NYJC 2026 FM TP - Linear Algebra Set 2 (Solutions)

2026 J2 H2 FM Linear Algebra (Set 2 – Solutions 1* The matrices A and B are defined as follows. cos sin , with 0sin cos   −= A and 10 01 = − B . The transformations 1T and 2T from 2  to 2  are defined by 1T: xx yy    →       A and 2T : . xx yy    →       B (a) By considering cos sin xr yr      =       for 0r and  0, 2 , describe the geometrical transformation 1 T , explaining your answer. [2] (b) Describe 2T geometrically. [1] Let matrix M be 11 322 .11 322  −   (c) Without using a calculator, state the smallest positive integer value of k such that kM gives the identity matrix and explain your answer. [2] (d) Given that 1 =, − C M B M find the Cartesian equations of the two lines through the origin which are invariant under the transformation given by the matrix C. [3] 1) RI_DHS_HCI_TMJC/Prelim/2023/02/Q2 (a) ( ) ( ) ( ) ( ) cos cos sin sin coscos cos sin cos sin sinsinA rrr rrr             −+    ==    ++     Hence 1T gives an anticlockwise rotation of angle  about the origin. M1 A1 (b) 10 01 xx yy      =    −−     2T is a reflection in the x-axis. B1 (c) M is obtained when we let 6  = in matrix A. Rotating a point about O for 2 radians brings it back to the original position. Since 2 12 , 6  = the smallest positive integer 12.k = M1 A1 (d) 1 1 1 1 1 133 102 2 2 2 ,1 1 0 1 1 1332 2 2 2 − −     −−    ==      −          C MBM hence C has eigenvalues 1 and 1.− Hence we have 11 1=Cx x and 22 1=−Cx x for eigenvectors 1x and 2,x M1

1 1 32where 1 2   =  x and 2 1 2 1 32 − =  x . So, the two invariant lines are 1 3 yx= and 3yx=− , which pass through the origin. A1 A1 2* (a) The matrix A has eigenvectors 1 2 1  − , 2 0 1    , and 0 1 0    with corresponding eigenvalues 1− , 3 and 1 respectively. Find A. [2] (b) The vectors 1 2 3,,y y y are defined as: 1 7 4 5  =− y , 2 1 8 13 − = y , 3 66 60 6  =− y . Given that Q is a 33 matrix, determine whether or not the vectors 1 2 3,,Qy Qy Qy are linearly independent, justifying your conclusion. [2] 2 ACJC\2017\Prelim\P1\Q3a & JPJC Prelim 9649/2022/01/Q10aii) and b) (a) 1 1 2 0 1 0 0 1 2 0 7 0 8 2 0 1 0 3 0 2 0 1 4 1 8 1 1 0 0 0 1 1 1 0 4 0 5 − −−           = − − = −           −      A M1 A1 (b)   ( ) 1 2 3 1 2 3 12 3 3 12 12 1 2 3 1 2 3 12 33 1 2 3 7 1 66 Since 4 8 60 0 , , is linearly dependent 5 13 6 The , and and not all zero such that + + + , and and not all zero n, Since ,                − − − = += + = =   =  + y y y Q y y y 0 Q y y y Q0 0 yyQ Q Q y0    1 2 3 , are linearly dependent. , yQyQy M1 A1 (c) (i) Show that the eigenvalues of the matrix 1 2 43 21 65 a a aa − =−− A are independent of a, where a is a non-zero constant. [3] (ii) Obtain, in terms of a, an eigenvector corresponding to each eigenvalue. [3] (iii) Hence find a matrix P and a diagonal matrix D such that 1 3 1( 2 )−−+=A I PDP . [2]

