Inverses of Matrices (Edexcel IGCSE Maths)

Inverse of a 2x2 matrix

Spec point — 5E

What is the inverse of a matrix?

• Only a square matrix has an inverse

• The inverse of a square matrix $A$is denoted as the matrix $A^{-1}$

  • The product of these matrices is the identity matrix, $\mathrm{AA}^{-1}=A^{-1}A=I$

How do I find the inverse of a 2x2 matrix?

• The method for finding the inverse of a $2\times2$ matrix is:

  • Switch the two entries on the top left-bottom right diagonal
  • Change the signs of the other two entries
  • Divide by the determinant $ad-bc$

$A = ( \begin{gathered} a b \\ c d \end{gathered} ) \Leftrightarrow A^{- 1} = \frac{1}{a d - b c} ( \begin{gathered} d - b \\ - c & a \end{gathered} )$

Exam Hint The formula for the inverse is on the course Formulae sheet, and will be provided for you in Paper 2 questions where it is needed.

Determinant questions can also appear in Paper 1, however. While these can usually be answered without having to use the inverse formula, it is still a good idea to know the formula going into the exam.

• Note that if the determinant, $ad-bc$, is equal to zero, then the term $\frac{1}{ad-bc}$ in the formula is not defined

  • If this is the case, then the matrix does not have an inverse
  • You will not need to deal with this situation on the exam

Worked Example $A = ( \begin{gathered} 1 3 \\ 2 7 \end{gathered} ) B = ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} )$

Show that $B$ is the inverse of $A$.

Answer:

Method 1

If you know the inverse formula, you can use it to calculate the inverse of $A$ directly

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

$\begin{gathered} A^{- 1} & = \frac{1}{1 \times 7 - 3 \times 2} ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} ) \\ = \frac{1}{7 - 6} ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} ) \\ = 1 ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} ) \\ = ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} ) \\ = B \end{gathered}$

Final answer: $B$is the inverse of$A$

Method 2

If you don't know the inverse formula, you can use the definition of the inverse

• I.e. if $A^{-1}$ is the inverse of $A$, then $\mathrm{AA}^{-1}=A^{-1}A=I$

So you can show that $B$ is the inverse of $A$ by showing

• either that $\mathrm{AB}=I$
• or that $\mathrm{BA}=I$

$\begin{gathered} \text{AB} = ( \begin{gathered} 1 3 \\ 2 7 \end{gathered} ) ( \begin{gathered} 7 - 3 \\ - 2 & 1 \end{gathered} ) \\ = ( \begin{gathered} 1 \times 7 + 3 \times ( - 2 ) & 1 \times ( - 3 ) + 3 \times 1 \\ 2 \times 7 + 7 \times ( - 2 ) & 2 \times ( - 3 ) + 7 \times 1 \end{gathered} ) \\ = ( \begin{gathered} 7 - 6 & - 3 + 3 \\ 14 - 14 & - 6 + 7 \end{gathered} ) \\ = ( \begin{gathered} 1 0 \\ 0 1 \end{gathered} ) \\ = I \end{gathered}$

Final answer: $\mathrm{AB}$is equal to the identity matrix, so$B$is the inverse of$A$