Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is

Monjur Alahi

Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is
A. `1`
B. `3`
C. `0`
D. `6`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - B
`(b)` Let `A=[{:(a,b,c),(p,q,r),(x,y,z):}]`
Given `{:(a+b+c=1),(p+q+r=1),(x+y+z=1):}`
`impliesA^(2)=[{:(a,b,c),(p,q,r),(x,y,z):}][{:(a,b,c),(p,q,r),(x,y,z):}]`
`=[{:(a^(2)+bp+cx,,ab+bq+cy,,ac+br+cz),(pa+qp+rx,,qb+q^(2)+ry,,pc+qr+rz),(xa+yp+zx,,xb+yq+zy,,xc+yr+z^(2)):}]`
Sum of elements of
`R_(1)=a^(2)+bp+cx+ab+bq+cy+ac+br+cz`
`=a(a+b+c)+b(p+q+r)+c(x+y+z)`
`=a+b+c=0`
Similarly sum of elements of
`R_(2)=p(a+b+c)+q(p+q+r)+r(x+y+z)`
`=p+q+r=1`
`R_(3)=x(a+b+c)+y(p+q+r)+z(x+y+z)`
`=x+y+z=1`
`:.` sum of element of `A^(2)` is `3`

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Matrix A ha s m rows and n+ 5 columns; matrix B has m rows and `11 - n` columns. If both AB and BA exist, then (A) AB and BA are square matrix (B) AB

If both AB and BA exist, then the number of columns of A is equal to the number of rows of B. Therefore, `n+5=m` (1) And the number of columns...

2 Answers 1 views
The matrix `A=[-5-8 0 3 5 0 1 2-]` is a. idempotent matrix b. involutory matrix c. nilpotent matrix d. none of these

`A^(2)=AxxA[(-5,-8,0),(3,5,0),(1,2,-1)]xx[(-5,-8,0),(3,5,0),(1,2,-1)]` `=[(25-24+0,40-40+0,0+0+0),(-15+15+0,-24+25+0,0+0+0),(-5+6-1,-8+10-2,0+0+1)]` `=[(1,0,0),(0,1,0),(0,0,1)]` Hence, the given matrix a is involutory.

2 Answers 1 views
Let `A` be a square matrix of order 3 such that adj. (adj. (adj. A)) `=[(16,0,-24),(0,4,0),(0,12,4)]`. Then find (i) `|A|` (ii) adj. A

We know that adj. (adj. A) `=|A|^(n-2)A`, where n is order of matrix. `:.` adj. (adj. (adj. A))`=|"adj. A"|^(n-2)` adj. A `=(|A|^(n-1))^((n-2))` adj. A For `n=3`, adj. (adj. (adj. A))`=|A|^(2)`...

2 Answers 1 views
If `A` and `B` are matrices of the same order, then `A B^T-B^T A` is a (a) skew-symmetric matrix (b) null matrix (c) unit matrix (d) symmetric matrix

Let `P=(AB^(T)-BA^(T))` `:. P^(T)=(AB^(T)-BA^(T))^(T)=(AB^(T))^(T)-(BA^(T))^(T)` `=(B^(T))^(T) (A)^(T)-(A^(T))^(T)B^(T)=BA^(T)-AB^(T)` `=-(AB^(T)-BA^(T))=-P` Hence, `(AB^(T)-BA^(T))` is a skew-symmetric matric.

2 Answers 1 views
Let B is an invertible square matrix and B is the adjoint of matrix A such that `AB=B^(T)`. Then

Given that `AB=B^(T)` and `B=` adj. A `implies A=B^(T)B^(-1)` `implies |A|=|B^(T)|xx|B^(-1)|=1` ...(i) Now, adj. `A=` adj `(B^(T) B^(-1))` `="adj"(B^(-1))xx"adj"(B^(T))` `=("adj B")^(-1)xx"adj"(B^(T))` `implies ("adj. B")^(T)=("adj. B")xx("adj. A")=("adj. B")B=|B|I` `implies ("adj. B")^(T)=|B|....

