Determinants and inverse matricesEdexcel A-Level Further Maths: Revision notes

For $\mathbf{A}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, the determinant is $\det\mathbf{A}=ad-bc$. It is written $|\mathbf{A}|$ or $\det\mathbf{A}$. Geometrically, the determinant is the area scale factor of the transformation: a shape of area $S$ becomes a shape of area $|\det\mathbf{A}|\,S$. If $\det\mathbf{A}<0$ the transformation also reverses orientation (like a reflection, which turns a clockwise label anticlockwise). Example: $\mathbf{A}=\begin{pmatrix}4&3\\2&5\end{pmatrix}$ has $\det\mathbf{A}=20-6=14$, so a triangle of area $5$ maps to area $70$ and keeps its orientation. A reflection matrix such as $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ has determinant $-1$: area unchanged, orientation reversed.

Expand along a row or column. Each entry is multiplied by its minor (the determinant of the $2\times2$ matrix left after deleting its row and column), with signs in the pattern $\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}$. For $\mathbf{C}=\begin{pmatrix}2&0&1\\1&3&-1\\0&2&4\end{pmatrix}$, expanding along row 1: $\det\mathbf{C}=2(12+2)-0+1(2-0)=30$. Choosing a row or column containing a zero saves work. For a transformation of 3-D space, $|\det\mathbf{C}|$ is the volume scale factor, so a solid of volume 8 maps to volume $240$. A negative determinant means the transformation reverses orientation (it includes a reflection).

A matrix is singular if $\det\mathbf{A}=0$ and non-singular if $\det\mathbf{A}\ne0$. A singular matrix has no inverse. A singular transformation collapses the plane onto a line (or a point), or 3-D space onto a plane, line or point, because its area (volume) scale factor is $0$. Information is lost, so it cannot be undone. To find when $\mathbf{B}=\begin{pmatrix}k&3\\4&k+1\end{pmatrix}$ is singular, solve $\det\mathbf{B}=0$: $k^2+k-12=(k+4)(k-3)=0$, so $k=-4$ or $k=3$.

For non-singular $\mathbf{A}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$: $\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix},\qquad \mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}.$ Swap the leading diagonal, change the signs of the other two entries, and divide by the determinant. For $\mathbf{A}=\begin{pmatrix}4&3\\2&5\end{pmatrix}$, $\mathbf{A}^{-1}=\frac{1}{14}\begin{pmatrix}5&-3\\-2&4\end{pmatrix}$. Useful properties: $(\mathbf{AB})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$ (reverse the order), $\det(\mathbf{AB})=\det\mathbf{A}\det\mathbf{B}$, and $\det\mathbf{A}^{-1}=\frac{1}{\det\mathbf{A}}$. Geometrically $\mathbf{A}^{-1}$ undoes the transformation $\mathbf{A}$.

The process for a non-singular $3\times3$ matrix $\mathbf{P}$:

1. Find $\det\mathbf{P}$ and check it is not zero.
2. Find the matrix of minors, then apply the sign pattern $+-+,-+-,+-+$ to get the matrix of cofactors.
3. Transpose the cofactor matrix (this is the adjugate).
4. Divide by $\det\mathbf{P}$: $\mathbf{P}^{-1}=\frac{1}{\det\mathbf{P}}\,\text{adj}\,\mathbf{P}$. For $\mathbf{P}=\begin{pmatrix}1&2&-1\\0&1&3\\2&-1&1\end{pmatrix}$ (determinant 18), the cofactors are $\begin{pmatrix}4&6&-2\\-1&3&5\\7&-3&1\end{pmatrix}$ and $\mathbf{P}^{-1}=\frac{1}{18}\begin{pmatrix}4&-1&7\\6&3&-3\\-2&5&1\end{pmatrix}$. You may use a calculator to find an inverse in an exam, but you should understand the process. The inverse reverses a transformation: if $\mathbf{P}\mathbf{x}=\mathbf{y}$ then $\mathbf{x}=\mathbf{P}^{-1}\mathbf{y}$.