Date | May Example question | Marks available | 4 | Reference code | EXM.2.AHL.TZ0.23 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Show that and Hence | Question number | 23 | Adapted from | N/A |
Question
Let A =(3143).
Let A2 + mA + nI = O where m, n∈Z and O = (0000).
Find the values of λ for which the matrix (A − λI) is singular.
Find the value of m and of n.
Hence show that I = 15A (6I – A).
Use the result from part (b) (ii) to explain why A is non-singular.
Use the values from part (b) (i) to express A4 in the form pA+ qI where p, q∈Z.
Markscheme
A − λI = (3−λ143−λ) A1
If A − λI is singular then det (A − λI) = 0 (R1)
det (A − λI) =(3−λ)2−4(=λ2−6λ+5) (A1)
Attempting to solve (3−λ)2−4=0 or equivalent for λ M1
λ = 1, 5 A1 N2
Note: Candidates need both values of λ for the final A1.
[5 marks]
(3143)2+m(3143)+n(1001)=(0000) A1
(3143)2=(1362413) (A1)
Forming any two independent equations M1
(eg 6+m=0, 13+3m+n=0 or equivalent)
Note: Accept equations in matrix form.
Solving these two equations (M1)
m=−6 and n=5 A1 N2
[5 marks]
A2 − 6A + 5I = O (M1)
5I = 6A − A2 A1
= A(6I − A) A1A1
Note: Award A1 for A and A1 for (6I − A).
I = 15A(6I − A) AG N0
Special Case: Award M1A0A0A0 only for candidates following alternative methods.
[5 marks]
METHOD 1
I = 15A(6I − A) = A × 15(6I − A) M1
Hence by definition 15(6I − A) is the inverse of A. R1
Hence A−1 exists and so A is non-singular R1 N0
METHOD 2
As det I = 1 (≠ 0), then R1
det 15A(6I − A) = 15 det A × det (6I − A) (≠ 0) M1
⇒ det A ≠ 0 and so A is non-singular. R1 N0
[3 marks]
METHOD 1
A2 = 6A − 5I (A1)
A4 = (6A − 5I)2 M1
= 36A2 − 60AI + 25I2 A1
= 36(6A − 5I) − 60A + 25I M1
= 156A − 155I (p = 156, q = −155) A1 N0
METHOD 2
A2 = 6A − 5I (A1)
A3 = 6A2 − 5A where A2 = 6A − 5I M1
= 31A − 30I A1
A4 = 31A2 − 30A where A2 = 6A − 5I M1
= 156A − 155I (p = 156, q = −155) A1 N0
Note: Do not accept methods that evaluate A4 directly from A.
[5 marks]