

`(:. AB =BA)`

`=A^2+2AB+B^2 ` proved.

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