Proving Matrix Relationships (Edexcel International A Level Further Maths)

Revision Note

Mark Curtis

Expertise

Maths

Proving Matrix Relationships

What is a matrix relationship?

  • A matrix relationship is an equation that connects different matrices

    • For example, if A, B and C are matrices then possible relationships are

      • AB = C

      • A + 2B = C

  • Remember that in general matrix multiplication is not commutative

    • AB BA

    • ABC ACBBAC ≠ …

    • CBC does not simplify to C2B

      • Since C(BC) ≠ C(CB)

What matrix relationships do I need to know?

  • If A and B are matrices and I is the identity matrix, you need to know the following:

    • AA-1 = I

    • A-1A = I

    • IA = AI = A

    • (AB)-1 = B-1A-1

      • The order reverses

How do I prove matrix relationships?

  • You need to carefully pre-multiply (on the left) or post-multiply (on the right) both sides by matrices or their inverses

    • To make B the subject of AB = C

      • A-1AB = A-1C (pre-multiply by A-1)

      • IB = A-1C (form the identity)

      • B = A-1C (simplify)

    • To make A the subject of AB = C

      • ABB-1 = CB-1 (post-multiply by B-1)

      • AI = CB-1 (form the identity)

      • A = CB-1 (simplify)

How do I prove that (AB)-1 = B-1A-1?

  • Start by multiplying AB by its inverse to form the identity

    • AB(AB)-1 = I

  • Then make (AB)-1 the subject

    • Pre-multiply by A-1 and simplify

      • A-1AB(AB)-1 = A-1I

      • IB(AB)-1=A-1

      • B(AB)-1=A-1

    • Then pre-multiply by B-1 and simplify

      • B-1B(AB)-1=B-1A-1

      • I(AB)-1=B-1A-1

      • (AB)-1=B-1A-1

Exam Tip

  • Learn the formula (AB)-1 =B-1A-1 as you are not given it in the Formulae Booklet.

  • Show lots of steps when doing matrix algebra.

    • Examiners want to see pre- or post-multiplying clearly.

Worked Example

Let bold Pbold Q and bold R be three matrices such that bold PQR to the power of negative 1 end exponent equals bold I where bold I is the identity matrix.

Prove that bold Q equals bold P to the power of negative 1 end exponent bold R.

You need to use matrix algebra to make bold Q the subject
One possible way is to remove bold P from the left by pre-multiplying both sides by bold P to the power of negative 1 end exponent

bold P to the power of negative 1 end exponent bold PQR to the power of negative 1 end exponent equals bold P to the power of negative 1 end exponent bold I

The bold P to the power of negative 1 end exponent bold P forms the identity matrix, bold I, on the left
The bold P to the power of negative 1 end exponent bold I simplifies to just bold P to the power of negative 1 end exponent on the right

bold IQR to the power of negative 1 end exponent equals bold P to the power of negative 1 end exponent

The bold IQR to the power of negative 1 end exponent on the left simplifies to just bold QR to the power of negative 1 end exponent

bold QR to the power of negative 1 end exponent equals bold P to the power of negative 1 end exponent

Now post-multiply both sides by bold R

bold QR to the power of negative 1 end exponent bold R equals bold P to the power of negative 1 end exponent bold R

The bold R to the power of negative 1 end exponent bold R on the left forms the identity matrix

bold QI equals bold P to the power of negative 1 end exponent bold R

The bold QI on the left simplifies to bold Q

bold Q equals bold P to the power of negative 1 end exponent bold R

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Mark Curtis

Author: Mark Curtis

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.