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

The elements of A are all multiples of 5. Sum of every pair of elements of A is divisible by 5. Therefore, we have to find the probability that B...

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.

Correct Answer - D `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|`

Correct Answer - 0.4 Given that `A A^(T)=4I` `implies |A|^(2)=4` or `|A|= pm 2` So `A^(T)=4A^(-1)=4 ("adj A")/(|A|)` `implies [(a_(11),a_(21),a_(31)),(a_(12),a_(22),a_(32)),(a_(13),a_(23),a_(33))]=4/(|A|)[(c_(11),c_(21),c_(31)),(c_(12),c_(22),c_(32)),(c_(13),c_(23),c_(33))]` Now `a_("ij")=4/(|A|) c_("ij")` `implies -2c_("ij")=4/(|A|) c_("ij")" "("as "a_("ij")+2c_("ij")=0)` `implies |A|=-2` Now...

Correct Answer - 4 det. `B=|(a_(11),a_(12)/3,a_(13)/3^(2)),(3a_(21),a_(22),a_(23)/3),(9a_(32),3a_(32),a_(33))|=3xx9|(a_(11),a_(12)/3,a_(13)/3^(2)),(a_(21),a_(22)/3,a_(23)/9),(a_(32),a_(32)/3,a_(33)/9)|` (Dividing `C_(2)` by 3 and `C_(3)` by 9) `=|(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(32),a_(32),a_(33))|` `="det. "A=2` Similarly, det, `C=` det. `A=2` `:.` det. `(BC)=` (det. B) (det. C) `=2xx2=4`

Correct Answer - D `|Q|=|(2^(2)a_(11),2^(3)a_(12),2^(2)a_(13)),(2^(3)a_(21),2^(4) a_(22),2^(5) a_(23)),(2^(4) a_(31),2^(5) a_(32),2^(6) a_(33))|` `=2^(2). 2^(3). 2^(4)|(a_(11),a_(12),a_(13)),(2a_(21),2a_(22),2a_(23)),(2^(2)a_(31),2^(2)a_(32),2^(2) a_(33))|` `=2^(9). 2. 2^(2) |(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))|` `=2^(12) |P|` `=2^(13)`

Correct Answer - C `(c )` `|A|=|{:(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33)):}|` `|B|=|{:(a_(11),k^(-1)a_(12),k^(-2)a_(13)),(ka_(21),a_(22),k^(-1)a_(23)),(k^(2)a_(31),ka_(32),a_(33)):}|` `=(1)/(k^(3))|{:(k^(2)a_(11),ka_(12),a_(13)),(k^(2)a_(21),ka_(22),a_(23)),(k^(2)a_(31),ka_(32),a_(33)):}|` `=|A|` `k_(1)|A|+k_(2)|B|=0` `:.k_(1)+k_(2)=0`

Correct Answer - A `(a)` Here `a_(ij)+a_(ji)=0impliesA^(T)=-A` and `b_(ij)-b_(ji)=0impliesB^(T)=B` and `A,B` are `3xx3` matrices, Hence `|A|=0`, `implies|A^(4)B^(3)|=0` gtbrgt `impliesA^(4)B^(3)` is singular matrix.

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...

Correct Answer - D `(d)` For symmetric matrix each place in upper triangle and leading diagonal can be filled in `7` ways. Then number of symmetric matrices are `7^(6)`. For skew...