Inverse Matrices (Edexcel IAL Further Maths)

Determinant of a 2x2 Matrix

Spec point — 5.4

What is the determinant?

• The determinant is a numerical value (positive or negative) calculated from elements of a square matrix

  • It is used to find the inverse of a matrix

• For a 2 × 2 matrix, $A$, the determinant, $\mathrm{det}(A)$ or $\left| A \right|$, is given by

$A = ( \begin{gathered} a b \\ c d \end{gathered} ) \Leftrightarrow \det A = \left| A \right| = a d - b c$

What properties of determinants do I need to know?

• If $\mathrm{det}(A)=0$ then $A$ is called a singular matrix

• If $\mathrm{det}(A)\neq0$ then $A$ is called a non-singular matrix

• The determinant of the identity matrix is 1

  • $\mathrm{det}(I)=1$

• The determinant of the zero matrix is 0

  • $\mathrm{det}(0)=0$

• You do not need to know the following rules, but they can be good checks in an exam:

  • $\mathrm{det}(\mathrm{AB})=\mathrm{det}(A)\times\mathrm{det}(B)$
  • $\det ( A^{- 1} ) = \frac{1}{\det ( A )}$

Worked Example Consider the matrix $A = ( \begin{gathered} 3 - 6 \\ p 7 \end{gathered} )$, where $p$ is a constant.

Given that $\mathrm{det}A=-3$, find the value of $p$.

Use that if $A = ( \begin{gathered} a b \\ c d \end{gathered} )$ then $\det A = \left| A \right| = a d - b c$ Work out the determinant algebraically

$\begin{gathered} \det ( A ) & = 3 \times 7 - ( - 6 ) \times p \\ = 21 + 6 p \end{gathered}$

Set this expression equal to -3 and solve for $p$

$\begin{gathered} 21 + 6 p & = - 3 \\ 6 p & = - 24 \end{gathered}$

$p=-4$

Inverse of a 2x2 Matrix

Spec point — 5.5

What is the inverse of a matrix?

• The inverse of a square matrix $A$is the matrix $A^{-1}$ which, when multiplied together (in either order), gives the identity matrix

  • $\mathrm{AA}^{-1}=A^{-1}A=I$
  • $A^{-1}$ has the same dimensions (order) as $A$

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

• To find the inverse of a 2 × 2 matrix:

  • Switch the two entries on leading diagonal (top-left to bottom right)
  • Change the signs of the other two entries
  • Divide by the determinant

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

• Note that you cannot divide by zero

  • If $\mathrm{det}A=0$, then $A$ is not invertible ( $A^{-1}$ does not exist)

    • $A$ is singular

  • If $\mathrm{det}A\neq0$, then $A$ is invertible

How do I find the inverse of a product of matrices?

• The inverse of a product of matrices is the product of the inverses of the matrices in reverse order

  • $( \text{AB} )^{- 1} = B^{- 1} A^{- 1}$

Exam Hint There are two ways to check whether your inverse matrix is correct:

• use a calculator (they can find inverses),
• or calculate $\mathrm{AA}^{-1}$ to see if you get the identity $I$.

Worked Example Let $P = ( \begin{gathered} 1 - 2 \\ 4 k \end{gathered} )$ where $k\neq-8$.

(a) Find $P^{-1}$, giving your answer in terms of $k$.

Use that $A = ( \begin{gathered} a b \\ c d \end{gathered} ) \Leftrightarrow A^{- 1} = \frac{1}{\det A} ( \begin{gathered} d - b \\ - c & a \end{gathered} )$ where $\mathrm{det}A=ad-bc$ First find $\det ( P )$

$\begin{gathered} \det ( P ) & = 1 \times k - 4 \times ( - 2 ) \\ = k + 8 \end{gathered}$

Now divide $P$ by $\det ( P )$, swap $a$ and $d$, and change signs of $b$ and $c$

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

You can also write this as$P^{- 1} = ( \begin{gathered} \frac{k}{k + 8} & \frac{2}{k + 8} \\ - \frac{4}{k + 8} & \frac{1}{k + 8} \end{gathered} )$

(b) Verify that $P^{-1}P=I$, where $I$ is the identity matrix.

Verify means check that it is true Substitute $P^{-1}$ and $P$ into the left-hand side and multiply out the matrices

$\begin{gathered} P^{- 1} P & = \frac{1}{k + 8} ( \begin{gathered} k 2 \\ - 4 & 1 \end{gathered} ) ( \begin{gathered} 1 - 2 \\ 4 k \end{gathered} ) \\ = \frac{1}{k + 8} ( \begin{gathered} k \times 1 + 2 \times 4 & k \times ( - 2 ) + 2 \times k \\ - 4 \times 1 + 1 \times 4 & - 4 \times ( - 2 ) + 1 \times k \end{gathered} ) \\ = \frac{1}{k + 8} ( \begin{gathered} k + 8 & 0 \\ 0 8 + k \end{gathered} ) \end{gathered}$* *

The $8+k$ is the same as $k+8$ Since $k\neq-8$ in the question, cancel the $( k + 8 )$'s

$P^{- 1} P = ( \begin{gathered} 1 0 \\ 0 1 \end{gathered} )$

The right-hand side is the 2 × 2 identity matrix

$P^{-1}P=I$

Even though$\mathrm{PP}^{-1}=I$is also true, this question only asks for the order shown