Q.3
T007, sio 07:50 107
Let P, = 1 = 0 1 0 , P,= 0 0 1, P, = 1 0 0
Lo 0 : 0 1 0 0 0 1
To 107.50 0 11 To o 17
P=10 0 1,Ps=100, P=10 10
100 1010000
12137
and X = PI 02 P!
k=1 13 2 1
where P denotes the transpose of the matrix P - Then which of the following options is/are correct?
(1) X is a symmetric matrix
(2) IX 1 = a 1. then a = 30
0 0
(3) The sum of diagonal entries of X is 18
(4) X - 301 is an invertible matrix
Answers
Answer:
Heya.......
Here's ur answer :
[1,2,3]
[1,2,3] Sol. P1 = P1
[1,2,3] Sol. P1 = P1 T = P1
[1,2,3] Sol. P1 = P1 T = P1 –1
[1,2,3] Sol. P1 = P1 T = P1 –1
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A =
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A = 3 2 1
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A = 3 2 1 1 0 2
[1,2,3] Sol. P1 = P1 T = P1 –1 & P2 = P2 T = P2 –1 P6 = P6 T = P6 –1 also AT = A where A = 3 2 1 1 0 2 2 1 3X =
6
k 1
T
k k
(P A P )
X = (P1A P1
T + .... + P6AP6
T)
Now XT = P1ATP1
T + .... + P6ATP6
T = X X is symmetric
Let B =
1
1
1
XB = P1AP1
TB + P2AP2
TB + .... + P6AP6
TB P1
TB = B
= P1AB + .... + P6AB
= (P1 + P2 + ... + P6)AB
= 30B
trace of X = tr(X) = tr(P1AP1
T
) + tr(P2AP2
T
) + ....
X
1
1
1
= 30
1
1
1
(X – 30)B = 0
|(X – 30)B| = 0
|X – 30I| |B| = 0
|X – 30I| = 0, |B| 0
Hope it helps!!!..... ✌
....
....This is Twilight Astro
Area of quadrilateral= side¹×side²×Sin theta
A=36×48×sin45°
A=1728×1÷√2
A=1221.88