Solving Matrix Equations (Edexcel IGCSE Maths)

Solving matrix equations with inverses

Spec point — 5C

• Inverses can be used to rearrange and solve equations with matrices

• This relies on the essential properties of inverses and of the identity matrix:

  • $\mathrm{AA}^{-1}=A^{-1}A=I$
  • $\mathrm{AI}=\mathrm{IA}=A$

• For example, to solve the matrix equation $\mathrm{AB}=C$ for $B$

  • First multiply both sides of the equation 'from the left' by $A^{-1}$

    • $A^{-1}\mathrm{AB}=A^{-1}C$

  • But $A^{-1}A=I$, so

    • $\mathrm{IB}=A^{-1}C$

  • And $\mathrm{IB}=B$, so

    • $B=A^{-1}C$

  • Then multiply together the two matrices on the right-hand side to find $B$ explicitly

• Similarly, you can solve the matrix equation $\mathrm{BA}=C$ for $B$

  • Though here you will need to multiply 'from the right' by $A^{-1}$

$\begin{gathered} BAA^{- 1} & = CA^{- 1} \\ \text{BI} = CA^{- 1} \\ B = CA^{- 1} \end{gathered}$

• This is similar to solving a regular equation by 'doing the same thing to both sides'

  • Except that order of multiplication matters with matrices

    • Multiplying 'from the left' is not the same as multiplying 'from the right'

  • For example, if you tried to solve $\mathrm{BA}=C$ for $B$ by multiplying 'from the left' by $A^{-1}$

    • you would get $A^{-1}\mathrm{BA}=A^{-1}C$
    • which does not simplify any further

Exam Hint Remember that a matrix and its inverse only 'collapse' to the identity matrix $I$ when they are next to each other in a multiplication.

• So $A^{-1}\mathrm{AB}=\mathrm{IB}=B$
• and $\mathrm{BAA}^{-1}=\mathrm{BI}=B$
• but $A^{-1}\mathrm{BA}$ does not simplify any further

Worked Example $P = ( \begin{gathered} 4 - 2 \\ - 2 & 2 \end{gathered} ) Q = ( \begin{gathered} k 1 \\ 3 0 \end{gathered} ) R = ( \begin{gathered} 10 4 \\ - 2 & - 2 \end{gathered} )$

where $k$ is a constant.

(a) Find $P^{-1}$.

Answer:

Use the inverse formula for a $2\times2$ matrix

• The inverse of $( \begin{gathered} a b \\ c d \end{gathered} )$ is $\frac{1}{a d - b c} ( \begin{gathered} d - b \\ - c & a \end{gathered} )$

$\begin{gathered} P^{- 1} & = \frac{1}{4 \times 2 - ( - 2 ) \times ( - 2 )} ( \begin{gathered} 2 2 \\ 2 4 \end{gathered} ) \\ = \frac{1}{8 - 4} ( \begin{gathered} 2 2 \\ 2 4 \end{gathered} ) \\ = 1 \frac{1}{4} ( \begin{gathered} 2 2 \\ 2 4 \end{gathered} ) \\ = ( \begin{gathered} 1 \frac{1}{4} \times 2 & 1 \frac{1}{4} \times 2 \\ 1 \frac{1}{4} \times 2 & 1 \frac{1}{4} \times 4 \end{gathered} ) \\ = ( \begin{gathered} 1 \frac{1}{2} & 1 \frac{1}{2} \\ 1 \frac{1}{2} & 1 \end{gathered} ) \end{gathered}$

$\begin{gathered} P^{- 1} & = ( \begin{gathered} 1 \frac{1}{2} & 1 \frac{1}{2} \\ 1 \frac{1}{2} & 1 \end{gathered} ) \end{gathered}$

(b) Given that $\mathrm{PQ}=R$ find the value of $k$.

Answer:

Use matrix algebra to solve the equation for $Q$

• Multiply both sides of the equation 'from the left' by $P^{-1}$

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

• By the definition of the inverse, $P^{-1}P=I$

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

• By the properties of the identity matrix, $\mathrm{IQ}=Q$

$Q=P^{-1}R$

That gives you a matrix 'formula' for $Q$

• Substitute in matrices $P^{-1}$ and $R$

$Q = ( \begin{gathered} 1 \frac{1}{2} & 1 \frac{1}{2} \\ 1 \frac{1}{2} & 1 \end{gathered} ) ( \begin{gathered} 10 4 \\ - 2 & - 2 \end{gathered} )$

• Carry out the multiplication

$\begin{gathered} Q = ( \begin{gathered} 1 \frac{1}{2} \times 10 + 1 \frac{1}{2} \times ( - 2 ) & 1 \frac{1}{2} \times 4 + 1 \frac{1}{2} \times ( - 2 ) \\ 1 \frac{1}{2} \times 10 + 1 \times ( - 2 ) & 1 \frac{1}{2} \times 4 + 1 \times - 2 \end{gathered} ) \\ = ( \begin{gathered} 5 - 1 & 2 - 1 \\ 5 - 2 & 2 - 2 \end{gathered} ) \\ = ( \begin{gathered} 4 1 \\ 3 0 \end{gathered} ) \end{gathered}$

$k=4$