(c) (i) Method 1: Using row operations Recall effect of row operations on determinant (LA1 Lecture Notes last page) 1 3 3 11 2 2 0 4 3 4 3 2 1 0 2 1 0 6 5 2 2 0 1 aa R R R a a a a a       − + → −= − − − − − − = ⎯⎯⎯⎯⎯ → − − = − − + − − AI 3 1 1 1 2 23 0 1 0 0 0 1 a C C C a    +→ − ⎯⎯⎯⎯→ − − = −− ( )( )( )2 1 1 0 1,1, 2      − − − − =  =− The eigenvalues of A are −1, 1, 2 which are independent of a. M1 M1 A1 Method 2: Direct Computation 1 2 43 0 2 1 0 65 a a aa    −− − =  − − = −− AI ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( )( ) 2 32 32 164 1 5 2 2 5 1 6 1 3 1 0 1147 2 2 0 2 2 0 1 2 0 The eigenvalues of are 1,1 and 2 which are independent of . 6 2 6 4 3 0 a aa a a a a                    + − − − − + − − − − − − − = −          −−  − − − − − − + = − + + − = − − + = + − = −  A M1 M1 A1 (ii) For  = −1, an eigenvector is 1 3 2 7 7 2 3 3 3 8 8 a a aa a −−               −  =− = −              −−        . For  = 1, an eigenvector is 2 12 4 5 6 1 222 3 4 2 a aa a aa a − − −            −  =− = −            −       . B1 B1 B1

For  = 2, an eigenvector is 1 6 2 2 1 11 1 0 022 3 4 2 a a a −                − = =              −        . (iii) Recall that if has eigenvalue and eigenvector Ax 1 1 has eigenvalue and eigenvector  −Ax has eigenvalue 1 and eigenvector (any non-zero vector is an eigenvector)Ix Let 7 1 1 30 8 2 2 aa  = − − P and ( ) ( ) ( ) 31 1 31 1 12531 82 2 0 0 1 0 0 0 2 0 0 27 0 000 0 2 −  +  = + =   + D Then ( ) 311 2−−+=A I PDP B1 for P B1 for D. 3 Let 32:T →  be a linear transformation defined by 2x xyTy x y zz  + = −+ for all 3 x y z     . (a) Find a basis for the range of T and explain why the dimension of the kernel of T is 1. [3] (b) Find a basis 1v for the kernel of T , where 1v is in the form 1 1 6 x y    and 11, xy   are constants to be determined. Hence, find a general solution 2 9 x Ty z   =  . [3] (c) Let 2 2 3 32 a ab a  = − + − v and 2 2 3 2 9 29 a a a  = − − −− v . Determine the conditions , ab   such that   3 1 2 3span , , =v v v  , explaining your answers clearly. [4]

4 TJC JC2 Prelim 9649/2019/02/Q2 (a) Observe that for 2x xyy x y zz  + = −+ A 1102 1 0 3 1 1 1 2 01 3  =→  −  − A Hence, 21,11        −    is a basis for the range of T . By the Rank and Nullity theorem, since ( ) 3dim 3 =  and ( )rank 2A = , ( )nullity 3 2 1A = − = . M1 A1 B1 (b) Solving 2 1 0 0 1 1 1 0 x y z      =    −     , 110 03 2001 3 x y z    =  −  3xz−= , 3 2 yz= Hence, 2 4 6  −   is a basis for ( )ker T . Observe that a particular solution is 1 0 8    . Hence, the general solution is 12 0 4 , 86 x y z  −           = +                  . M1 A1 B1 (c) For   3 1 2 3span , , =v v v  , the set  1 2 3, , v v v must be linearly independent since 3 linearly independent vectors will span 3  . Hence, the determinant of the matrix  1 2 3 v v v cannot be 0. Let 22 2 2 2 2 22 22 4 3 9 0 3 9 6 3 2 2 9 0 3 9 a a a a a b a a b a a a a a    −−    = − + − − → + −        − − − + −    X M1 M1

( ) ( )( ) ( )( ) ( ) ( )( ) 2 2 2 22 22 det 2 3 9 3 9 2 9 3 3 6 9 1 b a a a a a b a a ab =− + − − + − =− − + − − =− − − X Since ( )det 0 X , 3a , 1b . A1, A1