`(A) rarr (q,s), (B) rarr (p,t), (C) rarr (p,q,r,s), (D) rarr (q,s)`
(A) A diagonal matrix is commutative with every square
matrix, if it is scalar matrix, so every diagonal element is 4.
Therefore, `abs(A) = abs((4,0,0),(0,4,0),(0,0,4))= 64`
(B) ` abs((1-a, 1, 1),(1, 1-b, 1),(1,1,1-c))= 0`
Applying `R_(1) rarr R_(1) - R_(3) and R_(2) rarr R_(2) - R_(3)`, then
` abs((-a, 0, c),(0, -b, c),(1,1,1-c))= 0`
`rArr -a(-b+bc-c) - 0 + c(b) = 0`
`ab+ bc + ca = abc" " ...(i)`
Now, `AMge GM`
` rArr (ab+ bc+ ca)/3 ge (ab cdot bc cdot ca)^(1/3)` ltbrlt `rArr abc/3 ge (abc) ^(2/3)` [from eq. (i)]
`rArr (abc)^(1/3)ge 3 `
`therefore abcge 27 `
Hence, `lambda =27`
(C) `because A = [[a_(11), a_(12), a_(13) ],[a_(21), a_(22),a_(23)],[a_(31), a_(32),a_(33)]]`
Given, `sum_(k=1)^(3) a_(ik) = 9 lambda _(i), AAjin {1,2,3},`
`sum_(k=1)^(3) a_(kj) = 9 mu _(j), AAjin {1,2,3}` and
`a_(11) + a_(22) + a_(33) = 9 upsilon, ` where ` lambda _(i), mu _(j) , v in {1,2}`
Following types of matrices are possible:
`A = [[1,,],[,3,],[,,5]], B= [[2,,],[,3,],[,,4]], C = [[7,,],[,3,],[,,8]],`
`D = [[6,,],[,3,],[,,9]], E= [[1,,],[,6,],[,,2]], F = [[3,,],[,6,],[,,9]],`
`G= [[4,,],[,6,],[,,8]], H= [[5,,],[,6,],[,,7]], I = [[1,,],[,9,],[,,8]],`
`J= [[2,,],[,9,],[,,7]], K= [[3,,],[,9,],[,,6]], L = [[4,,],[,9,],[,,5]],`
Now, if we interchange 1 and 5 to obtain
`A_(1) = [[5,4,9],[7,3,8],[6,2,1]]`
Also, `A^(T) = [[1,8,9],[2,3,4],[6,7,5]]`
and `A_(1) ^(T) = [[5,7,6],[4,3,2],[9,8,1]]`
Then, from A we get four matrices `A, A_(1) ,A^(T),A_(1)^(T).`
Similarly, from `B, C, D, ..., K, L` we get 4 matrices.
Thus, total `12xx4 = 48` matrices. Hence, `lambda = 48`.
(D) For consistent, `abs((1,1,1),(c+2,c+4,6),((c+2)^(2),(c+4)^(2) , 36))=0`
Applying `C_(2) rarr C_(2) - C_(1)`, we get
`abs((1,0,1),(c+2,2,6),((c+2)^(2),4c+12 , 36))=0`
`rArr 2abs((1,0,1),(c+2,1,6),((c+2)^(2),2c+6 , 36))=0`
`rArr -122 c-0 + 1 [ (c+2 ) (2c+6)- (c+2)2^(2) ] =0`
`rArr c^(2) - 6c + 8 =0`
`rArr c= 2, 4`
` therefore c_(1) = 4, c_(2) = 2 `
` rArr c_(1)^(c_(2))= 4^(2) = 16`
