NYJC 2026 FM TP - Linear Algebra Set 4 (Solutions)

1. Let 2 2 4 3 3 2 05 a a b a a b ab −−  − − be the augmented matrix of a linear system. Using elementary row operations, simplify the augmented matrix and hence (a) find the values of a and b for the system to have a unique solution, [2] (b) give a geometrical interpretation for the system of equations when 0 and 5.ab =− [2] 2. In a four-pole electrical network, the output quantities 11 and EI are given in terms of the input quantities 22 and EI by 12 12 , where , , and are non-zero.EE ab a b c dII cd    =       Denoting by , ab cd   A show that 11 22 1 1 EI a EI c d  −   =     −     A . [2] Similarly, show that 11 22 1 1 EI IE d      =         A , where and  are constants to be determined in terms of a, b, c and d. [3] Given A is orthogonal, that is T =AA I , and 1,=A b and d are non-zero, find 12 and EE in terms of 12, , and .b d I I [5]

Qn Solutions 1 21 12 32 2 2 2 4 2 2 4 3 3 2 0 2 0 5 0 5 0 0 0 02 0 0 5 2 RR RR RR a a b a a b a a b a b a b a b a ab bb − + + − − −        − ⎯⎯⎯ → − −      −−     ⎯⎯⎯⎯ → − +− (a) 5Unique solution if there are 3 pivots. i.e. 0 and a b − (b) If 0 and 5, we have System is inconsistent. There is no solution. The system of equations consists of 3 planes with no common points. Triangular prism. 0 0 0 0 5 2 . 0 0 0 7 a a ab   −− −  = − Qn Solutions 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12 12 1 2 21 2 2 2 1 21 2 2 1 1 2 2 1 2 11 22 1 2 Sub 2 into 1 : 3 2 and 3 : 1 And si EE ab II cd cE acE bcIE aE bI cE I dII cE dI cE a I dI bcI aI ad bc I EI a ad bcc EI d     =       = + −−−−+   =    = − −−−−+    = − + = − − −−−−−  − −    =    −        ( ) 11 22 nce 1 shown 1 ad bc EI a EI c d =−  −    =     −       A A ( ) ( ) 12 12 1 2 21 2 2 2 1 21 2 2 4 5 EE ab II cd dE adE bdIE aE bI dI I cEI cE dI    =         = + −−−−+   =    = − −−−−+     ( ) ( ) ( ) ( ) ( ) 1 2 1 2 12 Sub 5 into 4 : 6 dE adE b I cE bI ad bc E = + − = + − −−−−−

( ) ( ) ( )11 22 11 22 6 and 5 : 1 a 1 1 nd E cb Ib ad bcd IE c EI b IE d c   −    =    −       =     −  =     =− A Method 1: 1 11 22 1 2 12 12 If is orthogonal, so and 1 Thus , 1Since 1 11 1 1 1 T T a c d b b d c a a d c b EI a EI c d Id Idb d IIbb dIIbb − == −   =    −    = =−  −    =     −     − =−  −  −+   =   = −+ A AA I A A AA Method 2: ( ) ( ) ( ) 22 22 22 22 10 01 11 1 2 03 T a b a c a b ac bd c d b d ac bd c d ab cd ac bd  ++    = = =      ++      + = −−−− +=  −−−− + = −−−−    AA ( )41 ad bc −= − = −−−A From (3), bdc a=− . Subst. into (4), 1bdad b a − − = 22 1 1 1 by (1) abd a d a ad  + =  = = Thus, from 3, cb=− 12 11 22 12 1 11 =11 d IIEI d bb EI ddb IIbb −+−    =−      −     −+