NYJC 2026 FM TP - Linear Algebra Set 1 (Solutions)

2026 J2 H2 FM Linear Algebra (Set 1) – Solutions 1* Do not use a calculator in answering this question. The matrix P is given by 5 3 1 9 3 2 1 5 1 1 1 1 − =− − P . (a) Find a matrix B such that 1 1 1 1 0 1 , 0 0 1 ab c − = BP where a, b and c are some constants to be determined. [3] (b) Hence solve the equation 4 1. 2 − = Px [3] (c) The equations below represent three planes 53 32 x y z d x y z e x y z f + − = + − = + + = where d, e and f are constants. What possible geometrical relationship(s) can you say about the three planes? Justify your answer. [2] 1) NJC/Prelim/2020/02/Q1 Recall “Elementary Matrices” We can see B as a matrix representing a sequence of elementary row operations that is performed on P, to get 1 1 1 1 01 0 0 1 ab c −   . We can find B by performing the same sequence of elementary row operations on the identity matrix.

( ) 13 3 3 1 2 3 1 3 5 1 5 3 1 9 1 0 0 3 2 1 5 0 1 0 1 1 1 1 0 0 1 1 1 1 1 0 0 1 3 2 1 5 0 1 0 5 3 1 9 1 0 0 1 1 1 1 0 0 1 0 1 4 8 0 1 3 0 2 6 14 1 0 5 RR RR RR RR  − − − −  − − − ⎯⎯⎯ → − − − ⎯⎯⎯ → − − − − − − 2 1 1 1 1 0 0 1 0 1 4 8 0 1 3 0 2 6 14 1 0 5 R− − ⎯⎯ → − − − − − 3 2 1 0.5 3 1 1 1 1 0 0 1 0 1 4 8 0 1 3 0 0 2 2 1 2 1 1 1 1 1 0 0 1 0 1 4 8 0 1 3 0 0 1 1 0.5 1 0.5 0 0 1 0 1 3 0.5 1 0.5 4, 8, 1 RR R a b c + − ⎯⎯⎯ → − − −− − ⎯⎯ → − − −−   = − − = =− =− B M1: R ow operations to get first two rows correct. A1: Matrix B A1: a, b, c values (b) 44 1 1 22 1 1 1 1 0 0 1 4 2 0 1 4 8 0 1 3 1 5 0 0 1 1 0.5 1 0.5 2 2 −−       =  =          −−              − = − =            − − −       Px BPx B x 2 4 3 13 12 1 1 1 1 2 1 1 0 0 4 0 1 4 8 5 0 1 0 4 13 0 0 1 1 2 0 0 1 1 2 1 0 0 4 9 0 1 0 4 13 0 0 1 1 2 RR RR RR − − − −        − ⎯⎯⎯ → −       − − − −    − ⎯⎯→ − −− M1 for multiplying by B and getting RHS. M1 writing augmented matrix.

14 24 34 4 9, 4 13, 2 xx xx xx + =− −= − =− 4 94 13 4Let , . Thus , . 2 t tx t t t t t −−  +=  =  −+   x  A1 for correct answer. (c) Method 1: 53 32 x y z d x y z e x y z f + − = + − = + + = 5 3 1 3 2 1 1 1 1 From (b), 5 3 1 1 0 0 3 2 1 0 1 0 1 1 1 0 0 1 RREF xd ye zf −         −=             −       − ⎯⎯⎯ →           Thus 5 3 1 3 2 1 1 1 1 −  − is invertible and hence the system of linear equations has a unique solution for any constants d, e and f, implying that the three planes have exactly one common point of intersection. Method 2 ( ) ( ) 3 Line of intersection for the first two planes is parallel to Line of intersection between 1st and 2nd plane not parallel to 3rd plane. S 5 3 3 2 2 11 2 1 2 0 11 = + −  + − =− + + −       ==           d i j k i j k i j k dn  o, three planes will intersect at a point. B1 for stating invertible B1 for one common point of intersection.

