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Date May Example question Marks available 2 Reference code EXM.1.AHL.TZ0.43
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Solve Question number 43 Adapted from N/A

Question

Find the values of a and b  given that the matrix  A = ( a 4 6 8 5 7 5 3 4 )  is the inverse of the matrix B = ( 1 2 2 3 b 1 1 1 3 ) .

[2]
a.

For the values of a and b  found in part (a), solve the system of linear equations

x + 2 y 2 z = 5 3 x + b y + z = 0 x + y 3 z = a 1.

[2]
b.

Markscheme

AB = I

(AB)11 = 1 ⇒ a – 12 + 6 = 1,    giving a  = 7          (A1) (C1)

(AB)22 = 1 ⇒ –16 + 5 b  + 7 = 1,    giving b  = 2          (A1) (C1)

[2 marks]

a.

the system is  B X = ( 5 0 6 ) where X = ( x y z ) .

Then,  X = A ( 5 0 6 ) = ( 7 4 6 8 5 7 5 3 4 ) ( 5 0 6 ) .           (M1)

Thus  x = 1 y = 2 z = 1           (A1) (C2)

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1—Number and algebra » AHL 1.14—Introduction to matrices
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Topic 1—Number and algebra

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