2 Answers 2 views
Let `A=[a_("ij")]` be `3xx3` matrix and `B=[b_("ij")]` be `3xx3` matrix such that `b_("ij")` is the sum of the elements of `i^(th)` row of A except `a

Correct Answer - C `A=[(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))]` `implies B=[(a_(12)+a_(13),a(11)+a_(13),a_(11)+a_(12)),(a_(22)+a_(23),a_(21)+a_(23),a_(21)+a_(22)),(a_(32)+a_(33),a_(31)+a_(33),a_(31)+a_(32))]` `implies X=A^(-1) B` `=1/(|A|)[(C_(11),C_(21),C_(31)),(C_(12),C_(22),C_(32)),(C_(13),C_(23),C_(33))]` `[(a_(12)+a_(13),a_(11)+a_(13),a_(11)+a_(12)),(a_(22)+a_(23),a_(21)+a_(23),a_(21)+a_(22)),(a_(32)+a_(33),a_(31)+a_(33),a_(31)+a_(32))]` `=1/(|A|) [(0,|A|,|A|),(|A|,0,|A|),(|A|,|A|,0)]=[(0,1,1),(1,0,1),(1,1,0)]` `implies |A^(-1)B|=2` or `|A^(-1)||B|=2` or `|B|=2|A|`

2 Answers 2 views
Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-1)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in it

Correct Answer - B `A^(n)-A^(n-2)=A^(2)-I implies A^(50)=A^(48)+A^(2)-I` Further, `A^(48)=A^(46)+A^(2)-I` `A^(46)=A^(44)+A^(2)-I` `{:(vdots,vdots,vdots,vdots):}` `(A^(4)=A^(2)+A^(2)-I)/(A^(50)=25 A^(2)-24I)` Here, `A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]` `implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]` `=[(1,0,0),(25,1,0),(25,0,1)]` `:. |A^(50)|=1` Also, `tr(A^(50))=1+1+1=3`. Further, `[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]` Similarly, `uu_(2)=[(0),(1),(0)]` and...

2 Answers 1 views
Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-2)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in it

Correct Answer - B `A^(n)-A^(n-2)=A^(2)-I implies A^(50)=A^(48)+A^(2)-I` Further, `A^(48)=A^(46)+A^(2)-I` `A^(46)=A^(44)+A^(2)-I` `{:(vdots,vdots,vdots,vdots):}` `(A^(4)=A^(2)+A^(2)-I)/(A^(50)=25 A^(2)-24I)` Here, `A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]` `implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]` `=[(1,0,0),(25,1,0),(25,0,1)]` `:. |A^(50)|=1` Also, `tr(A^(50))=1+1+1=3`. Further, `[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]` Similarly, `uu_(2)=[(0),(1),(0)]` and `uu_(3)=[(0),(0),(1)]implies...

2 Answers 1 views
Let `M` be a `2xx2` symmetric matrix with integer entries. Then `M` is invertible if The first column of `M` is the transpose of the second row of `M`

Correct Answer - C::D Let `M=[(a,c),(c,b)]" "("where "a, b, c, in I)` (1) If the first column of M is the transpose of the second row of M, then `[(a,c)]=[(c,b)]` `:....

2 Answers 1 views
A be a square matrix of order `2` with `|A| ne 0` such that `|A+|A|adj(A)|=0`, where `adj(A)` is a adjoint of matrix `A`, then the value of `|A-|A|adj

Correct Answer - D `(d)` Let `A=[{:(m,n),(p,q):}]`,`adj(a)=[{:(q,-n),(-p,m):}]` Let `|A|=d=mq-np` `|A+dadjA|=|{:(m+qd,n(1-d)),(p(1-d),q+md):}|=0` `impliesmq+m^(2)d+q^(2)d+mqd^(2)-np+2npd-npd^(2)=0` `implies(mq-np)+(mq0np)d^(2)+m^(2)d+q^(2)d+2mqd-2d^(2)=0` `implies(d+d^(3)-2d^(2))+d(m^(2)+q^(2)+2mq)=0` Now, `|A- d adj|A|=-(m+q)^(2)+4(mq-np)=4d=4`

2 Answers 1 views
X