2* The sets A and B are defined as follows: 3 : 0 , x A y x y z z  =  + − =  3 : 2 3 5 1 . x B y x y z z  =  + − =  (a) Show that A is a subspace of 3  but B is not a subspace of 3.  [3] (b) Determine if AB  is a subspace of 3.  [2] 2 DHS/Prelim/2017/01/Q1a) and b)i) modified a) Method 1: Direct Proof The zero vector is in A. Let , . Auv ( ) 1 1 1 1 1 1 1 1 1 0 0 0            + = +          − − −      = + = u v u v  Therefore A+uv and A is closed under vector addition. ( ) 11 1 1 0 11 aa        ==        −−     uu  where a is a scalar. Therefore aAu and A is closed under scalar multiplication. Therefore, A is a subspace of 3.  Method 2: Using Corollary 12 12 12 Let , , then 0 for 1, 2 i i i xx y y A x y z i zz        = = + − = =           uv ( ) ( )1 1 1 2 2 2 3 0 So, is a subspace of . + x y z x y z A A     + − + + − =  uv  The zero vector is not in B since ( ) ( ) ( )2 0 3 0 5 0 0 1+ − =  Therefore, B is a not a subspace of 3.  B1 B1 B1

3* (a) Given that A is an invertible (non-singular) 3 x 3 matrix, show that the first column of 1−A is the solution of the equation 1 0 0  = Ax . [1] (b) Give the equations whose solutions are the second and third columns of 1−A respectively. [1] (c) Using parts (a) and (b), find the inverse of P, where 1 01 0 0 1 ab c  = P for unknown constants a, b and c. [3] (d) Using the result in part (c), find the inverse of the matrix Q where 1 0 0 1 0 . 1 a bc  = Q [1] (b) 00 Both 2 , 0 01 0 0 0 However, 2 0 2 0 1 1 0 0 0 and 2 , 2 (i.e. ). Therefore 2 . 1 1 1 Since is not closed under ve B A B A B B A B AB                                +=                                                 3ctor addition, it is not a subspace of .  B1 B1 3 ASRJC Promo 9649/2020/Q2 (a) Let 11 12 13 1 21 22 23 31 32 33 a a a a a a a a a −  = A 11 12 13 11 21 22 23 21 31 32 33 31 11 00 00 a a a a a a a a a a a a              =  = =                   Ax x ie the solution is the first column of 1A− B1 (b) The equations whose solutions are the second and third columns of 1A− are 0 1 0  = Ax and 0 0 1  = Ax respectively. B1

(c) By (a) and (b), solutions to 1 0 0 0 , 1 , 0 0 0 1            ===                Px Px Px gives the 1st, 2nd and 3rd columns of 1−P respectively. 1 2 3 1 2 3 2 3 3 2 1 3 11 0 1 0 0 0 1 0 1 0 0, 0, 1 0 a b x cx x x ax bx x cx x x x x           =             ++        + =  = = =          1 1 first column of 0 0 −  = P 1 2 3 2 3 3 2 1 3 1 0 1 0, 1, 0 second column of 1 0 x ax bx x cx x x x a x a − ++        + =  = = =−          − = P 1 2 3 2 3 3 2 1 3 1 0 0 1, , 1 third column of 1 x ax bx x cx x x c x ac b x ac b c− ++        + =  = =− = −          − =− P Therefore 1 1 01 0 0 1 a ac b c− −− =− P B1 B1 B1 (d) 1 0 0 10 1 Ta bc  == QP ( ) ( ) 111 1 0 0 10 1 TT a ac b c −−−  = = = −  −− Q P P B1

4 Unsupported answers from a graphing calculator are not allowed in this question. In response to the government’s effort to promote food source diversification and local production, a local entrepreneur set up Apollo Poultry Farm. The farm has a total poultry rearing area of 200 m2. The entrepreneur plans to rear c chickens, d ducks, g geese and q quails to be sold every three months. The rearing space needed for each chicken, duck, goose and quail is 0.5 m 2, 1.5 m 2, 2 m 2 and 0.5 m 2 respectively. Each chicken, duck, goose and quail consumes 10 kg, 15 kg, 20 kg and 10 kg of poultry feed in three months. The labour cost of caring for each chicken, duck, goose and quail for three months is $5, $10, $15 and $10 respectively, which will be paid to a farm hand as salary. Every month, the entrepreneur has available 1000 kg of poultry feed and a budget of $800 to pay the farm hand. (a) Express the constraints in rearing space, feed and labour cost as inequalities in terms of c, d, g and q. [3] For the following parts, assume that the resources in rearing space, feed and labour