Proving Matrix Relationships (Edexcel IAL Further Maths)

Spec point — 5.5

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≠ ACB ≠ BAC ≠ …

  • 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 Hint

• 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 $P$, $Q$ and $R$ be three matrices such that $\mathrm{PQR}^{-1}=I$ where $I$ is the identity matrix.

Prove that $Q=P^{-1}R$.

You need to use matrix algebra to make $Q$ the subject One possible way is to remove $P$ from the left by pre-multiplying both sides by $P^{-1}$

$P^{-1}\mathrm{PQR}^{-1}=P^{-1}I$

The $P^{-1}P$ forms the identity matrix, $I$, on the left The $P^{-1}I$ simplifies to just $P^{-1}$ on the right

$\mathrm{IQR}^{-1}=P^{-1}$

The $\mathrm{IQR}^{-1}$ on the left simplifies to just $\mathrm{QR}^{-1}$

$\mathrm{QR}^{-1}=P^{-1}$

Now post-multiply both sides by $R$

$\mathrm{QR}^{-1}R=P^{-1}R$

The $R^{-1}R$ on the left forms the identity matrix

$\mathrm{QI}=P^{-1}R$

The $\mathrm{QI}$ on the left simplifies to $Q$

$Q=P^{-1}R$